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Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

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Samacheer Kalvi 11th Physics Kinematics Textual Questions Solved

Samacheer Kalvi 11th Physics Kinematics Multiple Choice Questions

Samacheer Kalvi 11th Physics Solution Chapter 2 Question 1.
Which one of the following Cartesian coordinate systems is not followed in physics?

Answer:

11th Physics Chapter 2 Book Back Answers Question 2.
Identify the unit vector in the following:
(a) \(\hat{i}+\hat{j}\)
(b) \(\frac{\hat{i}}{\sqrt{2}}\)
(c) \(\hat{k}-\frac{\hat{j}}{\sqrt{2}}\)
(d) \(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\)
Answer:
(d) \(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\)

11th Physics Kinematics Book Back Answers Question 3.
Which one of the following physical quantities cannot be represented by a scalar?
(a) Mass
(b) length
(c) momentum
(d) magnitude of acceleration
Answer:
(c) momentum

11th Physics Lesson 2 Book Back Answers Question 4.
Two objects of masses m₁ and m₂, fall from the heights h₁ and h₂ respectively. The ratio of the magnitude of their momenta when they hit the ground is [AIPMT 20121]
(a) \(\sqrt{\frac{h_{1}}{h_{2}}}\)
(b) \(\sqrt{\frac{m_{1} h_{1}}{m_{2} h_{2}}}\)
(c) \(\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)
(d) \(\frac{m_{1}}{m_{2}}\)
Answer:
(c) \(\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)

11th Physics 2nd Chapter Book Back Answers Question 5.
If a particle has negative velocity and negative acceleration, its speed
(a) increases
(b) decreases
(c) remains same
(d) zero
Answer:
(a) increases

Physics Class 11 Unit 2 Kinematics Question 6.
If the velocity is\(\overrightarrow{\mathrm{v}}\) – 2\(\hat{i}\) +t²\(\hat{j}\) – 9\(\overrightarrow{\mathrm{k}}\) , then the magnitude of acceleration at t = 0.5 s is
(a) 1 m s^-2
(b) 1 m
(c) zero
(d) -1 m s s^-2
Answer:
(a) 1 m s^-2

11th Physics 2nd Lesson Book Back Answers Question 7.
If an object is dropped from the top of a building and it reaches the ground at t = 4 s, then the height of the building is (ignoring air resistance) (g = 9.8 m s^-2).
(a) 77.3 m
(b) 78.4 m
(c) 80.5 m
(d) 79.2 m
Answer:
(b) 78.4 m

Samacheer Kalvi 11th Physics Solution Chapter 1 Question 8.
A ball is projected vertically upwards with a velocity v. It comes back to ground in time t. Which v -1 graph shows the motion correctly?[NSEP 00 – 01]

Answer:

Physics Chapter 2 Kinematics Question 9.
If one object is dropped vertically downward and another object is thrown horizontally from the same height, then the ratio of vertical distance covered by both objects at any instant is
(a) 1
(b) 2
(c) 4
(d) 0.5
Answer:
(a) 1

11th Physics Unit 2 Question 10.
A ball is dropped from some height towards the ground. Which one of the following represents the correct motion of the ball?

Answer:

Samacheer Kalvi Guru 11th Physics Question 11.
If a particle executes uniform circular motion in the xy plane in clockwise direction, then the angular velocity is in
(a) +y direction
(b) +z direction
(c) -z direction
(d) -x direction
Answer:
(c) -z direction

Samacheer Kalvi 11th Physics Question 12.
If a particle executes uniform circular motion, choose the correct statement [NEET 2016]
(a) The velocity and speed are constant.
(b) The acceleration and speed are constant.
(c) The velocity and acceleration are constant.
(d) The speed and magnitude of acceleration are constant.
Answer:
(d) The speed and magnitude of acceleration are constant.

Samacheer Kalvi 11th Physics Solution Book Question 13.
If an object is thrown vertically up with the initial speed u from the ground, then the time taken by the object to return back to ground is
(a) \(\frac{u^{2}}{2 g}\)
(b) \(\frac{u^{2}}{g}\)
(c) \(\frac{u}{2 g}\)
(d) \(\frac{2 u}{g}\)
Answer:
(d) \(\frac{2 u}{g}\)

Samacheer Kalvi Physics 11th Question 14.
Two objects are projected at angles 30° and 60° respectively with respect to the horizontal direction. The range of two objects are denoted as R_30° and R_60°– Choose the correct relation from the following:
(a) R_30° = R_60°
(b) R_30° = 4R_60°
(c) \(\mathrm{R}_{30^{\circ}}=\frac{\mathrm{R}_{60^{\circ}}}{2}\)
(d) R_30° = 2R_60°
Answer:
(a) R_30° = R_60°

11th Physics Samacheer Kalvi Question 15.
An object is dropped in an unknown planet from height 50 m, it reaches the ground in 2 s. The acceleration due to gravity in this unknown planet is
(a) g = 20 m s^-2
(b) g = 25 m s^-2
(c) g = 15 m s^-2
(d) g = 30 m s ^-2
Answer:
(a) g = 25 m s^-2

Samacheer Kalvi 11th Physics Kinematics Short Answer Questions

11th Physics Samacheer Kalvi Question 1.
Explain what is meant by Cartesian coordinate system?
Answer:
At any given instant of time, the frame of reference with respect to which the position of the object is described in terms of position coordinates (x, y, z) is called Cartesian coordinate system.

Physics Class 11 Samacheer Kalvi Question 2.
Define a vector. Give examples.
Answer:
Vector is a quantity which is described by the both magnitude and direction. Geometrically a vector is directed line segment.
Example – force, velocity, displacement.

Class 11 Physics Samacheer Kalvi Question 3.
Define a scalar. Give examples.
Answer:
Scalar is a property which can be described only by magnitude.
Example – mass, distance, speed.

Samacheer Kalvi 11th Physics Solution Chapter 3 Question 4.
Write a short note on the scalar product between two vectors.
Answer:
The scalar product (or dot product) of two vectors is defined as the product of the magnitudes of both the vectors and the cosine of the angle between them. Thus if there are two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) having an angle 0 between them, then their scalar product is defined as \(\overrightarrow{\mathrm{A}}\) • \(\overrightarrow{\mathrm{B}}\) = AB cos 0. Here, AB and are magnitudes of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\).

11th Samacheer Kalvi Physics Book Back Answers Question 5.
Write a short note on vector product between two vectors.
Answer:
The vector product or cross product of two vectors is defined as another vector having a magnitude equal to the product of the magnitudes of two vectors and the sine of the angle between them. The direction of the product vector is perpendicular to the plane containing the two vectors, in accordance with the right hand screw rule or right hand thumb rule. Thus, if\(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are two vectors, then their vector product is written as \(\overrightarrow{\mathrm{A}}\) × \(\overrightarrow{\mathrm{B}}\) which is a vector C defined by \(\overrightarrow{\mathrm{c}}\) = \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = (AB sin 0) \(\hat{n}\)
The direction \(\hat{n}\) of \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) , i.e., \(\overrightarrow{\mathrm{c}}\) is perpendicular to the plane containing the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\).

Samacheer Kalvi 11 Physics Solutions Question 6.
How do you deduce that two vectors are perpendicular?
Answer:
If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are perpendicular to each other than their scalar product \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) = 0 because cos 90° = 0. Then he vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are said to be mutually orthogonal.

Samacheer Kalvi 11th Physics Book Back Answers Question 7.
Define displacement and distance.
Answer:
Distance is the actual path length traveled by an object in the given interval of time during the motion. It is a positive scalar quantity. Displacement is the difference between the final and initial positions of the object in a given interval of time. It can also be defined as the shortest distance between these two positions of the object. It is a vector quantity.

Samacheer Kalvi 11th Physics Guide Pdf Question 8.
Define velocity and speed.
Answer:
Speed is defined as the ratio of total distance covered to the total time taken, it is a scalar quantity and always it is positive. Velocity is defined as the ratio of the displacement vector to the corresponding time interval. It is a vector quantity or it can also be defined as rate of change of displacement.

Question 9.
Define acceleration.
Answer:
Acceleration of a particle is defined as the rate of change of velocity or it can also be defined as the ratio of change in velocity to the given interval of time.

Question 10.
What is the difference between velocity and average velocity.
Answer:

Question 11.
Define a radian?
One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle.
1 rad = 57.295°

Question 12.
Define angular displacement and angular velocity.
Answer:
1. Angular displacement:
The angle described by the particle about the axis of rotation in a given time is called angular displacement.

2. Angular velocity:
The rate of change of angular displacement is called angular velocity.

Question 13.
What is non uniform circular motion?
If the speed of the object in circular motion is not constant, then we have non-uniform circular motion. For example, when the bob attached to a string moves in vertical circle, the speed of the bob is not the same at all time Whenever the speed is not same in circular motion, the particle will have both centripetal and tangential acceleration.

Question 14.
Write down the kinematic equations for angular motion.
Answer:
Kinematic equations for circular motion are –

1. \(\omega=\omega_{0}+\alpha t\)
2. \(\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}\)
3. \(\omega^{2}=\omega_{o}^{2}+2 \alpha \theta\)
4. \(\theta=\frac{\left(\omega_{0}+\omega\right)}{2} t\)

Here,
ω₀  = initial angular velocity
ω = final angular velocity
θ = angular displacement
α = angular acceleration
t = time.

Question 15.
Write down the expression for angle made by resultant acceleration and radius vector in the non uniform circular motion.
Answer:
The angle made by resultant acceleration and radius vector in the non uniform circular motion is –
\(\tan \theta=\frac{a_{t}}{\left(\frac{V^{2}}{r}\right)}\) or \(\theta=\tan ^{-1}\left(\frac{a_{t}}{\left(\frac{V^{2}}{r}\right)}\right)\)

Samacheer Kalvi 11th Physics Kinematics Long Answer Questions

Question 1.
Explain in detail the triangle law of addition.
Answer:
Let us consider two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) as shown in figure. To find the resultant of the two vectors we apply the triangular.

Law of addition as follows:
present the vectors A and by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.


To explain further, the head of the first vector \(\overrightarrow{\mathrm{A}}\) is connected to the tail of the second vect \(\overrightarrow{\mathrm{B}}\) Let O he the angle between\(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\). Then \(\overrightarrow{\mathrm{R}}\) is the resultant vector connecting the tail of the first vector \(\overrightarrow{\mathrm{A}}\) to the head of the second vector \(\overrightarrow{\mathrm{B}}\) The magnitude of \(\overrightarrow{\mathrm{R}}\). (resultant) given geometrically by the length of (OQ) and the direction of the resultant vector is the angle between \(\overrightarrow{\mathrm{R}}\). and \(\overrightarrow{\mathrm{A}}\). Thus we write
\(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{A}}\) +\(\overrightarrow{\mathrm{B}}\) \(\overrightarrow{\mathrm{OQ}}\) = \(\overrightarrow{\mathrm{OP}}\) + \(\overrightarrow{\mathrm{PQ}}\)

1. Magnitude of resultant vector:
The magnitude and angle of the resultant vector ar determined by using triangle law of vectors as follows.From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

From figure, let R is the magnitude of the resultant of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\).
cos θ = \(\frac { AN}{ B }\) ∴ AN = B cos θ and sinθ = \(\frac { BN}{ B }\) ∴BN = B sinθ
For ∆ OBN, we have OB² = ON² + BN²
⇒ R² = (A + B cos θ)² + (B sinθ)²
⇒ R² = A² + B² cos²θ + 2ABcosθ B² sin²θ
⇒ R² = A² + B²(cos²θ + sin²θ) + 2AB cos θ
⇒ R² = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

2. Direction of resultant vectors:
If 0 is the angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) then,
\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}\)
If R makes an angle α with \(\overrightarrow{\mathrm{A}}\) , then in AOBN,
tan α = \(\frac { BN}{ ON }\) = \(\frac { BN}{ OA + AN }\)
tan α = \(\frac { B sinθ }{ A + B cosθ}\) ⇒ α = \(\tan ^{-1}\left(\frac{B \sin \theta}{A+B \cos \theta}\right)\)

Question 2.
Discuss the properties of scalar and vector products.
Answer:
Properties of scalar product of two vectors are:
(1) The product quantity \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\) is always a scalar. It is positive if the angle between the vectors is acute (i.e., < 90°) and negative if the angle between them is obtuse (i.e. 90°<0< 180°).

(2) The scalar product is commutative, i.e. \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) ≠ \(\overrightarrow{\mathrm{B}}\). \(\overrightarrow{\mathrm{A}}\)

(3) The vectors obey distributive law i.e. \(\overrightarrow{\mathrm{A}}\)(\(\overrightarrow{\mathrm{B}}\) + \(\overrightarrow{\mathrm{C}}\)) = \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\) + \(\overrightarrow{\mathrm{A}}\) .\(\overrightarrow{\mathrm{C}}\)
(4) The angle between the vectors θ = \(\cos ^{-1}\left[\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{\mathrm{AB}}\right]\)

(5) The scalar product of two vectors will be maximum when cos θ = 1, i.e. θ = 0°, i.e., when the vectors are parallel;
\((\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}})_{\max }=\mathrm{AB}\)

(6) The scalar product of two vectors will be minimum, when cos θ = -1, i.e. θ = 180°.
\((\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}})_{\min }=-\mathrm{AB}\) when the vectors are anti-parallel.

(7) If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are perpendicular to each other than their scalar product \(\overrightarrow{\mathrm{A}}\) .\(\overrightarrow{\mathrm{B}}\) = 0, because cos 90° 0. Then the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are said to be mutually orthogonal.

(8) The scalar product of a vector with itself is termed as self-dot product and is given by (\(\overrightarrow{\mathrm{A}}\))² = \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{A}}\) = AA cos 0 = A². Here angle 0 = 0°.
The magnitude or norm of the vector \(\overrightarrow{\mathrm{A}}\) is |\(\overrightarrow{\mathrm{A}}\)| = A = \(\sqrt{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}}\).

(9) In case of a unit vector \(\hat{n}\)
\(\hat{n}\) . \(\hat{n}\) = 1 x 1 x cos 0 = 1. For example, \(\hat{i}\) – \(\hat{i}\) = \(\hat{j}\) . \(\hat{j}\) = \(\hat{k}\) . \(\hat{k}\) = 1.

(10) In the case of orthogonal unit vectors, \(\hat{i}\),\(\hat{j}\) and \(\hat{k}\),
\(\hat{i}\) . \(\hat{j}\) = \(\hat{j}\).\(\hat{k}\) = \(\hat{k}\) . \(\hat{i}\)= 1.1 cos 90° = 0

(11) In terms of components the scalar product of \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) can be written as
\(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\) = (A_x\(\hat{i}\) + A_y\(\hat{j}\) + A_z\(\hat{k}\)).(B_x\(\hat{i}\) + B_y\(\hat{j}\) + B_z\(\hat{k}\))
= A _xB_x + A_yB_y+ A_zB_z, with all other terms zero.
The magnitude of vector | \(\overrightarrow{\mathrm{A}}\) | is given by
| \(\overrightarrow{\mathrm{A}}\) | = A = \(\sqrt{\mathrm{A}_{x}^{2}+\mathrm{A}_{y}^{2}+\mathrm{A}_{z}^{2}}\)

Properties of vector product of two vectors are:
(1) The vector product of any two vectors is always another vector whose direction is perpendicular to the plane containing these two vectors, i.e., orthogonal to both the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\), even though the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) may or may not be mutually orthogonal.

(2) The vector product of two vectors is not commutative, i.e., \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) ≠ \(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\). But,
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)=-\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\).
Here it is worthwhile to note that |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| =
|\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)| = AB sin 0 i.e., in the case of the product vectors \(\overrightarrow{\mathrm{B}}\)=-\(\overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\), the magnitudes are equal but directions are opposite to each other.

(3) The vector product of two vectors will have maximum magnitude when sin 0 = 1, i.e., 0 = 90° i.e., when the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are orthogonal to each other.
\((\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})_{\mathrm{max}}=\mathrm{AB} \hat{n}\) = AB \(\hat{n}\)

(4) The vector product of two non-zero vectors will be minimum when sin θ = 0, i.e θ = 0° or 180°
\((\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})_{\min }=0\)
i. e., the vector product of two non – zero vectors vanishes, if the vectors are either parallel or anti parallel.

(5) The self – cross product, i.e., product of a vector with itself is the null vector
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{A}}\) = AA sin 0° \(\hat{n}\) = \(\overrightarrow{\mathrm{0}}\) In physics the null vector 0 is simply denoted as zero.

(6) The self – vector products of unit vectors are thus zero.
\(\hat{i}\) x \(\hat{i}\) = \(\hat{ j}\) x \(\hat{j}\) = \(\hat{k}\) x \(\hat{k}\) = 0

(7) In the case of orthogonal unit vectors, \(\hat{i}\), \(\hat{j}\). \(\hat{k}\) , in accordance with the right hand screw rule:
\(\hat{i}\) x \(\hat{j}\) = \(\hat{k}\), \(\hat{j}\) x \(\hat{k}\) = \(\hat{i}\) and \(\hat{k}\) x \(\hat{i}\) = \(\hat{j}\)

Also, since the cross product is not commutative,
\(\hat{j}\) x \(\hat{i}\) = –\(\hat{k}\), \(\hat{k}\) x \(\hat{j}\) = –\(\hat{i}\) and \(\hat{i}\) x \(\hat{k}\) = \(\hat{j}\)

(8) In terms of components, the vector product of two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is –

Note that in the \(\hat{j}^{\mathrm{th}}\) component the order of multiplication is different than \(\hat{i}^{\mathrm{th}}\) and \(\hat{k}^{\mathrm{th}}\) components.

(9) If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) form adjacent sides in a parallelogram, then the magnitude of |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| will give the area of the parallelogram as represented graphically in figure.

(10) Since we can divide a parallelogram into two equal triangles as shown in the figure, the area of a triangle with \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) as sides is \(\frac { 1 }{ 2 }\) |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| . This is shown in the Figure. A number of quantities used in Physics are defined through vector products. Particularly physical quantities representing rotational effects like torque, angular momentum, are defined through vector products.

Question 3.
Derive the kinematic equations of motion for constant acceleration.
Answer:
Consider an object moving in a straight line with uniform or constant acceleration ‘a’. Let u be the velocity of the object at time t = 0, and v be velocity of the body at a later time t.

Velocity – time relation:
(1) The acceleration of the body at any instant is given by the first derivative of the velocity with respect to time,
a = \(\frac {dv}{dt}\) or dv = a dt
Integrating both sides with the condition that as time changes from 0 to t, the velocity changes from u to v. For the constant acceleration,

(2) The velocity of the body is given by the first derivative of the displacement with respect to time.
v = \(\frac {ds}{dt}\) or ds = vdt
and since v = u + at,
we get ds = (u+ at ) dt
Assume that initially at time t = 0, the particle started from the origin. At a later time t, the particle displacement is s. Further assuming that acceleration is time-independent, we have

Velocity – displacement relation:

(3) The acceleration is given by the first derivative of velocity with respect to time.
a = \(\frac {dv}{dt}\) = \(\frac {dv}{ds}\) = \(\frac {ds}{dt}\) = \(\frac {dv}{ds}\) v [since ds/dt = v] where s is displacement traverse
This is rewritten as a = \(\frac{1}{2} \frac{d v^{2}}{d s}\) or ds = \(\frac{1}{2 a} d\left(v^{2}\right)\) Integrating the above equation, using the fact when the velocity changes from u² to v², displacement changes from 0 to s, we get

We can also derive the displacement 5 in terms of initial velocity u and final velocity v. From equation we can
write,
at = v – u
Substitute this in equation, we get

Question 4.
Derive the equations of motion for a particle (a) falling vertically (b) projected vertically.
Answer:
’Equations of motion for a particle falling vertically downward from certain height. Consider an object of mass m falling from a height h. Assume there is no air resistance. For convenience, let us choose the downward direction as positive y – axis as shown in the figure. The object experiences acceleration ‘g’ due to gravity which is constant near the surface of the Earth. We can use kinematic equations to explain its motion. We have The acceleration \(\overrightarrow{\mathrm{a}}\) = g \(\hat{i}\)
By comparing the components, we get,
Equations of motion for a particle thrown vertically upwards,
a_x = 0, a_x = 0, a_y = g Let us take for simplicity, a_y = a = g

If the particle is thrown with initial velocity ‘u’ downward which is in negative y – axis, then velocity and position at of the particle any time t is given by
v = u + gt
v = ut + \(\frac {1}{2}\) – gt²
The square of the speed of the particle when it is at a distance y from the hill – top, is v² = u² + 2 gy
Suppose the particle starts from rest.
Then u = 0
Then the velocity v, the position of the particle and v² at any time t are given by (for a point y from the hill – top)
v = gt …………(i)
y = \(\frac {1}{2}\) – gt² …………(ii)
v² = 2gy …………(iii)
The time (t = T) taken by the particle to reach the ground (for which y = h), is given by using equation (ii),
h = \(\frac {1}{2}\) – gT² …………(iv)
T = \(\sqrt{\frac{2 h}{g}}\) …………(v)
The equation (iv) implies that greater the height (h), particle takes more time (T) to reach the ground. For lesser height (h), it takes lesser time to reach the ground. The speed of the particle when it reaches the ground (y = h) can be found using equation (iii), we get,
\(v_{\text {ground }}=\sqrt{2 g h}\) …………(vi)
The above equation implies that the body falling from greater height (h) will have higher velocity when it reaches the ground. The motion of a body falling towards the Earth from a small altitude (h<<R), purely under the force of gravity is called free fall. (Here R is radius of the Earth).

case (ii):
A body thrown vertically upwards:
Consider an object of mass m thrown vertically upwards with an initial velocity u. Let us neglect the air friction. In this case we choose the vertical direction as positive y axis as shown in the figure, then the acceleration a = -g (neglect air friction) and g points towards the negative y axis. The kinematic equations for this motion are,
The velocity and position of the object at any time t are,

v = u – gt ……………(vii)
s = ut – \(\frac {1}{2}\) – gt² …………..(viii)
The velocity of the object at any position y (from the point where the object is thrown) is
v² = u² – 2gy …………..(ix)

Question 5.
Derive the equation of motion, range and maximum height reached by the particle thrown at an oblique angle 9 with respect to the horizontal direction.
Answer:
This projectile motion takes place when the initial velocity is not horizontal, but at some angle with the vertical, as shown in Figure.
(Oblique projectile)
Examples:

• Water ejected out of a hose pipe held obliquely.
• Cannon fired in a battle ground.

Consider an object thrown with initial velocity at an angle θ with the horizontal.
Then,
\(\overrightarrow{\mathrm{u}}\) = u_x î + u_y\(\hat{j}\) .
where u_x = u cos θ is the horizontal component and u_y = u sin θ the vertical component of velocity. Since the acceleration due to gravity is in the direction opposite to the direction of vertical component u_y , this component will gradually reduce to zero at the maximum height of the projectile. At this maximum height, the same gravitational force will push the projectile to move downward and fall to the ground. There is no acceleration along the x direction throughout the motion. So, the horizontal component of the velocity (u_x = u cos θ) remains the same till the object reaches the ground. Hence after the time t, the velocity along horizontal motion v_x = u_x + a_xt = u_x = u cos θ. The horizontal distance travelled by projectile m time t is s_x = \(u_{x} t+\frac{1}{2} a_{x} t^{2}\)
Here, s_x = x, u_x = u cos θ, a_x = 0

Thus, x = u cos θ or t = \(\frac {x}{u cos θ}\) ……..(i)
Next, for the vertical motion v_y= u_y + a_yt
Here u_y = u sin θ, a_y = -g (acceleration due to gravity acts opposite to the motion).
Thus, v_y= u sin θ – gt
The vertical distance traveled by the projectile in the same time t is
Here, s_y = y, u_y = u sin θ, a_x = -g. Then
y = u sinθ t – \(\frac {1}{2}\) – gt² ………..(ii)
Substitute the value of t from equation (i) in equation (ii), we have .

Thus the path followed by the projectile is an inverted parabola Maximum height (h_max): The maximum vertical distance travelled by the projectile during the journey is called maximum height. This is determined as follows:
For the vertical part of the motion.
\(v_{y}^{2}=u_{y}^{2}+2 a_{y} s\)
Here, u_y = u sin θ, a = -g, s = h_max, and at the maximum height v_y = 0
Hence, (0)² = u² sin² θ = 2 gh_max or \(h_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}\)

Time of flight (T_f):
The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight. This time of flight is the time taken by the projectile to go from point O to B via point A as shown
we know that s_y = y = 0 (net displacement in y-direction is zero),
u_y = u sin θ, a_y = -g , t = T_f Then

Horizontal range (R):
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). This is found easily since the horizontal component of initial velocity remains the same. We can write Range R = Horizontal component of velocity x time of flight = u cos θ x \(\mathrm{T}_{f}=\frac{u^{2} \sin 2 \theta}{g}\) The horizontal range directly depends on the initial speed (u) and the sine of angle of projection (θ). It inversely depends on acceleration due to gravity ‘g’.
For a given initial speed u, the maximum possible range is reached when sin 2θ is maximum, sin 2θ = 1. This implies 2θ = π/2 or θ = \(\frac {π}{4}\) This means that if the particle is projected at 45 degrees with respect to horizontal, it attains maximum range, given by.
\(\mathrm{R}_{\max }=\frac{u^{2}}{g}\) ………..(vi)

Question 6.
Derive the expression for centripetal acceleration.
Answer:
In uniform circular motion the velocity vector turns continuously without changing its magnitude (speed), as shown in figure.

Note that the length of the velocity vector is not changed during the motion, implying that the speed remains constant. Even though the velocity is tangential at every point in the circle, the acceleration is acting towards the center of the circle. This is called centripetal acceleration. It always points towards the center of the circle. This is shown in the figure.

The centripetal acceleration is derived from a simple geometrical relationship between position and velocity vectors.

Let the directions of position and velocity vectors shift through the same angle θ in a small interval of time ∆t, as shown in figure. For uniform circular motion, r = \(\left|\vec{r}_{1}\right|\) = \(\left|\vec{r}_{2}\right|\) and v = \(\left|\vec{v}_{1}\right|\) = \(\left|\vec{v}_{2}\right|\). If the particle moves from position vector \(\vec{r}_{1}\) to \(\vec{r}_{2}\), the displacement is given by ∆\(\overrightarrow{\mathrm{r}}\) = \(\vec{r}_{2}\) – \(\vec{r}_{1}\) and the change in velocity from \(\vec{v}_{1}\) to\(\vec{v}_{2}\) is given by ∆\(\overrightarrow{\mathrm{v}}\) = \(\vec{v}_{2}\) – \(\vec{v}_{1}\),. The magnitudes of the displacement ∆r and of ∆v satisfy the following relation. \(\frac {∆r}{r}\) = \(\frac {-∆v}{v}\) = θ Here the negative sign implies that ∆v points radially inward, towards the center of the circle.

For uniform circular motion v = cor, where co is the angular velocity of the particle about center. Then the centripetal acceleration can be written as.
a = -ω²r

Question 7.
Derive the expression for total acceleration in the non uniform circular motion.
Answer:
If the speed of the object in circular motion is not constant, then we have non-uniform circular motion. For example, when the bob attached to a string moves in vertical circle, the speed of the bob is not the same at all time. Whenever the speed is not same in circular motion, the particle will have both centripetal and tangential acceleration as shown in the figure.

The resultant acceleration is obtained by vector sum of centripetal and tangential acceleration Since centripetal acceleration is \(\frac{v^{2}}{r}\), the magnitude of this resultant acceleration is given by –\(\dot{a}_{\mathrm{R}}=\sqrt{a_{t}^{2}+\left(\frac{v^{2}}{r}\right)^{2}}\)
This resultant acceleration makes an angle 0 with the radius vector as shown in figure.
This angle is given by tan θ = \(\frac{a_{t}}{\left(v^{2} / r\right)}\)

Samacheer Kalvi 11th Physics Kinematics Numerical Questions

Question 1.
The position vector of the particle has length 1 m and makes 30° with the x-axis. What are the lengths of the x and y – components of the position vector?
Answer:
Given,
Length of position vector = 1 m
Angle made with x axis = 30
Solution:
Length of X component (OB) = OA cos θ
= 1 x cos 30°
= \(\frac{\sqrt{3}}{2}\) (or) 0.87 m
Length of Y component (AB) = OA sin θ = 1 x sin 30° = \(\frac { 1 }{ 2 }\) = 0.5 m.

Question 2.
A particle has its position moved from \(\vec{r}_{1}\) = 3\(\hat{i}\) + 4\(\hat{j}\) to r2 = \(\hat{i}\) + 2\(\hat{i}\). Calculate the displacement vector (∆\(\overrightarrow{\mathrm{r}}\) ) and draw the \(\vec{r}_{1}\), \(\vec{r}_{2}\) and ∆\(\overrightarrow{\mathrm{r}}\) vector in a two dimensional Cartesian coordinate system.
Answer:
Given,
Position vectors \(\vec{r}_{1}\) = 3\(\hat{i}\) + 4\(\hat{j}\)
\(\vec{r}_{1}\) = \(\hat{i}\) + 2\(\hat{j}\)
Solution:
Displacement vector:
∆r= \(\vec{r}_{2}\) – \(\vec{r}_{1}\) = (1 – 3)\(\hat{i}\) + (2 – 4) \(\hat{j}\)
∆r = -2\(\hat{i}\) -2\(\hat{j}\) = -2(\(\hat{i}\) + \(\hat{j}\))

Question 3.
Calculate the average velocity of the particle whose position vector changes from \(\vec{r}_{1}\) = 5\(\hat{i}\) + 6\(\hat{j}\) to \(\vec{r}_{2}\) = 2\(\hat{i}\) + 3 \(\hat{j}\) in a time 5 seconds.
Answer:
Given,
Position vectors of a particle
\(\vec{r}_{1}\) = 5\(\hat{i}\) + 6\(\hat{j}\),
\(\vec{r}_{2}\) = 2\(\hat{i}\) + 35\(\hat{j}\)
time(t) = 5s
Solution:

Question 4.
Convert the vector \(\overrightarrow{\mathrm{r}}\) = 3\(\hat{i}\) + 2\(\hat{j}\) into a unit vector.
Answer:
Given:
Position vector\(\hat{r}\) = 3\(\hat{i}\) + 2\(\hat{j}\)
Solution:

Question 5.
What are the resultants of the vector product of two given vectors given by \(\overrightarrow{\mathrm{A}}\) = 4\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\) and \(\overrightarrow{\mathrm{B}}\) = 5\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\) ?
Answer:
Given,
Vectors \(\overrightarrow{\mathrm{A}}\) = 4\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\)
\(\overrightarrow{\mathrm{B}}\) = 5\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\)
Solution:

Question 6.
object at an angle such that the horizontal range is 4 times of the maximum height. What is the angle of projection of the object?
Answer:
Give,
Horizontal range = 4H_max
Solution:

Question 7.
The following graphs represent velocity – time graph. Identity what kind of motion a particle undergoes in each graph.

Answer:
(a) At all the points, slope of the graph is constant.
∴ \(\overrightarrow{\mathrm{a}}\) = constant

(b) No change in magnitude of velocity with respect to time
∴ \(\overrightarrow{\mathrm{v}}\) = constant

(c) Slope of this graph is greater than graph (a) but constant
∴ \(\overrightarrow{\mathrm{a}}\) = constant but greater than the graph (a)

(d) At each point slope of the curve increases.
∴ \(\overrightarrow{\mathrm{a}}\) is a variable and object is in accelerated motion.

Question 8.
The following velocity – time graph represents a particle moving in the positive x-direction. Analyse its motion from 0 to 7 s. Calculate the displacement covered and distance travelled by the particle from 0 to 2 s.

Answer:
As per graph,
(a) From 0 to 1.5 s the particle moving in a opposite direction.

• From 1.5 s to 2 s the particle is moving with increasing velocity.
• From 2 s to 5 s velocity of the particle is constant of magnitude 1 ms ^-1
• From 5 s to 6 s velocity of the particle is decreasing.
• From 6 s to 7 s the particle is at rest.

(b) Distance covered by the particle – Area covered under (v -t) graph

Displacement of the particle

Question 9.
A particle is projected at an angle of θ with respect to the horizontal direction. Match the following for the above motion.
(a) v_x – decreases and increases
(b) v_y  – remains constant
(c) Acceleration – varies
(d) Position vector – remains downward
Answer:
(a) v_x = remains constant
(b) v_y = decreases and increases
(c) a = remains downward
(d) r = varies

Question 10.
A water fountain on the ground sprinkles water all around it. If the speed of the water coming out of the fountain is v, calculate the total area around the fountain that gets wet.
Answer:
Given,
Speed of water = v
Solution:
Water comes from a fountain can be taken as projectile and the distance covered is maximum range of projectile i.e. θ = 45°.
Range of the particle (R_max) = \(\frac{v^{2}}{g}\) sin 2θ = \(\frac{v^{2}}{g}\)
here, R_max is radius of the area covered.

Question 11.
The following table gives the range of a particle when thrown on different planets. All the particles are thrown at the same angle with the horizontal and with the same initial speed. Arrange the planets in ascending order according to their acceleration due to gravity, (g value).

Answer:
Range = \(\frac{v^{2}}{g}\) sin 2θ ∴ g α \(\frac { 1 }{ range }\)
Ascending order of the planet with respect to their “g” is Mercury, Mars, Earth, Jupiter.

Question 12.
The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B. Then the angle between A and B is
(a) 30°
(b) 45°
(c) 150°
(d) 120°
Answer:
Given:
Resultant of \(\overrightarrow{\mathrm{A}}\) & \(\overrightarrow{\mathrm{B}}\) is perpendicular to \(\overrightarrow{\mathrm{A}}\) and magnitude of resultant (C) = \(\frac { 1 }{ 2 }\) \(\overrightarrow{\mathrm{B}}\) and α = 90°
Solution:
(i) Magnitude of resultant:

(ii) direction of resultant:

Question 13.
Compare the components for the following vector equations
(a) T\(\hat{j}\) -mg\(\hat{j}\) = ma\(\hat{j}\)
(b) \(\overrightarrow{\mathrm{T}}\) + \(\overrightarrow{\mathrm{F}}\) = \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
(c) \(\overrightarrow{\mathrm{T}}\) – \(\overrightarrow{\mathrm{F}}\) = \(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)
(d) T\(\hat{j}\) + mg\(\hat{j}\)= ma\(\hat{j}\)
Answer:
Components of the vectors
(a) T – mg = ma
(b) \(\overline{\mathrm{T}}_{x}+\overline{\mathrm{F}}_{x}\) = \(\overline{\mathrm{A}}_{x}+\overline{\mathrm{B}}_{x}\) (or) \(\overline{\mathrm{T}}_{y}+\overline{\mathrm{F}}_{y}=\overline{\mathrm{A}}_{y}+\overline{\mathrm{B}}_{y}\)
(c) \(\overline{\mathrm{T}}_{x}-\overline{\mathrm{F}}_{x}=\overline{\mathrm{A}}_{x}+\overline{\mathrm{B}}_{x}\)  (or) \(\overline{\mathrm{T}}_{y}-\overline{\mathrm{F}}_{y}=\overline{\mathrm{A}}_{y}+\overline{\mathrm{B}}_{y}\)
(d) T + mg = ma

Question 14.
Calculate the area of the triangle for which two of its sides are given by the vectors A = 5\(\hat{i}\) – 3\(\hat{j}\), B = 4\(\hat{i}\) + 6\(\hat{j}\) .
Answer:
Solution:
Area of the triangle = \(\frac { 1 }{ 2 }\) |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{A}}\)|

Question 15.
If Earth completes one revolution in 24 hours, what is the angular displacement made by Earth in one hour? Express your answer in both radian and degree.
Answer:
Given,
time period of earth = 24 hours
Solution:
Earth covers 360° in 24 hours
∴Angular displacement m 1 hour = \(\frac { 360° }{ 24 }\) = 15° (or) \(\frac { π }{ 12 }\)
Angular displacement in radian = \(\frac { 15° }{ 57.295° }\) = 0.262 rad

Question 16.
An object is thrown with initial speed 5 ms ^-1 with an angle of projection 30°. What is the height and range reached by the particle?
Answer:
Given,
Initial speed (u) = 5 ms^-1
Angle of projection θ = 30°
Solution:

Question 17.
A foot – ball player hits the ball with speed 20 ms^-1 with angle 30° with respect to horizontal direction as shown in the figure. The goal post is at a distance of 40 m from him. Find out whether ball reaches the goal post.

Answer:
Given:
Initial speed (u) = 20 ms^-1
Angle of projection (θ) = 30°
The distance of the goal post = 40 m
Solution:
Range of the projectile

The distance of goal post is 40 m. But the range of the ball is 35.35 m only. So ball will not reach the goal post.

Question 18.
If an object is thrown horizontally with an initial speed 10 ms ^-1 from the top of a building of height 100 m. What is the horizontal distance covered by the particle?
Answer:
Given,
Initial speed =10 ms^-1
Height of the building (h) = 100 m
Range = ?
Solution:
Range of the object = R = \(u \sqrt{\frac{2 h}{g}}\) = 10\(\sqrt{\frac{200}{9.8}}\) = 45.1 m
R = 45 m.

Question 19.
An object is executing uniform circular motion with an angular speed of \(\frac { π }{ 12 }\) radian per second. At t = 0 the object starts at angle θ = 0. What is the angular displacement of the particle after 4 s?
Answer:
Given:
Angular speed = \(\frac { π }{ 12 }\) rad/ sec
Solution:
Angular speed = \(\frac { Angular displacement}{ time taken }\)
Angular displacement = \(\frac { π }{ 12 }\) x 4 = \(\frac { π }{ 12 }\) = 60°

Question 20.
Consider the x-axis as representing east, the v – axis as north and z – axis as vertically upwards. Give the vector representing each of the following points.
(a) 5 m north east and 2 m up
(b) 4 m south east and 3 m up
(c) 2 m north west and 4 m up
Answer:
Given,
Solution:
(a) Length along X – axis = 5 cos 45° = \(\frac{5}{\sqrt{2}}\)m
Length along Y- axis = 5 sin 45° = \(\frac{5}{\sqrt{2}}\)m
Length along Z – Axis = 2 m
In vector rotation = \(\frac{5}{\sqrt{2}}\)\(\hat{i}\) + \(\frac{5}{\sqrt{2}}\)\(\hat{j}\) + 2\(\hat{k}\) = \(\frac{5(\hat{i}+\hat{j})}{\sqrt{2}}\) + 2\(\hat{k}\)

(b) Length along X = 4 cos 45° = \(\frac{4}{\sqrt{2}}\)m
Length along Y = 4 sin 45° = \(\frac{4}{\sqrt{2}}\)m
Length along Z-axis = 3 m
In vector rotation = \(\frac{4}{\sqrt{2}}\)\(\hat{i}\) – \(\frac{4}{\sqrt{2}}\)\(\hat{i}\) + 3k = 4(\(\hat{i}\) – \(\hat{j}\)) \(\sqrt{2}+3 \hat{k}\)

(c) Length along X = – 2 cos 45° = \(\sqrt{2}+3 \hat{k}\) = \(\frac{2}{\sqrt{2}} m=\sqrt{2} m\)
Length along Y = 2 sin 45° = \(\frac{2}{\sqrt{2}} m=\sqrt{2} m\)
length along Z = 4 m
∴ In vector rotation = \(-\sqrt{2} \hat{i}+\sqrt{2} \hat{j}+4 \hat{k}\)

Question 21.
The Moon is orbiting the Earth approximately once in 27 days, what is the angle transformed by the Moon per day?
Answer:
Given,
period of moon = 27 days
Solution:
i.e. in 27 days moon covers 360°
In one day angle traversed by moon = \(\frac { 360° }{ 2H }\) = 13.3°

Question 22.
An object of mass m has angular acceleration a = 0.2 rad s2. What is the angular displacement covered by the object after 3 second? (Assume that the object started with angle zero with zero angular velocity).
Answer:
Given,
Angular acceleration = α = 0.2 rad s^-2
Time = 3s
Initial velocity = 0
Solution:

Samacheer Kalvi 11th Physics Kinematics Additional Questions Solved

Samacheer Kalvi 11th Physics Kinematics Multiple Choice Questions 

Question 1.
The radius of the earth was measured by –
(a) Newton
(b) Eratosthenes
(c) Galileo
(d) Ptolemy
Answer:
(b) Eratosthenes

Question 2
The branch of mechanics which deals with the motion of objects without taking force into account is –
(a) kinetics
(b) dynamics
(c) kinematics
(d) statics
Answer:
(c) kinematics

Question 3.
If the coordinate axes (x, y, z) are drawn in anticlockwise direction then the co-ordinate system is known as –
(a) Cartesian coordinate system
(b) right handed coordinate system
(c) left handed coordinate system
(d) cylindrical coordinate system
Answer:
(b) right handed coordinate system

Question 4.
The dimension of point mass is –
(a) 0
(b) 1
(c) 2
(d) kg
Answer:
(a) 0

Question 5.
If an object is moving in a straight line then the motion is known as –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(a) linear motion

Question 6.
An athlete running on a straight track is an example for the whirling motion of a stone attached to’a string is a –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(a) linear motion

Question 7.
The whirling motion of a stone attached to a string is a –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(b) circular motion

Question 8.
Spinning of the earth about its own axis is known as –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(d) rotational motion

Question 9.
If an object executes a to and fro motion about a fixed point, is an example for –
(a) rotational motion
(b) vibratory motion
(c) circular motion
(d) curvilinear motion
Answer:
(b) vibratory motion

Question 10.
Vibratory motion is also known as –
(a) circular motion
(b) rotational motion
(c) oscillatory motion
(d) spinning
Answer:
(c) oscillatory motion

Question 11.
The motion of satellite around the earth is an example for –
(a) circular motion
(b) rotational motion
(c) elliptical motion
(d) spinning
Answer:
(a) circular motion

Question 12.
An object falling freely under gravity close to earth is –
(a) one dimensional
(b) circular motion
(c) rotational motion
(d) spinning motion
Answer:
(a) one dimensional

Question 13.
Motion of a coin on a carrom board is an example of –
(a) one dimensional motion
(b) two dimensional motion
(c) three dimensional motion
(d) none
Answer:
(b) two dimensional motion

Question 15.
A bird flying in the sky is an example of –
(a) one dimensional motion
(b) two dimensional motion
(c) three dimensional motion
(d) none
Answer:
(c) three dimensional motion

Question 16.
Example for scalar is –
(a) distance
(b) displacement
(c) velocity
(d) angular momentum
Answer:
(a) distance

Question 17.
Which of the following is not a scalar?
(a) Volume
(b) angular momentum
(c) Relative density
(d) time
Answer:
(b) angular momentum

Question 18.
Vector is having –
(a) only magnitude
(b) only direction
(c) bot magnitude and direction
(d) either magnitude or direction
Answer:
(c) both magnitude and direction

Question 19.
“norm” of the vector represents –
(a) only magnitude
(b) only direction
(c) both magnitude and direction
(d) either magnitude or direction
Answer:
(a) only magnitude

Question 20.
If two vectors are having equal magnitude and same direction is known as –
(a) equal vectors
(b) col-linear vectors
(c) parallel vectors
(d) on it vector
Answer:
(a) equal vectors

Question 21.
The angle between two collinear vectors is / are –
(a) 0°
(b) 90°
(c) 180°
(d) 0° (or) 180°
Answer:
(d) 0° (or) 180°

Question 22.
The angle between parallel vectors is –
(a) 0°
(b) 90°
(c) 180°
(d) 0° (or) 180°
Answer:
(a) 0°

Question 23.
The angle between anti parallel vectors is –
(a) 0°
(b) 90°
(c) 180°
(d) 0° (or) 180°
Answer:
(c) 180°

Question 24.
Unit vector is –
(a) having magnitude one but no direction
(b) \(A \widehat{A}\)
(c) \(\frac{\widehat{A}}{A}\)
(d) |A|
Answer:
(c) \(\frac{\widehat{A}}{A}\)

Question 25.
A unit vector is used to specify –
(a) only magnitude
(b) only direction
(c) either magnitude (or) direction
(d) absolute value
Answer:
(b) only direction

Question 26.
The angle between any two orthogonal unit vectors is –
(a) 0°
(b) 90°
(c) 180°
(d) 360°
Answer:
(b) 90°

Question 27.
If \(\hat{n}\) is a unit vector along the direction of \(\overrightarrow{\mathrm{A}}\), the \(\hat{n}\) is-
(a) \(\overrightarrow{\mathrm{A}}\) A
(b) n x A
(c) \(\overrightarrow{\mathrm{A}} / \mathrm{A}\)
(d \(\overrightarrow{\mathrm{A}}\) |A|
Answer:
(c) \(\overrightarrow{\mathrm{A}} / \mathrm{A}\)

Question 28.
The magnitude of a vector can not be-
(a) positive
(b) negative
(e) zero
(cl) 90
Answer:
(b) negative

Question 29.
If R = P + Q, then which of the following is true?
(a) P > Q
(b) Q >P
(c) P = Q
(d) R > P, Q
Answer:
(d) R > P, Q

Question 30.
A force of 3 N and 4 N are acting perpendicular to an object, the resultant force is-
(a) 9 N
(b) 16 N
(c) 5 N
(d) 7 N
Answer:
(c) 5 N

Question 31.
Torque is a-
(a) scalar
(b) vector
(c) either scalar (or) vector
(d) none
Answer:
(6) vector

Question 32.
The resultant of \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\) acts along x – axis. If A = 2\(\hat{i}\) – 3 \(\hat{j}\) + 2\(\hat{k}\) then B is-
(a) -2\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)
(b) 3\(\hat{j}\) – 2\(\hat{k}\)
(c) -2\(\hat{i}\) -3 \(\hat{j}\)
(d) -2\(\hat{i}\) – 2\(\hat{k}\)
Answer:
(b) 3\(\hat{j}\) – 2\(\hat{k}\)

Question 33.
The angle between (\(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)) and (\(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)) can be –
(a) only 0°
(b) only 90°
(c) between 0° and 90°
(d) between 0° and 180°
Answer:
(d) between 0° and 180°

Question 34.
If a vector \(\overrightarrow{\mathrm{A}}\) = 3\(\hat{i}\) + 2\(\hat{j}\) then what is 4 A-
(a) 12\(\hat{i}\) + 8\(\hat{j}\)
(b) 0.75\(\hat{i}\) + 0.5\(\hat{j}\)
(c) 3\(\hat{i}\) + 2\(\hat{j}\)
(d) 7\(\hat{i}\) + 6\(\hat{j}\)
Answer:
(a) 12\(\hat{i}\) + 8\(\hat{j}\)

Question 35.
If P = mV then the direction of P along-
(a) m
(b) v
(c) both (a) and (b)
(d) neither m nor v
Answer:
(b) v

Question 36.
The scalar product \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\) is equal to-
(a) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
(b) AB sin θ
(c) AB cos θ
(d) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
Answer:
(c) AB cos θ

Question 37.
The scalar product \(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\)is equal to-
(a) \(\overrightarrow{\mathrm{A}}\) +\(\overrightarrow{\mathrm{B}}\)
(b) \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\)
(c) AB sin θ
(d) (\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)
Answer:
(b) \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\)

Question 38.
The scalar product of two vectors will be maximum when θ is equal to –
(a) 0°
(b) 90°
(c) 180°
(d) 270°
Answer:
(a) 0°

Question 39.
The scalar product of two vectors will be minimum. When θ is equal to –
(a) 0°
(b) 45°
(c) 180°
(d) 60°
Answer:
(c) 180°

Question 40.
The vectors A and B to be mutually orthogonal when –
(a) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\) = 0
(b) \(\overrightarrow{\mathrm{A}}\) –\(\overrightarrow{\mathrm{B}}\) = 0
(c) \(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\) = 0
(d) \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = 0
Answer:
(c) \(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\) = 0

Question 41.
The magnitude of the vector is –
(a) A²
(b) \(\sqrt{\mathrm{A}^{2}}\)
(c) \(\sqrt{\mathrm{A}}\)
(d) \(\sqrt[3]{\mathrm{A}}\)
Answer:
(b) \(\sqrt{\mathrm{A}^{2}}\)

Question 42.
\(\hat{i}\) .\(\hat{j}\) is –
(a) 0
(b) I
(c) ∞
(d) none
Answer:
(a) 0

Question 43.
If \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are two vectors, which are acting along x, y respectively, then \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) lies along –
(a) x
(b) y
(c) z
(d) none
Answer:
(c) z

Question 44.
The direction of \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) is given by-
(a) right hand screw rule
(b) right hand thumb rule
(c) both (a) and (b)
(d) neither (a) and (b)
Answer:
(c) both (a) and (b)

Question 45.
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) is –
(a) AB cos θ
(b) AB sin θ
(c) AB tan θ
(d) AB sec θ
Answer:
(b) AB sin θ

Question 46.
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) isequal to –
(a) \(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)
(b) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
(c) –\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)
(d) \(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)
Answer:
(c) –\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)

Question 47.
The vector product of any two vectors gives a –
(a) vector
(b) scalar
(e) tensor
(d) col-linear
Answer:
(a) vector

Question 48.
|\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| is equal to –
(a) -|\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)|
(b) |\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)|
(c) -|\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)|
(d) \( \frac{\overline{\mathrm{A}} \times \overline{\mathrm{B}}}{|\overline{\mathrm{A}} \times \overline{\mathrm{B}}|}\)
Answer:
(b) |\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)|

Question 49.
The vector product of two vectors will have maximum magnitude when θ is equal to –
(a) 0°
(b) 90°
(c) 180°
(d) 360°
Answer:
(b) 90°

Question 50.
The vector product of two non-zero vectors will be minimum when O is equal to –
(a) 0°
(b) 180°
(e) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(e) both (a) and (b)

Question 51.
The product of a vector with itself is equal to –
(a) 0
(b) 1
(c) ∞
(d) A²
Answer:
(a) 0

Question 52.
\(\hat{i}\) x \(\hat{i}\) is –
(a) 0
(b) 1
(c) ∞
(d) \(\hat{j}\)
Answer:
(a) 0

Question 53.
\(\hat{i}\) x \(\hat{j}\) is –
(a) \(\hat{i}\)
(b) \(\hat{j}\)
(c) \(\hat{k}\)
(d) \(\overrightarrow{\mathrm{z}}\)
Answer:
(c) \(\hat{k}\)

Question 54.
\(\hat{j}\) x \(\hat{i}\) is –
(a) –\(\hat{i}\)
(b) –\(\hat{j}\)
(c) –\(\hat{k}\)
(d) \(\overrightarrow{\mathrm{z}}\)
Answer:
(c) –\(\hat{k}\)

Question 55.
If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) form adjacent sides of parallelogram, then the magnitude of |\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)| will give of parallelogram –
(a) length
(b) area
(c) volume
(d) diagonal
Answer:
(b) area

Question 56.
If \(\overrightarrow{\mathrm{P}}\) – \(\overrightarrow{\mathrm{Q}}\) then which of the following is incorrect. –
(a) \(\hat{P}\) = \(\hat{Q}\)
(b) |\(\hat{P}\)| = |\(\hat{Q}\)|
(c) P\(\hat{Q}\) = Q\(\hat{A}\)
(d) \(\hat{P}\) \(\hat{Q}\) = PQ
Answer:
(d) \(\hat{P}\) \(\hat{Q}\) = PQ

Question 57.
The momentum of a particle is \(\overrightarrow{\mathrm{P}}\) = cos θ \(\hat{i}\) + sin θ \(\hat{j}\) . The angle between momentum and the force acting on a body is –
(a) 0°
(b) 45°
(c) 90°
(d) 180°
Answer:
(c) 90°

Question 58.
A and B are two vectors, if A and B are perpendicular to each other then –
(a) \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = 0
(b) \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = l
(c) \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) = 0
(d) \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) = \(\overrightarrow{\mathrm{A}}\)\(\overrightarrow{\mathrm{B}}\)
Answer:
(c) \(\overrightarrow{\mathrm{A}}\) \(\overrightarrow{\mathrm{B}}\) = 0

Question 59.
The angle between two vectors -3\(\hat{i}\) + 6\(\hat{k}\) and 2\(\hat{i}\) + 3\(\hat{j}\) + \(\hat{k}\) is –
(a) 0°
(b) 45°
(c) 60°
(d) 90°
Answer:
(d) 90°

Question 60.
The radius vector is 2\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) while linear momentum is 2\(\hat{i}\) + 3\(\hat{j}\) + \(\hat{k}\) Then the angular momentum is
(a) -2\(\hat{i}\) + 4\(\hat{k}\)
(b) 4\(\hat{i}\) – 8\(\hat{k}\)
(c) 2\(\hat{k}\) – 4\(\hat{j}\) + 2\(\hat{k}\)
(d) 4\(\hat{i}\) – 8\(\hat{j}\)
Answer:
(a) -2\(\hat{i}\) + 4\(\hat{k}\)

Question 61.
Which of the following cannot be a resultant of two vectors of magnitude 3 and 6?
(a) 3
(b) 6
(c) 10
(d) 7
Answer:
(c) 10

Question 62.
Twelve forces each of magnitude 10 N acting on a body at an angle of 30° with other forces then their resultant is-
(a) 10 N
(b)120 N
(c) \(\frac{10}{\sqrt{3}}\)
(d) zero

Question 63.
Two forces are in the ratio of 3 : 4. The maximum and minimum of their resultants are in the ratio is –
(a) 4 : 3
(b) 3 : 4
(c) 7 : 1
(d) 1 : 7
Answer:
(c) 7 : 1

Question 64.
If | \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) | = |\(\overrightarrow{\mathrm{P}}\) | + |\(\overrightarrow{\mathrm{Q}}\)|. The angle between the vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) is –
(a) 0°
(b) 180°
(c) 60°
(d) 90°
Answer:
(a) 0°
|\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) | = |\(\overrightarrow{\mathrm{P}}\) | + |\(\overrightarrow{\mathrm{Q}}\)|
Square on both sides and the resultant becomes
P² + Q² + 2PQ cos θ = P² + Q² + 2PQ cos θ = 1
θ = 0

Question 65.
If |\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) = |\(\overrightarrow{\mathrm{P}}\) | — |\(\overrightarrow{\mathrm{P}}\)|, then the angle between the vectors \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\)
(a) 0°
(b) 90°
(c) 180°
(d) 360°
Answer:
(c) |\(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\)| = |\(\overrightarrow{\mathrm{P}}\) | |\(\overrightarrow{\mathrm{P}}\)| ‘
Square on both side, and the resultant becomes
P² + Q² + 2PQ cos θ = P² + Q² – 2PQ .
cos θ = -1
θ = 180°

Question 66.
If |\(\overrightarrow{\mathrm{P}}\) x \(\overrightarrow{\mathrm{Q}}\)| = |\(\overrightarrow{\mathrm{P}}\) . \(\overrightarrow{\mathrm{Q}}\)| then angle between \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) then angle between P and Q will be –
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Answer:
(c) |\(\overrightarrow{\mathrm{P}}\) x \(\overrightarrow{\mathrm{Q}}\)| = |\(\overrightarrow{\mathrm{P}}\) . \(\overrightarrow{\mathrm{Q}}\)| Expand the terms
PQ sinθ = PQ cos θ
tan θ = 1
θ = 45°

Question 67.
If | \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) | = |\(\overrightarrow{\mathrm{P}}\) | |\(\overrightarrow{\mathrm{Q}}\)|, then angle between \(\overrightarrow{\mathrm{P}}\) and \(\overrightarrow{\mathrm{Q}}\) will be –
(a) 0°
(b) 45°
(c) 90°
(d) 180°
Answer:
(c) | \(\overrightarrow{\mathrm{P}}\) + \(\overrightarrow{\mathrm{Q}}\) | = |\(\overrightarrow{\mathrm{P}}\)| |\(\overrightarrow{\mathrm{Q}}\) |
Square on both side, and the resultants become,
P² + Q² + 2PQ cos 0 = P² + Q² – 2PQ cos θ 4PQ cos θ = 0
θ = 90°

Question 68.
If A and B are the sides of triangle, then area of triangle –
(a) \(\frac{1}{2}|\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}|\)
(b) \(\frac{1}{2}|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|\)
(c) AB sin θ
(d) AB cos θ
Answer:
(b) \(\frac{1}{2}|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|\)

Question 69.
A particle moves in a circular path of radius 2 cm. If a particle completes 3 rounds, then the distance and displacement of the particle are –
(a) 0 and 37.7
(b) 37.7 and 0
(c) 0 and 0
(d) 37.7 and 37.7
Answer:
(b) Radius = 2 cm
Circumference of the circle = 2nr = 4n cm
Distance covered in 3 rounds = 127r cm = 37.7 cm
Initial and final positions are same
∴ Displacement = 0

Question 70.
If rx and r2 are position vectors, then the displacement vector is –
(a) \(\vec{r}_{1} \times \vec{r}_{2}\)
(b) \(\vec{r}_{1} \cdot \vec{r}_{2} \)
(c) \(\vec{r}_{1}+\vec{r}_{2}\)
(d) \(\vec{r}_{2}+\vec{r}_{1} \)
Answer:
(d) \(\vec{r}_{2}+\vec{r}_{1} \)

Question 71.
The ratio of the displacement vector to the corresponding time interval is –
(a) average speed
(b) average velocity
(c) instantaneous speed
(d) instantaneous velocity
Answer:
(b) average velocity

Question 72.
The ratio of total path length travelled by the particle in a time interval –
(a) average speed
(b) average velocity
(c) instantaneous speed
(d) instantaneous velocity
Answer:
(a) average speed

Question 73.
The product of mass and velocity of a particle is –
(a) acceleration
(b) force
(c) torque
(d) momentum
Answer:
(d) momentum

Question 74.
The area under the force, displacement curve is –
(a) potential energy
(b) work done.
(c) impulse
(d) acceleration
Answer:
(b) work done

Question 75.
The area under the force, time graph is –
(a) momentum
(b) force
(c) work done
(d) impulse
Answer:
(d) impulse

Question 76.
The unit of momentum is –
(a) kg ms^-1
(b) kg ms^-2
(c) kg m²s^-1
(d) kg^-1 m² s^-1
Answer:
(b) kg ms^-2

Question 77.
The slope of the position – time graph will give –
(a) displacement
(b) velocity
(c) acceleration
(d) force
Answer:
(d) force

Question 78.
The area under velocity-time graph gives-
(a) positive
(b) negative
(c) either positive (or) negative
(d) zero
Answer:
(c) either positive (or) negative

Question 79.
The magnitude of distance is always-
(a) positive
(b) negative
(c) either positive (or) negative
(d) zero
Answer:
(a) positive

Question 80.
If two objects A and B are moving along a straight line in the same direction with the velocities v_A and v_B respectively, then the relative velocity is-
(a) v_A + v_B
(b) v_A – v_B
(c) v_A v_B
(d) v_A / v_B
Answer:
(b) V_A – V_B

Question 81.
If two objects A and B are moving along a straight line in the opposite direction with the velocities V_A and V_B respectively, then relative velocity is-
(a) V_A + V_B
(b) V_A – V_B
(c) V_{A .} V_B
(d) V_A / V_B
Answer:
(a) V_A + V_B

Question 82.
If two objects moving with a velocities of V_A and V_B at an angle of 0 between them, the relative velocity is –

Answer:

Question 83.
A person moving horizontally with velocity \(\overrightarrow{\mathrm{V}_{m}}\) The relative velocity of rain with respect to the person is –
(a) \(\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{\mathrm{m}}\)
(b) \(\sqrt{\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{m}}\)
(c) \(\mathrm{V}_{\mathrm{R}}-\mathrm{V}_{m}\)
(d) \(\sqrt{\mathrm{v}_{\mathrm{R}}^{2}+\mathrm{V}_{m}^{2}}\)
Answer:
(d) \(\sqrt{\mathrm{v}_{\mathrm{R}}^{2}+\mathrm{V}_{m}^{2}}\)

Question 84.
A person moving horizontally with velocity \(\overrightarrow{\mathrm{V}_{m}}\) . Rain falls vertically with velocity \(\overrightarrow{\mathrm{V}_{R}}\) To save himself from the rain, he should hold an umbrella with vertical at an angle of –
(a) \(\tan ^{-1}\left(\frac{V_{R}}{V_{m}}\right)\)
(b) \(\tan ^{-1}\left(\frac{V_{m}}{V_{R}}\right)\)
(c) \(\tan \theta=\mathrm{V}_{m}+\mathrm{V}_{\mathrm{R}}\)
(d) \(\tan ^{-1}\left(\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{m} / \mathrm{V}_{\mathrm{R}}-\mathrm{V}_{m}\right)\)
Answer:
(b) \(\tan ^{-1}\left(\frac{V_{m}}{V_{R}}\right)\)

Question 85.
A car starting from rest, accelerates at a constant rate x for sometime after which it decelerates at a constant rate v to come to rest. If the total time elapsed is t, the maximum velocity attained by the car is given by –

Answer:

Question 86.
A car covers half of its journey with a speed of 10 ms^-1 and the other half by 20 ms^-1. The average speed of car during the total journey is –
(a) 70 ms^-1
(b) 15 ms^-1
(c) 13.33 ms^-1
(d) 7.5 ms^-1
Answer:
(c) Let x is the total distance
Time to cover 1st half = \(\frac{x / 2}{10}\)
Time to cover 2nd half = \(\frac{x / 2}{20}\)
Average speed =

Question 87.
A swimmer can swim in still water at of 10 ms^-1 While crossing a river his average speed is 6 ms^-1. If he crosses the river in the shortest possible time, what is the speed of flow of water?
(a) 16 ms^-1
(b) 4 ms^-1
(c) 60 ms^-1
(d) 8 ms^-1
Answer:
(d) The resultant velocity of swimmer must be perpendicular to speed of water to cross the river in a shortest time
∴ \(v_{s}^{2}=v^{2}+v_{w}^{2}\)
\(v_{w}^{2}=v_{s}^{2}-v^{2}\) = 100 – 36 = 64
∴ V = 8 m/s^-1

Question 88.
A 100 m long train is traveling from North to South at a speed of 30 ms^-1. A bird is flying from South to North at a speed of 10^-1. How long will the bird take to, cross the train?
(a) 3 s
(b) 2.5 s
(c) 10 s
(d) 5 s
Answer:
(b) Length of train = 100 m
Relative velocity = 30 + 10 = 40 ms^-1
Time taken to cross the train (t) = \(\frac {distance}{ R.velocity }\) = \(\frac { 100 }{ 40 }\) = 2.5 s

Question 89.
The first derivative of position vector with respect to time is –
(a) velocity
(b) acceleration
(c) force
(d) displacement
Answer:
(a) velocity

Question 90.
The second derivative of position vector with respect to time is –
(a) velocity
(b) acceleration
(c) force
(d) displacement
Answer:
(b) acceleration

Question 91.
The slope of displacement-time graph gives –
(a) velocity
(b) acceleration
(c) force
(d) displacement
Answer:
(a) velocity

Question 92.
The slope of velocity-time graph gives –
(a) velocity
(b) acceleration
(c) force
(d) displacement
Answer:
(b) acceleration

Question 93.
The position vector of a particle is \(\vec{r}\) = 4t²\(\hat{i}\) + 2t\(\hat{j}\) + 3t\(\hat{k}\) The acceleration of a particle is having only –
(a) X – component
(b) Y – component
(c) Z – component
(d) X – Y component
Answer:
(a) X – component
4t²\(\hat{i}\) + 2t\(\hat{j}\) + 3t\(\hat{k}\)
\(\vec{v}\) = \(\frac{d \vec{r}}{dt}\) = 8t\(\hat{i}\) + 2\(\hat{j}\)
a = \(\frac{d^{2} r}{d t^{2}}\) = 8\(\hat{i}\) a is having only X-component.

Question 94.
The position vector of a particle is \(\vec{r}\) = 4t²\(\hat{i}\) + 2t\(\hat{j}\) + 3t\(\hat{k}\). The speed of the particle at t = 5 s is –
(a) 42 ms^-1
(b) 3s
(c) 3 ms^-1
(d) 40 ms^-1
Answer:
(a) 42 ms^-1
\(\vec{r}\) = 4t²\(\hat{i}\) + 2t\(\hat{j}\) + 3t\(\hat{k}\)
Speed v = — = \(\frac{d \vec{r}}{dt}\) = 8t\(\hat{i}\) + 2\(\hat{j}\)
at t = 5 s v = 40 + 2 = 42

Question 95.
An object is moving in a straight line with uniform acceleration a, the velocity-time relation is –
(a) u = v + at
(b) v = u + at
(c) v² = u² + a²t²
(d) v² – u² = at
Answer:
(b) v = u + at

Question 96.
An object is moving in a straight line with uniform acceleration, the displacement-time relation is –
(a) S = \(u t^{2}+\frac{1}{2} a t^{2}\)
(b) S = \(u t-\frac{1}{2} a t^{2}\)
(c) S = \(u t+\frac{1}{2} a t^{2}\)
(d) S = \(u t-a t^{2}\)
Answer:
(c) S = \(u t+\frac{1}{2} a t^{2}\)

Question 97.
An object is moving in a straight line with uniform acceleration, the velocity-displacement reflation is –
(a) V = u + 2as
(b) S = ut + -at
(c) V² = u² – 2as
(d) V² = u² + 2as
Answer:
(d) V² = u² + 2as

Question 98.
For free-falling body, its initial velocity is –
(a) 0
(b) 1
(c) ∞
(d) none
Answer:
(a) 0

Question 99.
An object falls from a height h (h<<R), the speed of the object when it reaches the ground is –
(a) \(\frac{1}{2} g t^{2}\)
(b) \(\sqrt{g t}\)
(c) gh
(d) \(\sqrt{2 g h}\)
Answer:
(d) \(\sqrt{2 g h}\)

Question 100.
An object falls from a height h (h<< R) the time taken by an object to reaches the ground is –
(a) \(\frac{1}{2} g t^{2}\)
(b) \(\sqrt{2 g h}\)
(c) \(\sqrt{\frac{2 h}{g}}\)
(d) \(\sqrt{\frac{2 g}{h}}\)
Answer:
(d) \(\sqrt{\frac{2 g}{h}}\)

Question 101.
In the absence of air resistance, horizontal velocity of the projectile is –
(a) always negative
(b) equal to ‘g’
(c) directly proportional to g
(d) a constant
Answer:
(d) a constant

Question 102.
In the horizontal projection, the range of the projectile is –
(a) \(\sqrt{\frac{2 h}{g}}\)
(b) \(u \sqrt{\frac{h}{g}}\)
(c) \(u \sqrt{\frac{2 h}{g}}\)
(d) \(u \sqrt{\frac{g}{2 h}}\)
Answer:
(c) \(u \sqrt{\frac{2 h}{g}}\)

Question 103.
In oblique projection, maximum height attained by the projectile is –
(a) \(\frac { t }{ u cos θ }\)
(b) \(\frac { u cos θ }{ 2g }\)
(c) \(\frac { 2g }{ u cos θ }\)
(d) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
Answer:
(d) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)

Question 104.
In oblique projection time of flight of a projectile is –
(a) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
(b) \(\frac { 2u cos θ }{ g }\)
(c) \(\frac{u^{2} \sin 2 \theta}{g}\)
(d) \(\frac{u^{2}}{g}\)
Answer:
(b) \(\frac { 2u cos θ }{ g }\)

Question 105.
In oblique projection horizontal range of the projectile is –
(a) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
(b) \(\frac { 2u cos θ }{ g }\)
(c) \(\frac{u^{2} \sin 2 \theta}{g}\)
(d) \(\frac{u^{2}}{g}\)
Answer:
(a) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)

Question 106.
In oblique projection, maximum horizontal range of the projectile is –
(a) \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
(b) \(\frac { 2u cos θ }{ g }\)
(c) \(\frac{u^{2} \sin 2 \theta}{g}\)
(d) \(\frac{u^{2}}{g}\)
Answer:
(d) \(\frac{u^{2}}{g}\)

Question 107.
One radian is equal to –
(a) \(\frac {π}{ 180 }\) degree
(b) 60°
(c) 57.295°
(d) 53.925°
Answer:
(c) 57.295°

Question 108.
In relation between linear and angular velocity is –
(a) ω = vr
(b) ω = \(\frac {v }{ r }\)
(c) ω = \(\frac { r}{ v }\)
(d) ω = \(\frac { r }{ ω }\)
Answer:
(b) ω = \(\frac {v }{ r }\)

Question 109.
Centripetal acceleration is given by –
(a) \(\frac{v^{2}}{r}\)
(b) \(-\frac{v^{2}}{r}\)
(c) \(\frac{r}{v^{2}}\)
(d) \(-\frac{r}{v^{2}}\)
Answer:
(b) \(-\frac{v^{2}}{r}\)

Question 110.
In uniform circular motion –
(a) Speed changes but velocity constant
(b) Velocity changes but speed constant
(c) both speed and velocity are constant
(d) both speed and velocity are variable
Answer:
(b) Velocity changes but speed constant

Question 111.
In non – uniform circular motion, the resultant acceleration is given by –

Answer:

Question 112.
In non – uniform circular motion, the resultant acceleration makes an angle with the radius vector is –

Answer:

Question 113.
A compartment of an uniformly moving train is suddenly detached from the train and stops after covering some distance. The distance covered by the compartment and distance covered by the train in the given time –
(a) both will be equal
(b) second will be half of first
(c) first will be half of second
(d) none
Answer:
(c) first will be half of second

Question 114.
An object is dropped from rest. Its v – t graph is –

Answer:

Question 115.
When a ball hits the ground as free fall and renounces but less than its original height? Which is represented by –

Answer:

Question 116.
Which of the following graph represents the equation y = mx – C?

Answer:

Question 117.
Which of the following graph represents the equation y = mx + C?

Answer:

Question 118.
Which of the following graph represents the equation y = mx?

Answer:

Question 119.
Which of the following graph represents the equation y = -mx + C?

Answer:

Question 120.
Which of the following graph represents the equation y = kx²?

Answer:

Question 121.
X = -ky² is represented by –

Answer:

Question 122.
X = ky² is represented by –

Answer:

Question 123.
y = kx² is represented by –

Answer:

Question 124.
X °∝\(\frac { 1 }{ Y }\) (or) XY = constant is represented by –

Answer:

Question 125.
y = e^-kx is represented by –

Answer:

Question 126.
Y = 1 – e^-kx is represented by –

Answer:

Question 127.
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 is represented by –

Answer:

Question 128.
Let y =f(x) is a function. Its maxima (or) minima can be obtained by –
(a) y = 0
(b) f(x) = 0
(c) \(\frac {dy}{dx}\) = 0
(d) \(\frac{d^{2} y}{d x^{2}}\) = 0
Answer:
(c) \(\frac {dy}{dx}\) = 0

Question 129.
A particle at rest starts moving in a horizontal straight line with uniform acceleration. The ratio of the distance covered during the fourth and the third second is –
(a) \(\frac {4}{ 3 }\)
(b) \(\frac { 26 }{ 9 }\)
(e) \(\frac { 7}{ 5 }\)
(d) 2
Answer:
(c) The distance travelled during nth second –
S_n  = u + \(\frac { 1 }{ 2 }\) a (2n -1)
Distance travelled during 4th second S₁ = \(\frac { 1 }{ 2 }\) (8 – 1)
Distance travelled during 3rd second S₂ = \(\frac { 1 }{ 2 }\) a(6 – 1)
\(\frac{\mathrm{S}_{1}}{\mathrm{S}_{2}}\) = \(\frac { 7 }{ 5 }\)

Question 130.
The distance travelled by a body, falling freely from rest in t = 1s, t = 2s and t = 3s are in the ratio of –
(a) 1 : 2 : 3
(h) 1 : 3 : 5
(c) 1 : 4 : 9
(d) 9 : 4 : 1
Answer:
(c) The distance travelled by a free falling body S = \(\frac { 1 }{ 2 }\) gt²
∴ S α t²
∴ S₁ : S₂  : S₃ : 1² : 2² : 3² = 1 : 4 : 9.

Question 131.
The displacement of the particle along a straight line at time ¡ is given by X = a + ht + ct² where a, b, c are constants. The acceleration of the particle is-
(a) a
(b) b
(c) c
(d) 2c
Answer:
(d) X = a + bt + ct²
\(\frac { dX }{ dt }\) = v = b + 2ct
Acceleration = \(\frac{d^{2} X}{d t^{2}}\) = 2c.

Question 132.
Two bullets are fired at an angle of θ and (90 – θ) to the horizontal with same speed. The ratio of their times of flight is –
(a) 1 : 1
(b) 1: tan θ
(c) tan θ : 1
(d) tan² θ : 1
Answer:
(c) Time of flight t_f =  \(\frac { 2x sinθ}{ 9 }\)
t_f α sinθ
∴ \(\frac{t_{f_{1}}}{t_{f_{2}}}\) = \(\frac { sinθ }{sin (90 – θ) }\) = \(\frac { sinθ }{cos θ }\) = tanθ
\(t_{f_{1}}: t_{f_{2}}\) = tan θ : 1

Question 133.
A particle moves along a circular path under the action of a force. The work done by the force is –
(a) positive and non zero
(b) zero
(c) negative and non-zero
(d) none
Answer:
(b) zero

Question 134.
For a particle, revolving in a circle with speed, the acceleration of the particle is –
(a) along the tangent
(b) along the radius
(c) along its circumference
(d) zero
Answer:
(b) along the radius

Question 135.
A gun fires two bullets with same velocity at 60° and 30° with horizontal. The bullets strike at the same horizontal distance. The ratio of maximum height for the two bullets is in the ratio of –
(a) 1 : 2
(b) 3 : 1
(c) 2 : 1
(d) 1 : 3
Answer:
(b) 3 : 1
Max height attained h_max = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
∴ h_max α sin² θ i.e h_max α \(\frac { 1-cos2θ}{ 2 }\)
\(\frac{h_{\max 1}}{h_{\max } 2}\) = \(\frac{3 / 2}{1 / 2}\) = 3

Question 136.
A ball is thrown vertically upward. it is a speed of lo m/s. When it has reached one half of its maximum height. I-low high does the ball rise? (g = 10 ms^-2)
(a) 5 m
(b) 7 m
(c) 10 m
(d) 12 m
Answer:
(c) 10 m

Question 137.
A car moves from X to Y with a uniform speed V_n  and returns to Y with a uniform speed V_d The average speed for this round trip is –

Answer:

Question 138.
Two projectiles of same mass and with same velocity are thrown at an angle of 60° and 30° with the horizontal then which of the following will remain same?
(a) time of flight
(b) range of projectile
(c) maximum height reached
(d) all the above
Answer:
(b) range of projectile

Question 139.
A n object of mass 3 kg is at rest. Now a force of \(\overrightarrow{\mathrm{F}}\) = 6 t²\(\hat{i}\) + 4t\(\hat{j}\) is applied on the object, then the velocity of object at t = 3 second is –
(a) 18\(\hat{i}\) + 3 \(\hat{j}\)
(b) 18\(\hat{i}\) + 6\(\hat{j}\)
(c) 3 \(\hat{i}\) + 18\(\hat{j}\)
(d) 18 \(\hat{i}\) + 4\(\hat{j}\)
Answer:
(b) F = 6 t²\(\hat{i}\) + 4t\(\hat{j}\)

Question 140.
The angle for which maximum height and horizontal range are same for a projectile is –
(a) 32°
(b) 48°
(c) 76°
(d) 84°
Answer:
(c) 76°
H_max = horizontal range
\(\frac{u^{2} \sin ^{2} \theta}{2 g}\) = \(\frac{u^{2} \sin 2 \theta}{g}\)
\(\frac{\sin ^{2} \theta}{2}\) = 2 sin θ cos θ = sin θ = 4 cos θ tan θ = 4 θ = 76°

Question 141.
A bullet is dropped from some height, when another bullet is fired horizontally from the same height. They will hit the ground –

(a) depends upon mass of bullet
(b) depends upon the observer
(c) one after another
(d) simultaneously
Answer:
(d) simultaneously

Question 142.
From this velocity – time graph, which of the following is correct?
(a) Constant acceleration
(b) Variable acceleration
(c) Constant velocity
(d) Variable acceleration
Answer:
(b) Variable acceleration

Question 143.
When a projectile is at its maximum height, the direction of its velocity and acceleration are –
(a) parallel to each other
(b) perpendicular to each other
(c) anti – parallel to each other
(d) depends on its speed
Answer:
(b) perpendicular to each other

Question 144.
At the highest point of oblique projection, which of the following is correct?
(a) velocity of the projectile is zero
(b) acceleration of the projectile is zero
(c) acceleration of the projectile is vertically downwards
(d) velocity of the projectile is vertically downwards
Answer:
(c) acceleration of the projectile Is vertically downwards

Question 145.
The range of the projectile depends –
(a) The angle of projection
(b) Velocity of projection
(c) g
(d) all the above
Answer:
(d) all the above

Question 146.
A constant force is acting on a particle and also acting perpendicular to the velocity of the particle. The particle describes the motion in a plane. Then –
(a) angular displacement is zero
(b) its velocity is zero
(c) it velocity is constant
(d) it moves in a circular path
Answer:
(d) it moves in a circular path

Question 147.
If a body moving in a circular path with uniform speed, then –
(a) the acceleration is directed towards its center
(b) velocity and acceleration are perpendicular to each other
(c) speed of the body is constant but its velocity is varying
(d) all the above
Answer:
(d) all the above

Question 148.
A body is projected vertically upward with the velocity y = 3\(\hat{i}\) + 4\(\hat{j}\) ms-¹. The maximum height attained by the body is (g 10 ms^-2).
(a) 7 m
(b) 1.25 m
(c) 8 m
(d) 0.08 m
Answer:
(b) v = 3\(\hat{i}\) + 4\(\hat{j}\)
H_max = \(\frac{v^{2} \sin ^{2} \theta}{2 g}\) = \(\frac{v^{2}}{2 g}\) [ θ = 90 ]
v = \(\sqrt{9+16}\)  = \(\sqrt{25}\)
v² = 25
H_max = \(\frac{25}{20}\)  = \(\frac { 5 }{ 4 }\) = 1.25 m

Samacheer Kalvi 11th Physics Kinematics Short Answer Questions – I (1 Mark)

Question 1.
What is frame of reference?
Answer:
In a coordinate system, the position of an object is described relative to it, then such a coordinate system is called as frame of reference.

Question 2.
What are the types of motion?
Answer:

• Linear motion
• Circular motion
• Rotational motion
• Vibratory motion.

Question 3.
What is linear motion? Give example.
Answer:
An object is said to be in linear motion if it moves in a straight line.
Example – an athlete running on a straight track.

Question 4.
What is circular motion? Give example.
Answer:
Circular motion is defined as a motion described by an object traversing a circular path.
Example – The whirling motion of a stone attached to a string.

Question 5.
What is rotational motion? Give example.
Answer:
During a motion every point in the object traverses a circular path about an axis except the points located on the axis, is called as rotational motion.
Example – Spinning of the earth about its own axis.

Question 6.
What is vibratory motion? Give example.
Answer:
If an object or particle executes a to and fro motion about a fixed point, it is said to be in vibratory motion.
Example – Vibration of a string on a guitar.

Question 7.
What is one dimensional motion? Give example.
Answer:
One dimensional motion is the motion of a particle moving along a straight line.
Example –  Motion of a train along a straight railway track.

Question 8.
What is two dimensional motion? Give example.
Answer:
If a particle moving along a curved path in a plane, then it is said to be in two dimensional motion.
Example – Motion of a coin on a carrom board.

Question 9.
What is three dimensional motion? Give example.
Answer:
If a particle moving in used three dimensional space, then the particle is said to be in three dimensional motion.
E.g. A bird flying in the sky.

Question 10.
Write about the properties of components of vectors.
Answer:
If two vectors \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{B}}\) are equal, then their individual components are also equal. then their individual components are also equal.
Let\(\overline{\mathrm{A}}\) = \(\overline{\mathrm{B}}\)
then A_x  \(\hat{i}\) + A_y \(\hat{j}\) + A_z \(\hat{k}\) = B_x \(\hat{i}\) + B_y \(\hat{j}\) + B_z \(\hat{k}\)
i.e. A_x = B_x, A_y =  B_y  = A_z = B_z

Question 11.
Give an example for scalar product of two vectors.
Answer:
The work done by a force \(\overrightarrow{\mathrm{F}}\) to move an object through a small displacement \(\overrightarrow{\mathrm{dr}}\) then
Work done W = \(\overrightarrow{\mathrm{F}}\) .\(\overrightarrow{\mathrm{dr}}\) (or) W = F dr cos θ

Question 12.
Give any three example for vector product of two vectors.
Answer:

1. Torque \(\overrightarrow{\mathrm{t}}\) = \(\overrightarrow{\mathrm{r}}\) x \(\overrightarrow{\mathrm{F}}\). Where i is force and \(\overrightarrow{\mathrm{F}}\) is force and \(\overrightarrow{\mathrm{r}}\) position vector of a particle.
2. Angular momentum \(\overrightarrow{\mathrm{L}}\) = \(\overrightarrow{\mathrm{r}}\) x \(\overrightarrow{\mathrm{P}}\) where \(\overrightarrow{\mathrm{P}}\) is the linear momentum.
3. Linear velocity \(\overrightarrow{\mathrm{v}}\) = \(\overrightarrow{\mathrm{ω}}\) x \(\overrightarrow{\mathrm{r}}\) where \(\overrightarrow{\mathrm{ω}}\) is angular velocity.

Question 13.
What is position vector?
Answer:
It is vector which denotes the position of a particle at any instant of time, with respect to some reference frame or coordinate system.
The position \(\overrightarrow{\mathrm{r}}\) vector of the particle at a point P is given by
\(\overrightarrow{\mathrm{r}}\) = x\(\hat{i}\) + y\(\hat{j}\) + z\(\hat{k}\)
where x, y and z are components of \(\overrightarrow{\mathrm{r}}\).

Question 14.
Write a note an momentum.
Answer:
Momentum of a particle is defined as product of mass with velocity. It is denoted as \(\overrightarrow{\mathrm{p}}\) Momentum is also a vector quantity
\(\overrightarrow{\mathrm{r}}\) = m\(\overrightarrow{\mathrm{v}}\)
The direction of momentum is also in the direction of velocity, and the magnitude of momentum is equal to product of mass and speed of the particle.
p = mv
In component form the momentum can be written as
p_x\(\hat{i}\) + p_y\(\hat{j}\)+ p_z\(\hat{k}\) = mv_x\(\hat{i}\) + mv_y\(\hat{j}\) + mv_z\(\hat{k}\)
Here,
p_x = x component of momentum and is equal to mv_x
P_x = y component of momentum and is equal to mv_y
P_x = z component of momentum and is equal to mv_z

Question 15.
“Displacement vector is basically a position vector”. Comment on it.
Answer:
This statement is almost correct only. Because the displacement vector also gives the position of a point just like a position vector. The difference between these two vectors is p. The displacement vector gives the position of a point with respect to a point other than origin but position vector gives the position of a point with respect to origin.

Question 16.
Will two dimensional motion with an acceleration only in one dimension?
Answer:
Yes. In oblique projection, the acceleration is acting vertically downward but the object follows a parabolic path.

Question 17.
A foot ball is kicked by a player with certain angle to the horizontal. Is there any point at which velocity is perpendicular to its acceleration.
Answer:
Yes. At its maximum height in the parabolic path vertical velocity is zero but due to horizontal component, velocity acts along horizontally, but acceleration of the

Question 18.
Give any two examples for parallelogram law of vectors.
Answer:

• the flight of a bird
• working of a sling.

Question 19.
Why does rubber ball bounce greater heights on hills than in plains?
Answer:
The maximum height attained by the projectile is inversely proportional to acceleration due to gravity. At greater height, acceleration due to gravity will be lesser than plains. So ball can bounce higher in hills than in plains.

Question 20.
Is it possible for body to have variable velocity but constant speed? Give example.
Answer:
Yes, it is possible. In horizontal circular motion the speed of a particle is always constant but due to the variation in direction continuously, the velocity of a particle varies.

Question 21.
What is relative velocity?
Answer:
When two objects are moving with different velocities, then the velocity of one object with respect to another object is called relative velocity of an object with respect to another.

Question 22.
What is average acceleration?
Answer:
The average acceleration is defined as the ratio of change in velocity over the time interval
a_avg = \(\frac{\Delta \overrightarrow{\mathrm{v}}}{\Delta t}\) It is a vector quantity.

Question 23.
Write a note an instantaneous acceleration.
Answer:
Instantaneous acceleration or acceleration of a particle at time ‘t’ is given by the ratio of change in velocity over ∆t, as ∆t approaches zero.
Acceleration
In other words, the acceleration of the particle at an instant t is equal to rate of change of velocity

(1) Acceleration is a vector quantity. Its SI unit is ms^-2 and its dimensional formula is [M°L¹ T^-2]

(2) Acceleration is positive if its velocity is increasing, and is negative if the velocity is decreasing. The negative acceleration is called retardation or deceleration.

Question 24.
Write an acceleration in terms of its component?
(Or)
Show that the acceleration is the second derivative of position vector with respect to time.
Answer:
in terms of components, we can write,


are the components of instantaneous acceleration. Since each component of velocity is the derivative of the corresponding coordinate, we can express the components a_x, a_y, and a_z as

Then the acceleration vector \(\overrightarrow{\mathrm{a}}\) it self is

Thus acceleration is the second derivative of position vector with respect to time.

Question 25.
What are the examples of projectile motion?
Answer:

1. An object dropped from window of a moving train.
2. A bullet fired from a rifle.
3. A ball thrown in any direction.
4. A javelin or shot put thrown by an athlete.
5. A jet of water issuing from a hole near the bottom of a water tank.

Question 26.
Explain projectile motion.
Answer:
A projectile moves under the combined effect of two velocities.

• A uniform velocity in the horizontal direction, which will not change provided there is no air resistance.
• A uniformly changing velocity (i.e., increasing or decreasing) in the vertical direction.

There are two types of projectile motion:

• Projectile given an initial velocity in the horizontal direction (horizontal projection)
• Projectile given an initial velocity at an angle to the horizontal (angular projection)

To study the motion of a projectile, let us assume that,

• Air resistance is neglected.
• The effect due to rotation of Earth and curvature of Earth is negligible.
• The acceleration due to gravity is constant in magnitude and direction at all points of the motion of the projectile.

Question 27.
What is time of flight?
Answer:
The time taken for the projectile to complete its trajectory or time taken by the projectile to hit the ground is called time of flight.

Question 28.
Under what condition is the average velocity equal the instantaneous velocity?
Answer:
When the body is moving with uniform velocity.

Question 29.
Draw Position time graph of two objects, A & B moving along a straight line, when their relative velocity is zero.

Question 30.
Suggest a situation in which an object is accelerated and have constant speed.
Answer:
Uniform Circular Motion.

Question 31.
Two balls of different masses are thrown vertically upward with same initial velocity. Maximum heights attained by them are h₁ and h₂ respectively what is h₁/h₂?
Answer:
Same height,
∴  h₁/h₂ = 1

Question 32.
A car moving with velocity of 50 kmh^-1 on a straight road is ahead of a jeep moving with velocity 75 kmh^-1 would the relative velocity be altered if jeep is ahead of car?
Answer:
No change.

Question 33.
Which of the two – linear velocity or the linear acceleration gives the direction of motion of a body?
Answer:
Linear velocity.

Question 34.
Will the displacement of a particle change on changing the position of origin of the coordinate system?
Answer:
Will not change.

Question 35.
If the instantaneous velocity of a particle is zero, will its instantaneous acceleration be necessarily zero?
Answer:
No. (highest point of vertical upward motion under gravity).

Question 36.
A projectile is fired with Kinetic energy 1 KJ. If the range is maximum, what is its Kinetic energy, at the highest point?
Answer:
Here \(\frac { 1 }{ 2 }\) mv² =1kJ=1000 J, θ = 45°
At the highest point, K.E. = \(\frac { 1 }{ 2 }\) m(v cos 0)² = \(\frac { 1 }{ 2 }\)\(\frac{m v^{2}}{2}\) = \(\frac {1000}{ 2 }\) = 500 J.

Question 37.
Write an example of zero vector.
Answer:
The velocity vectors of a stationary object is a zero vectors.

Question 38.
State the essential condition for the addition of vectors.
Answer:
They must represent the physical quantities of same native.

Question 39.
When is the magnitude of (\(\overline{\mathrm{A}}\) + \(\overline{\mathrm{B}}\)) equal to the magnitude of (\(\overline{\mathrm{A}}\) – \(\overline{\mathrm{B}}\))?
Answer:
When \(\overline{\mathrm{A}}\) is perpendicular to \(\overline{\mathrm{B}}\).

Question 40.
What is the maximum number of component into which a vector can be resolved?
Answer:
Infinite.

Question 41.
A body projected horizontally moves with the same horizontal velocity although it moves under gravity Why?
Answer:
Because horizontal component of gravity is zero along horizontal direction.

Question 42.
What is the angle between velocity and acceleration at the highest point of a projectile motion?
Answer:
90°.

Question 43.
When does

• height attained by a projectile maximum?
• horizontal range is maximum?

Answer:

• Height is maximum at θ = 90
• Range is maximum at θ = 45.

Question 44.
What is the angle between velocity vector and acceleration vector in uniform circular motion?
Answer:
90°.

Question 45.
A particle is in clockwise uniform circular motion the direction of its acceleration is radially inward. If sense of rotation or particle is anticlockwise then what is the direction of its acceleration?
Answer:
Radial in ward.

Question 46.
A train is moving on a straight track with acceleration a. A passenger drops a stone. What is the acceleration of stone with respect to passenger?
Answer:
\(\sqrt{a^{2}+g^{2}}\) where g = acceleration due to gravity.

Question 47.
What is the average value of acceleration vector in uniform circular motion over one cycle?
Answer:
Null vector.

Question 48.
Does a vector quantity depends upon frame of reference chosen?
Answer:
No.

Question 49.
What is the angular velocity of the hour hand of a clock?
Answer:
ω = \(\frac {2π}{ 12 }\) = \(\frac { π }{6}\) rad h^-1

Question 50.
What is the source of centripetal acceleration for earth to go round the sun?
Answer:
Gravitation force of sun.

Question 51.
What is the angle between (\(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{A}}\) ) and (\(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{A}}\)) ?
Answer:
90°

Samacheer Kalvi 11th Physics Kinematics Short Answer Questions – II (2 Marks)

Question 1.
What are positive and negative acceleration in straight line motion?
Solution:
If speed of an object increases with time, its acceleration is positive. (Acceleration is in the direction of motion) and if speed of an object decreases with time its acceleration is negative (Acceleration is opposite to the direction of motion).

Question 2.
Can a body have zero velocity and still be accelerating? If yes gives any situation.
Solution:
Yes, at the highest point of vertical upward motion under gravity.

Question 3.
The displacement of a body is proportional to t³, where t is time elapsed. What is the nature of acceleration –  time graph of the body?
Solution:
As a α t³ ⇒ s = kt³
Velocity, V = \(\frac { ds }{ dt }\) = 3 kt³
Acceleration, a = \(\frac { dv }{ dt }\) = 3 kt³
i.e., a α t
⇒ motion is uniform, acceleration motion, a – t graph is straight-line.

Question 4.
Suggest a suitable physical situation for the following graph.

Solution:
A ball thrown up with some initial velocity rebounding from the floor with reduced speed after each hit.

Question 5.
An object is in uniform motion along a straight line, what will be position time graph for the motion of object, if (i) x₀ = positive, v = negative is constant.
(i) x₀ = positive, v = negative is |\(\vec{v}\) | constant.
(ii) both x₀ and v are negative |\(\vec{v}\) | is constant.
(iii) x₀ = negative, v = positive |\(\vec{v}\) | is constant.
(iv) both x₀ and v are positive |\(\vec{v}\) | is constant where x₀ is position at t = 0.
Solution:

Question 6.
A cyclist starts from centre O of a circular park of radius 1 km and moves along the path OPRQO as shown. If he maintains constant speed of 10 ms^-1. What is his acceleration at point R in magnitude & direction?

Solution:
Centripetal acceleration, a_c = \(\frac{v^{2}}{r}\) = \(\frac{10^{2}}{1000}\) = 0.1 m/s² along RO.

Question 7.
What will be the effect on horizontal range of a projectile when its initial velocity is doubled keeping angle of projection same?
Solution:
\(\frac{u^{2} \sin 2 \theta}{g}\) ⇒ R α u²
Range comes four times.

Question 8.
The greatest height to which a man can throw a stone is h. What will be the greatest distance upto which he can throw the stone?
Solution:
Maximum height:
H = \(\frac{u^{2} \sin ^{2} \theta}{g}\) ⇒ H_max = \(\frac{u^{2}}{2g}\) = h (at θ = 90)
Maximum range R_max = \(\frac{u^{2}}{g}\) = 2h

Question 9.
A person sitting in a train moving at constant velocity throws a ball vertically upwards. How will the ball appear to move to an observer.

• Sitting inside the train
• Standing outside the train

Solution:

• Vertical straight line motion
• Parabolic path.

Question 10.
A gunman always keep his gun slightly tilted above the line of sight while shooting. Why?
Solution:
Because bullet follow Parabolic trajectory under constant downward acceleration.

Question 11.
Is the acceleration of a particle in circular motion not always towards the center. Explain.
Solution:
No acceleration is towards the center only in case of uniform circular motion.

Samacheer Kalvi 11th Physics Kinematics Short Answer Questions – III (3 Marks)

Question 1.
Draw
(a) acceleration – time
(b) velocity – time
(c) Position – time graphs representing motion of an object under free fall. Neglect air resistance.
Solution:
The object falls with uniform acceleration equal to ‘g’

Question 2.
The velocity time graph for a particle is shown in figure. Draw acceleration time graph from it.

Solution:

Question 3.
For an object projected upward with a velocity v₀, which comes back to the same point after some time, draw
(i) Acceleration – time graph
(ii) Position – time graph
(iii) Velocity time graph
Solution:

Question 4.
The acceleration of a particle in ms² is given by a = 3t² + 2t + 2, where time t is in second. If the particle starts with a velocity v = 2 ms^-1 at t = 0, then find the velocity at the end of 2s.
Solution:

Question 5.
At what angle do the two forces (P + Q) and (P – Q) act so that the resultant is \(\sqrt{3 P^{2}+Q^{2}}\)?
Solution:
Use
R = \(\sqrt{3 P^{2}+Q^{2}}\)
R = \(\sqrt{3 \mathrm{P}^{2}+\mathrm{Q}^{2}}\)
A = P + Q
B = P – Q
solve, θ = 60°

Question 6.
A car moving along a straight highway with speed of 126 km h 1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?
Solution:
Initial velocity of car,
u = 126 kmh^-1 = 126 x \(\frac {5}{18}\) ms^-1 = 35 ms^-1 ………(i)
Since, the car finally comes to rest, v = 0
Distance covered, s = 200 m, a = ?, t = ?
v² = u² – 2as
or a = \(\frac{v^{2}-u^{2}}{2 s}\) ………..(ii)
substituting the values from eq. (i) in eq . (ii), we get
a = \(\frac{0-(35)^{2}}{2 \times 200}\) = \(\frac{0-(35)^{2}}{2 \times 200}\)
= \(-\frac{46}{16}\) ms^-2 = -3.06 ms^-2
Negative sign shows that acceleration in negative which is called retardation, i.e., car is uniformly retarded at – a = 3.06 ms^-2.
To find t, let us use the relation
v = u + at
t = \(\frac {v-u}{ a }\)
use a = -3.06 ms^-2, v = 0, u = 35 ms^-1
∴ t = \(\frac {v-u}{ a }\) = \(\frac {0-35}{ -3.06 }\) = 11.44 s
∴ t = 11.44 sec

Samacheer Kalvi 11th Physics Kinematics Long Answer Questions
Question 1.
Explain the types of motion with example.
Answer:
(a) Linear motion:
An object is said to be in linear motion if it moves in a straight line.
Examples:

• An athlete running on a straight track
• AA particle falling vertically downwards to the Earth.

(b) Circular motion:
Circular motion is defined as a motion described by an object traversing a circular path.
Examples:

• The whirling motion of a stone attached to a string.
• The motion of a satellite around the Earth.
• These two circular motions are shown in figure.

(c) Rotational motion:
If any object moves in a rotational motion about an axis, the motion is called ‘rotation’. During rotation every point in the object transverses a circular path about an axis, (except the points located on the axis).
Examples:

• Rotation of a disc about an axis through its center
• Spinning of the Earth about its own axis.
• These two rotational motions are shown in figure

(d) Vibratory motion:
If an object or particle executes a to-and-fro motion about a fixed point, it is said to be in vibratory motion. This is sometimes also called oscillatory motion.
Examples:

• Vibration of a string on a guitar
• Movement of a swing
• These motions are shown in figure

Other types of motion like elliptical motion and helical motion are also possible.

Question 2.
What are the different types of vectors? ,
Answer:
1. Equal vectors:
Two vectors A and B are said to be equal when they have equal magnitude and same direction and represent the same physical quantity

(a) Collinear vectors:
Col-linear vectors are those which act along the same line. The angle between them can be 0° or 180°.

(i) Parallel vectors:
If two vectors A and B act in the same direction along the same line or on parallel lines, then the angle between them is 0°. Geometrical representation of parallel vectors.

(ii) Anti-parallel vectors:
Two vectors A and B are said to be anti – parallel when they are in opposite directions along the same line or on parallel lines. Then the angle between them is 180°.

2. Unit vector:
A vector divided by its magnitude is a unit vector. The unit vector for \(\overrightarrow{\mathrm{A}}\) is denoted by \(\widehat{A}\) . It has a magnitude equal to unity or one.
Since, \(\widehat{A}\) = \(\frac{\bar{A}}{A}\) we can write \(\overrightarrow{\mathrm{A}}\) = A\(\widehat{A}\)
Thus, we can say that the unit vector specifies only the direction of the vector quantity.

3. Orthogonal unit vectors:
Let \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) be three unit vectors which specify the directions along positive x-axis, positive y-axis and positive z-axis respectively. These three unit vectors are directed perpendicular to each other, the angle between any two of them is 90°.\(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) and are examples of orthogonal vectors. Two vectors which are perpendicular to each other are called orthogonal vectors as shown in the figure.

Question 3.
Explain the concept of relative velocity in one and two dimensional motion.
Answer:
When two objects A and B are moving with different velocities, then the velocity of one object A with respect to another object B is called relative velocity of object A with respect to B.

Case I:
Consider two objects A and B moving with uniform velocities V_A and V_B, as shown, along straight tracks in the same direction \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\), \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) with respect to ground.
The relative velocity of object A with respect to object B is \(\overrightarrow{\mathrm{V}}_{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\).
The relative velocity of object B with respect to object A is \(\overrightarrow{\mathrm{V}}_{\mathrm{BA}}\) = \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) Thus, if two objects are moving in the same direction, the magnitude of relative velocity of one object with respect to another is equal to the difference in magnitude of two velocities.

Case II.
Consider two objects A and B moving with uniform velocities \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\) along the same straight tracks but opposite in direction.

The relative velocity of an object A with respect to object B is –
\(\overrightarrow{\mathrm{V}}_{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) – (-\(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\)) = \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) + \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\)

The relative velocity of an object B with respect to object A is
\(\overrightarrow{\mathrm{V}}_{\mathrm{AB}}\) = –\(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) = – (\(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) + \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\))
Thus, if two objects are moving in opposite directions, the magnitude of relative velocity of one object with respect to other is equal to the sum of magnitude of their velocities.

Case III.
Consider the velocities \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) at an angle θ between their directions. The relative velocity of A with respect to B, \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\)
Then, the magnitude and direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) is given by \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = \(\sqrt{\vec{v}_{\mathrm{A}}^{2}+\vec{v}_{\mathrm{B}}^{2}-2 v_{\mathrm{A}} v_{\mathrm{B}} \cos \theta}\) and tan β = \(\frac{v_{\mathrm{B}} \sin \theta}{v_{\mathrm{A}}-v_{\mathrm{B}} \cos \theta}\) (Here β is angle between (\(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) and \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\))
\(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) cos θ .

(i) When θ = 0, the bodies move along parallel straight lines in the same direction, We have \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = (\(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) – \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\)) in the direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) . Obviously \(\overrightarrow{\mathrm{v}}_{\mathrm{BA}}\) = (\(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) + \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\)) in the direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\).

(ii) When θ = 180°, the bodies move along parallel straight lines in opposite directions,
We have \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = (\(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\)+ \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\)) in the direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{A}}\) . Similarly, vBA = (vB + vA) in the direction of \(\overrightarrow{\mathrm{v}}_{\mathrm{B}}\) .

(iii) If the two bodies are moving at right angles to each other, then 0 = 90°. The magnitude of the relative velocity of A with respect to B = \(\overrightarrow{\mathrm{v}}_{\mathrm{AB}}\) = \(\sqrt{v_{\mathrm{A}}^{2}+v_{\mathrm{B}}^{2}}\).

(iv) Consider a person moving horizontally with velocity \(\overrightarrow{\mathrm{V}}_{\mathrm{M}}\) . Let rain fall vertically with velocity \(\overrightarrow{\mathrm{V}}_{\mathrm{R}}\) . An umbrella is held to avoid the rain. Then the relative velocity of the rain with respect to the person is,

which has magnitude
\(\overrightarrow{\mathrm{V}}_{\mathrm{RM}}\) = \(\overrightarrow{\mathrm{V}}_{\mathrm{R}}\) – \(\overrightarrow{\mathrm{V}}_{\mathrm{M}}\)
And direction 0 = tan^-1\(\left(\frac{V_{\mathrm{M}}}{\mathrm{V}_{\mathrm{R}}}\right)\) with the vertical as shown in figure.

Question 4.
Shows that the path of horizontal projectile is a parabola and derive an expression for
1. Time of flight
2. Horizontal range
3. resultant relative and any instant
4. speed of the projectile when it hits the ground?
Answer:
Consider a projectile, say a ball, thrown horizontally with an initial velocity \(\vec{u}\) from the top of a tower of height h. As the ball moves, it covers a horizontal distance due to its uniform horizontal velocity u, and a vertical downward distance because of constant acceleration due to gravity g. Thus, under the combined effect the ball moves along the path OPA. The motion is in a 2 – dimensional plane. Let the ball take time t to reach the ground at point A, Then the horizontal distance travelled by the ball is x(t) = x, and the vertical distance travelled is y(t) = y.

We can apply the kinematic equations along the x direction and y direction separately. Since this is two-dimensional motion, the velocity will have both horizontal component u_x and vertical component u_y.

Motion along horizontal direction:
The particle has zero acceleration along x direction. So, the initial velocity ux remains constant throughout the motion. The distance traveled by the projectile at a time t is given by the equation x = u_xt +\(\frac { 1 }{ 2 }\) at² . Since a = 0 along x direction, we have x = u_xt ……….(1)

Motion along downward direction:
Here u_y = 0 (initial velocity has no downward component), a = g (we choose the + ve y – axis in downward direction), and distance y at time t.
From equation, y = u_xt +\(\frac { 1 }{ 2 }\) at² we get
y = \(\frac { 1 }{ 2 }\) at² …………..(2)

Substituting the value oft from equation (i) in equation (ii) we have

y = Kx²
where K = \(\frac{g}{2 u_{x}^{2}}\) is constant
Equation (iii) is the equation of a parabola. Thus, the path followed by the projectile is a parabola.

1. Time of Flight:
The time taken for the projectile to complete its trajectory or time taken by the projectile to hit the ground is called time of flight. Consider the example of a tower and projectile. Let h be the height of a tower. Let T be the time taken by the projectile to hit the ground, after being thrown horizontally from the tower.

We know that s_y = u_yt + \(\frac { 1 }{ 2 }\) at² for vertical motion. Here .s_y = h, t = T, u_y = 0 (i.e., no initial vertical velocity). Then h = \(\frac { 1 }{ 2 }\) gt² or T = \(\sqrt{\frac{2 h}{g}}\) Thus, the time of flight for projectile motion depends on the height of the tower, but is independent of the horizontal velocity of projection. If one ball falls vertically and another ball is projected horizontally with some velocity, both the balls will reach the bottom at the same time. This is illustrated in the Figure

2. Horizontal range:
The horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground is called horizontal range.
For horizontal motion, we have
s_x = u_xt + \(\frac { 1 }{ 2 }\) at²
Here,s_x = R (range), u_x = u, a = 0 (no horizontal acceleration) T is time of flight. Then horizontal range = uT
Since the time of flight T = \(\sqrt{\frac{2 h}{g}}\) we substitute this and we get the horizontal range of the particle as R = u \(\sqrt{\frac{2 h}{g}}\)
The above equation implies that the range R is directly proportional to the initial velocity u and inversely proportional to acceleration due to gravity g.

3. Resultant Velocity (Velocity of projectile at any time):
At any instant t, the projectile has velocity components along both x-axis and y-axis. The resultant of these two components gives the velocity of the projectile at that instant t, as shown in figure. The velocity component at any t along horizontal (x-axis)
is V_x = U_x + a_xt
Since, u_x = u, ax = 0 , we get
u_x = u a_x = 0 we get
v_x = u
The component of velocity along vertical direction (y – axis) is v_y = u_y + a_yt
Since, u_y= 0, a_y = g, we get
V_y = gt
Hence the velocity of the particle at any instant is –
v = u\(\hat{i}\) + g\(\hat{j}\)
The speed of the particle at any instant t is given by
v = \(\sqrt{v_{x}^{2}+v_{y}^{2}}\)
v= \(\sqrt{u^{2}+g^{2} t^{2}}\)

4. Speed of the projectile when it hits the ground:
When the projectile hits the ground after initially thrown horizontally from the top of tower of height h, the time of flight is –
t = \(\sqrt{\frac{2 h}{g}}\)
The horizontal component velocity of the projectile remains the same i.e v_x = u.
The vertical component velocity of the projectile at time T is
v = gT = g \(\sqrt{\frac{2 h}{g}}\) = \(\sqrt{\frac{2gh}\)
The speed of the particle when it reaches the ground is
v = \(\sqrt{u^{2}+2 g h}\).

Question 5.
Derive the relation between Tangential acceleration and angular acceleration.
Answer:
Consider an object moving along a circle of radius r. In a time ∆t, the object travels in an arc distance As as shown in figure. The corresponding angle subtended is ∆θ
The ∆s can be written in terms of ∆θ
∆s = r∆θ ………(i)
in a time ∆t, we have
\(\frac { ∆s }{ ∆t}\) = t \(\frac { ∆θ }{ ∆t}\) …………(ii)
¡n the limit ∆t – 0, the above equation becomes
\(\frac { ds }{ dt}\) = rω …………….(iii)
Here \(\frac { ds }{ dt}\) is linear speed (y) which is tangential to the circle and co is angular speed.
So equation (iii) becomes.
v _r = rω …….(iv)

which gives the relation between linear speed and angular speed.
Eq. (iv) is true only for circular motion. In general the relation between linear and angular velocity is given by
\(\vec{v}\) = \(\vec{\omega} \times \vec{r}\) ………..(v)
For circular motion eq. (y) reduces to eq. (iv) since \(\overrightarrow{\mathrm{ω}}\) and \(\overrightarrow{\mathrm{r}}\)are perpendicular to each other.
Differentiating the eq. (iv) with respect to time, we get (since r is constant)
\(\frac { dv }{ dt}\) = \(\frac { rdv }{ dt}\) = rα
Here \(\frac { dv }{ dt}\) Is the tangential acceleration and is denoted as a_t = \(\frac {dω}{ dt}\)is the angular acceleration
α. Then eq. (v) becomes
a_t = rα ………..(vii)

Samacheer Kalvi 11th Physics Kinematics Numerical Questions

Question 1.
The V – t graphs of two objects make angle 30° and 60° with the time axis. Find the ratio of their accelerations.
Solution:
\(\frac{a_{1}}{a_{2}}\) = \(\frac{tan 30}{tan 60}\) = \(\frac{1 / \sqrt{3}}{\sqrt{3}}\) = \(\frac{1}{3}\) = 1 : 3.

Question 2.
When the angle between two vectors of equal magnitudes is 2n/?>, prove that the magnitude of the resultant is equal to either.
Solution:
R = \(\left(\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cos \theta\right)^{1 / 2}\) = \(\left(p^{2}+p^{2}+2 p \cdot p \cos \frac{2 \pi}{3}\right)\) = \(\left[2 p^{2}+2 p^{2}\left(\frac{-1}{2}\right)\right]^{1 / 2}\) = p.

Question 3.
If \(\overline{\mathrm{A}}\) = 3\(\hat{i}\) + 4\(\hat{j}\) and \(\overline{\mathrm{B}}\) = 7\(\hat{i}\) + 24 \(\hat{j}\), find a vector having the same magnitude as \(\overline{\mathrm{B}}\) and parallel to \(\overline{\mathrm{A}}\).
Solution:
\(|\overrightarrow{\mathrm{A}}|=\sqrt{3^{2}+4^{2}}=5\)
also \(|\overrightarrow{\mathrm{B}}|=\sqrt{7^{2}+24^{2}}=25\)
desired vector = \(|\overrightarrow{\mathrm{B}}|\) \(\widehat{A}\) = 25 x \(\frac{3 \hat{i}+4 \hat{j}}{5}\) = 5(3\(\hat{j}\) + 4\(\hat{j}\)) = 15 \(\hat{i}\) + 20\(\hat{j}\).

Question 4.
What is the vector sum of n coplanar forces, each of magnitude F, if each force makes an angle of \(\frac {2 π}{ n }\) with the preceding force?
Solution:
Resultant force is zero.

Question 5.
A car is moving along X- axis. As shown in figure it moves from O to P in 18 seconds and return from P to Q in 6 second. What are the average velocity and average speed of the car in going from

• O to P
• From O to P and hack to Q

Solution:

• O to P, Average velocity =20 ms^-1
• O to P and back to Q

Average velocity = 10 ms^-1
Average speed = 20 ms^-1

Question 6.
On a 60 km straight road, a bus travels the first 30 km with a uniform speed of 30 kmh^-1 . How fast must the bus travel the next 30 km so as to have average speed of 40 kmh^-1 for the entire trip?
Solution:

Question 7.
The displacement x of a particle varies with time as x = 4t² – 15t + 25. Find the position, velocity and acceleration of the particle at t = O.
Solution:
Position, x = 25 m
Velocity = \(\frac { dx}{dt}\)8t – 15
t = 0, v = 0 – 15 = -15 m/s
acceleration, a = \(\frac { dx}{dt}\) = 8 ms^-2

Question 8.
A driver takes 0.20 second to apply the breaks (reaction time). If he is driving car at a speed of 54 kmh^-1 and the breaks cause a deceleration of 6.0 ms^-2. Find the distance travelled by car after he sees the need to put the breaks.
Solution:
(distance covered during 0.20 s) + (distance covered until rest)
= (15 x 0.25) + [18.75] = 21.75 m.

Question 9.
From the top of a tower 100 m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velocity of 25 m/s. Find when and where the two balls will meet? (g = 9.8 m/s)
Solution:
For the ball dropped from the top
x = 4.9 t²…….(i)
For the ball thrown upwards
100 – x =25t – 4.9 t² …….(ii)
From eq. (i) and (ii)
t = 4s  x = 78.4 m.

Question 10.
A ball thrown vertically upwards with a speed of 19.6 ms^-1 from the top of a tower returns to the earth in 6s. Find the height of the tower, (g = 9.8 m/s)
Solution:
s = ut + \(\frac { 1 }{ 2 }\) at²
-h = 19. 6 x 6 + \(\frac { 1 }{ 2 }\) x (-9.8) x (6)²
h = 58.8 m.

Question 11.
Two town A and B are connected by a regular bus service with a bus leaving in either direction every T min. A man cycling with a speed of 20 kmh^-1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed do the buses ply of the road?
Solution:
V =40 krnh^-1 and T = 9 min.

Question 12.
A motorboat is racing towards north at 25 kmh^-1 and the water current in that region is 10 kmh^-1 in the direction of 60° east of south. Find the resultant velocity of the boat.
Solution:
V= 21.8 kmh^-1
angle with north, θ = 23.4°.

Question 13.
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft position 10 second apart is 30°, what is the speed of the aircraft?
Solution:
Speed = 182.2 ms^-1

Question 14.
A boat is moving with a velocity (3\(\hat{i}\) -4\(\hat{j}\)) with respect to ground. The water in river is flowing with a velocity (-3\(\hat{i}\) – 4\(\hat{j}\)) with respect to ground. What is the relative velocity of boat with respect to river?
Solution:
\(\overrightarrow{\mathrm{V}}_{\mathrm{BW}}\)= \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\).

Question 15.
A hiker stands on the edge of a clift 490 m above the ground and throws a stone horizontally with an initial speed of 15 ms^-1. Neglecting air resistance, find the time taken by the stone to reach the ground and the speed with which it hits the ground. (g 9.8 ms^-2)
Solution:
time = 10 seconds
V = \(\sqrt{\mathrm{V}_{s}^{2}+\mathrm{V}_{Y}^{2}}\) = \(\sqrt{15^{2}+98^{2}}\) = 99.1 m/s^-1

Question 16.
A bullet fired at an angle of 30° with the horizontal hits the ground 3 km away. By adjusting the angle of projection, can one hope to hit the target 5 km away ? Assume that the muzzle speed to be fixed and neglect air resistance.
Solution:
Maximum Range = 3.46 km
So it is not possible.

Question 17.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 seconds, what is the magnitude and direction of acceleration of the stone?
Solution:
\(\frac { 88 }{ 25 }\) rad s^-1, \(\frac { 2π}{ T}\) = \(\frac { 2πN}{t}\)
a = 991.2 cm s^-2

Question 18.
A cyclist is riding with a speed of 27 kmh^-1 . As he approaches a circular turn on the road of radius 30 m^-2, he applies brakes and reduces his
speed at the constant rate 0.5 ms^-2. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Solution:
a^c = \(\frac{v^{2}}{r}\) = 0.7ms^-2
a^t = 0.5 ms^-2
a = \(\sqrt{a^{2}-a^{2}}\) = 0.86 ms^-2
If O is the angle between the net acceleration and the velocity of the cyclist, then
0= tan^-1 \(\left(\frac{a_{c}}{a_{\mathrm{T}}}\right)\) = tan^-1 = 54°28′

Question 19.
If the magnitude of two vectors are 3 and 4 and their scalar product is 6, find angle between them and also find \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\)|
Solution:
\(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\) = AB cos θ
|\(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\)| = AB sin θ
or 6 = (3 x 4) cos θ = 3 x 4 x 60°
or θ = 60°
= 3 x 4 x \(\frac{\sqrt{3}}{2}\) = \(6 \sqrt{3}\)

Question 20.
Find the value ofA so that the vector \(\overrightarrow{\mathrm{A}}\) = 2 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\) and \(\overrightarrow{\mathrm{B}}\) = 4\(\hat{i}\) -2 \(\hat{j}\) +2\(\hat{k}\) are perpendicular to each other.
Solution:
\(\overrightarrow{\mathrm{A}} \perp \overrightarrow{\mathrm{B}}\)
⇒ \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}\)
⇒ t = 3

Question 21.
The velocity time graph of a particle is given by –

•  Calculate distance and displacement of particle from given v – t graph.
• Specify the time for which particle undergone acceleration, retardation and moves with constant velocity.
• Calculate acceleration, retardation from given v – t graph.
• Draw acceleration-time graph of given v – t graph.

Solution:
(i) distance = area of ∆OAB + area of trapezium BCDE = 12 + 28 = 40 m
(ii) displacement area of ∆OAB – area of trapezium BCDE = 12 – 28 = – 16 m

• times acc. \((0 \leq t \leq 4) \) and \((12\leq t \leq 16) \)
• retardation \((4\leq t \leq 8) \)
• constant velocity \((8\leq t \leq 12) \)
  

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Tamilnadu Samacheer Kalvi 10th Social Science Geography Solutions Chapter 1 India: Location, Relief and Drainage

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India: Location, Relief and Drainage TEXTUAL EXERCISE

I. Choose the correct answer.

10th Social Geography 1st Lesson Question 1.
The north-south extent of India is
(a) 2,500 km
(b) 2,933 km
(c) 3,214 km
(d) 2,814 km
Answer:
(c) 3,214 km

India Location Relief And Drainage Question 2.
The Southern most point of India is:
(a) Andaman
(b) Kanyakumari
(c) Indira Point
(d) Kavaratti
Answer:
(c) Indira Point

India – Location, Relief And Drainage Questions And Answers Question 3.
The extent of Himalayas in the east-west is about
(a) 2,500 km
(b) 2,400 km
(c) 800 km
(d) 2,200 km
Answer:
(a) 2,500 km

India Location Relief And Drainage Pdf Question 4.
……………. River is known as ‘Sorrow of Bihar’.
(a) Narmada
(b) Godavari
(c) Kosi
(d) Damodar
Answer:
(c) Kosi

India Location Relief And Drainage Questions And Answers Question 5.
Deccan Plateau covers an area of about ……. sq.km.
(a) 8 lakh
(b) 6 lakh
(c) 5 lakh
(d) 7 lakh
Answer:
(d) 7 lakh

India Location, Relief And Drainage Notes Question 6.
A landmass bounded by sea on three sides is referred to as:
(a) Coast
(b) Island
(c) Peninsula
(d) Strait
Answer:
(c) Peninsula

South Indian Rivers Are East Flowing Give Reason Question 7.
The Palk Strait and Gulf of Mannar separates India from ………
(a) Goa
(b) West Bengal
(c) Sri Lanka
(d) Maldives
Answer:
(c) Sri Lanka

India Location, Relief And Drainage Pdf Question 8.
The highest peak in South India is:
(a) Ooty
(b) Kodaikanal
(c) Anaimudi
(d) Jindhagada
Answer:
(c) Anaimudi

10th Geography Samacheer Kalvi Question 9.
………… Plains are formed by the older alluviums.
(a) Bhabar
(b) Tarai
(c) Bhangar
(d) Khadar
Answer:
(c) Bhangar

Samacheer Kalvi 10th Geography Book Question 10.
Pulicat Lake is located between the states of:
(a) West Bengal and Odisha
(b) Karnataka and Kerala
(c) Odisha and Andhra Pradesh
(d) Tamil Nadu and Andhra Pradesh
Answer:
(d) Tamil Nadu and Andhra Pradesh

II. Match the following.

Answers:
1. (c)
2. (a)
3. (e)
4. (b)
5. (d)

III. Give Reasons.

10th Social Geography Unit 1 Question 1.
The Himalayas are called young fold mountains.
Answer:
They were formed by earth movements which affected the relief of the earth in the last phase of its geological history. Because of the young age which is evident from the striking contrast in relief, Himalayan ranges are called young fold mountains.

10th Samacheer Kalvi Social Question 2.
North Indian Rivers are perennial.
Answer:
North Indian Rivers have their origin from the snow-covered Himalayas. As these rivers have water throughout the year they are referred to as perennial rivers.

Question 3.
Chottanagpur Plateau is rich in mineral resources.
Answer:
Chottanagpur Plateau is a store house of mineral resources such as mica, bauxite, copper, limestone, iron ore and coal.

Question 4.
The great Indian desert is called Marusthali.
Answer:
In Sanskrit Marusthali means ‘dead land’. It is the region of the actual desert with severe arid climate and very low vegetation. This region has different types of sand dunes.

Question 5.
The Eastern states are called seven sisters.
Answer:
Mizoram and Manipur are connected to the rest of India through Basak Village in Assam. And
it is due to this interdependence, they were given the sobriquet. It is a well known fact that the
states of Arunachal Pradesh, Assam, Meghalaya, Manipur, Mizoram, Nagaland and Tripura were named the seven sisters in 1972.

Question 6.
The river Godavari is often referred as Vridha Ganga.
Answer:
“Vridha” means old. It originates and flow through the Peninsular plateau, the oldest among the physiographic divisions of India. Like River Ganga,Godavari is the longest with an area of 3.13 lakh km2, among the peninsular rivers. Hence the river is often referred as “Vridha Ganga”.

IV. Distinguish between the following

Question 1.
Himalayan rivers and Peninsular rivers.
Answer:

Question 2.
Western Ghats and Eastern Ghats.
Answer:

Question 3.
Himadri and Himachal.
Answer:

Question 4.
Western Coastal Plains and Eastern Coastal Plains.
Answer:

V. Answer in brief.

Question 1.
Name the neighbouring countries of India.
Answer:
India shares its land boundaries with Pakistan in the west, Afghanistan in the north-west, China, Nepal and Bangladesh in the east.
Our Southern neighbours across the sea consists of the two island countries, namely Sri Lanka and Maldives.

Question 2.
Give the importance of lST.
Answer:
The longitudinal extent of India between the West and the East is about 30° from Gujarat to Arunachal Pradesh. This longitudinal difference make a difference of nearly 2 hours in local time between Gujarat in the West and Arunachal Pradesh in the East. In order to avoid the time difference between the places 1ST is calculated. The Indian Standard Time is calculated based on 82°30′ East longitude.

Question 3.
Write a short note on Deccan Plateau.
Answer:

• The Deccan Plateau is a triangular landmass that lies to the south of the river Narmada.
• This is the largest physiographic division of our country.
• It covers an area of about 16 lakh sq.km (about half of the total area of the country)
• It is an old rocky plateau region.
• The topography consists of a series of plateaus and hill ranges interspersed with river valleys.
• It is higher in the west and slopes gently eastwards.
• The Western Ghats and the Eastern Ghats mark the western and the eastern edges of the Deccan plateau respectively.
• Aravalli hills mark the north-western boundary of the plateau region.
• Its northern and north-eastern boundaries are marked by the Bundelkhand upland, Kaimur and Rajmahal hills.
• Western Ghats lie parallel to the Western coast. They are continuous and can be crossed through passes only like Thai, Bhor and the Pal Ghats. It is higher then the Eastern Ghats. It cause orographic rain by facing the rain-bearing moist winds to rise along the western slopes of the Ghats. The height of the Western Ghats progressively increases from north to south. The highest peak include the Anaimudi (2,695 metres) and the DodaBetta (2,637 metres).
• The Eastern Ghats stretch from the Mahanadi valley to the Nilgiris in the south. The Eastern Ghats are discontinuous and irregular and dissected by rivers draining into the Bay of Bengal. Mahendragiri (1,501 metres) is the highest peak in the Eastern Ghats. Shevroy Hills and the Javadi Hills are located to the southeast of the Eastern Ghats. The famous hill stations of Udagamandalam, popularly Known as Ooty and the Kodaikanal are located here.
• One of the distinct features of the peninsular plateau is the black soil area known as Deccan Trap. This is of volcanic region hence the rocks are igneous. These rocks have denuded over time and are responsible for the formation of black soil. ‘

Question 4.
State the west-flowing rivers of India.
Answer:
Narmadha, Tapti and Mahi are the west-flowing rivers of India. They flow into the Arabian Sea through Gulf of Cambay.

Question 5.
Write a brief note on the island group of Lakshadweep.
Answer:

• It lies close to the Malabar coast of Kerala.
• This group of islands is composed of small coral islands.
• Earlier they were known as Laccadive, Minicoy and Amindive. In 1973, these were named as Lakshadweep.
• It covers small area of 32 sq.km. Kavaratti island is the administrative headquarters of Lakshadweep.
• This island group has great diversity of flora and fauna. The Pitti island which is uninhabited, has a bird sanctuary.

VI. Answer in a paragraph.

Question 1.
Explain the divisions of the Northern Mountains and its importance to India.
Answer:

1. Northern mountains are the youngest and the loftiest mountain chains in the world.
2. It stretches for a distance of2500km from the Indus gorge in the West to Brahmaputra gorge in the East.
3. The major divisions of the Northern mountains are:
   • The Trans Himalayas
   • Himalayas
   • Eastern or Purvanchal hills

(i) The Trans Himalayas:

• It lies in Jammu and Kashmir and Tibetian plateau.
• It is also known as Western Himalayas. As its areal extent is more in Tibet, it is also known as Tibetean Himalayas.
• The rocks of this region are of Thethys sediments and contain fossils bearing marine sediments.
• The prominent ranges of this division are Zaskar, Ladakh, Kailash and Karakoram.

(ii) The Himalayas:

• It is formed by the uplifted compression of the Thethys sea due to tectonic forces.
• It has three parallel ranges.

(a) The Greater Himalayas (Himadri)
(b) The Lesser Himalayas (Himachal)
(c) The Outer Himalayas (Siwaliks)

(a) The Greater Himalayas or Himadri:

• The most continuous range.
• Almost all the lofty peaks of the Himalayas are located in this range.
  Eg: Mt. Everest (8,848m) and Kanchenjunga (8,586m).
• It is the region of permanent snow cover.
• It has many glaciers like Gangothri, Yamunothri and Siachen.

(b) The Lesser Himalayas or Himachal:

• It is the middle range of Himalayas.
• Made up of rocks like slate, limestone and quartzite.
• Important ranges pir panjal, Dhauladhar and Mahabharat.
• Familiar for hill stations – Shimla, Mussourie Nainital, Almora, Ranikhet and Daijeeling.

(c) The Outer Himalayas or Siwaliks:

• It is the most discontinuous range dissected by the Himalayan rivers.
• The longitudinal valleys found between the Siwaliks and the Himachal are called Duns in the West and Duars in the East. Eg: Dehradun.
• This range is ideal for the development of settlements.

(iii) The Eastern Himalayas or Purvanchal Hills:

• They are the Eastern off shoots of Himalayas.
• Most of these hills are located along the border of India and Myanmar.
• Some of the important hills are Patkai Bum, Naga hills, Manipur hills, Mizo hills, Garo hills, Khasi hills and Jaintia hills.
• Collectively known as Purvanchal hills.

Importance:

• Forms as the natural barrier to the Sub continent.
• Source for many perennial rivers such as Indus, Ganges and Brahmaputra.
• Paradise of tourists due to its natural beauty.
• Renowned for rich bio-diversity.
• Many hill stations and Pilgrim centres like Amamath, Kedamath, Badrinath and Vaishnavidevi temple are located.
• Natural climatic barrier prevents the cold winds from Central Asia. Blocks the South west monsoon winds and causes heavy rainfall to North India.

Question 2.
Give an account on the major peninsular rivers of India.
Answer:

• The rivers in South India are called the Peninsular rivers.
• Most of the rivers originate from the Western Ghats.
• These are seasonal rivers, (non-perennial)
• They have a large seasonal fluctuation in volume of water as they are solely fed by rain.
• These rivers flow in valleys with steep gradients.
• Based on the direction of flow, the Peninsular rivers are divided into the west flowing and east flowing rivers.

East Flowing Rivers:

1. Mahanadi:

1. It originates near Sihawa in Raipur district of Chattisgarh and flows through Odisha.
2. Its length is 851 km.
3. Seonath, Telen, Sandur and lb are its major tributaries.
4. The main stream of Mahanadi gets divided into several distributaries such as Paika, Birupa, Chitartala, Genguti and Nun.
5. The Mahanadi empties its water in Bay of Bengal.

2. Godavari:

1. Godavari is the longest river (1,465 km) with an area of 3.13 lakh km2 among the Peninsular rivers.
2. It is also called Vridha Ganga.
3. It originates in Nasik district of Maharashtra, a portion of Western Ghats.
4. It flows through the states of Telangana and Andhra Pradesh before joining Bay of Bengal.
5. Puma, Penganga, Pranitha, Indravati, Tal and Salami are its major tributaries.
6. The river near Rajahmundry gets divided into two Channels called Vasistha and Gautami and forms one of the largest deltas in India.
7. Kolleru, a fresh water lake is located in the deltaic region of the Godavari.

3. Krishna:

1. The river Krishna originates from a spring at a place called Mahabaleshwar in the Western Ghats of Maharashtra.
2. Its length is 1,400 km and an area of 2.58 lakh sq km.
3. It is the second longest Peninsular river Bhima, Peddavagu, Musi, Koyna and Thungab hadra are the major tributaries of this river.
4. It also flows through Andhra Pradesh and joins in Bay of Bengal, at Hamasaladevi.

4. Kaveri:

1. The river Kaveri originates at Talakaveri, Kudagu hills of Karnataka.
2. Its length is 800 km.
3. The river Kaveri is called Dhakshin Ganga or Ganga of south.
4. Harangi, Hemavati, Kabini, Bhavani, Arkavathy, Noyyal, Amaravathi etc are the main tributaries of the river Kaveri.
5. In Karnataka the river bifurcates twice, forming the sacred islands of Srirangapatnam and Sivasamudram.
6. While entering Tamil Nadu, the Kaveri continues through a series of twisted wild gorges until it reaches Hogenakkal Falls and flows through a straight, narrow gorge near Salem.
7. The Kaveri breaks at Srirangam Island with two channels, river Coleroon and Kaveri.
8. At last, it empties into the Bay of Bengal at Poompuhar.

West Flowing Rivers:

1. Narmada:

1. It rises in Amarkantak Plateau in Madhya Pradesh at an elevation of about 1057 m and flows for a distance of about 1,312 km.
2. It covers an area of 98,796 sq km and forms 27 km long estuary before outfalling into the Arabian Sea through the Gulf of Cambay.
3. It is the largest among the west flowing rivers of Peninsular India.
4. Its principal tributaries are Burhner, Halon, Heran, Banjar, Dudhi, Shakkar, Tawa, Bama and Kolar.

2. Tapti:

1. The Tapti is one of the major rivers of Peninsular India with the length of about 724 km.
2. It covers an area of 65,145 sq km.
3. Tapti river rises near Multai in the Betul district of Madhya Pradesh at an elevation of about 752 m.
4. It is one of only the three rivers in Peninsular India that run from east to west – the others being the Narmada and the Mahi.
5. The major tributaries are Vaki, Gomai, Arunavati, Aner, Nesu, Buray, Panjhra and Bori.
6. It out falls into the Arabian Sea through the Gulf of Cambay.

3. Give a detailed account on the basin of the Ganga.

1. The Ganga River system is the largest drainage system of India.
2. The river Ganga is 2,480 km long.
3. It rises in the Gangotri glacier in the Himalayas at a height of 6000 metres.
4. It cuts deep gorges through the Siwalik range and enters into the plain at Haridwar.
5. The Ganga plain occupies an area of about 3,37,000 covering the states of Uttar Pradesh, Bihar and West Bengal.
6. The river Yamuna rises in Yamunotri glacier. After flowing for a distance of about 1300 km, it joins Ganga on its right bank at Allahabad.
7. The rivers Chambal, Betwa, Son and Ken rise in the Deccan Plateau and join Ganga on its right bank.
8. The Ghandak, the Gomati, the Ghaghara and the Kosi join the Ganga on its left bank.
9. It is covered by thick alluvial sediments.
10. The Ganga plains slopes gently from Haryana and drains into Bay of Bengal.
11. It is covered by thick alluvial sediments.
12. The largest distributary of Ganga is Hooghly.
13. Most of the Ganga Delta lies in Bangladesh.
14. The Seaward of the Ganga Delta has tidal estuaries, sand banks and islands known as Sunderbans.

VII. Map exercises: Mark the following in the outline map of India

Question 1.
Major mountain ranges – Karakoram, Ladakh, Zaskar, Aravalli, Western Ghats, Eastern Ghats.
Answer:

Question 2.
Major rivers — Indus, Ganga, Brahmaputra, Narmada, Tapti, Mahanadi, Godavari, Krishna & Kaverl.
Answer:

Question 3.
Major plateaus – Malwa, Chotanagpur, Deccan.
Answer:

VIII. Activities

Question 1.
Observe the Peninsular Plateau map of India and mark the major plateau divisions of India.
Answer:

Question 2.
Prepare a table showing the major West flowing and East flowing rivers of peninsular India.
Answer:

Question 3.
Assume that you are travelling from West Bengal to Gujarat along the beautiful coasts of India. Find out the states which you would pass through.
I will pass through the states of Odisha, Andhra Pradesh, Tamil Nadu, Kerala, Karnataka and Maharashtra.
Answer:

Question 4.
Find out the states through which the river Ganga flows.
The river Ganga flows through the states of Uttarakhand, Uttar Pradesh, Bihar, Jharkhand, and West Bengal.
Answer:

Question 5.
Prepare a table showing the major rivers in India and find out it’s tributaries, origin, length and area.
Answer:

Find Out:

Question 1.
The number of Union Territories along the Western Coast and Eastern Coast?
Answer:

1. Along Western coast – Four – Diu, Daman, Dadra and Nagar Haveli, Lakshadweep islands.
2. Along the Eastern coast -Two – Puducherry and Andaman and the Nicobar Islands.

Question 2.
Area wise which is the smallest and largest state?
Answer:
Largest state: Rajasthan
Smallest State: Goa

Question 3.
The states which do not have an international border or lie on the coast.
Answer:
The States that do not share an International boundary or lie on the coast are Haryana, Madhya Pradesh, Jharkhand and Chattisgarh and Telangana.
Union Territories – Chandigarh and Delhi.

Question 4.
Classify into four groups each having common frontiers with
i) Pakistan
ii) China
iii) Myanmar and
iv) Bangladesh
Answer:
Pakistan: Jammu and Kashmir, Punjab, Rajasthan and Gujarat.
China: Jammu and Kashmir, Himachal Pradesh, Sikkim and Arunachal Pradesh.
Myanmar: Arunachal Pradesh, Nagaland, Manipur, Mizoram.
Bangladesh: Bihar, West Bengal, Jharkhand, Assam, Meghalaya and Tripura.

Question 5.
Find the Hill stations which are located in Himalayan Mountains.
Answer:
Shimla, Mussourie, Nainital, Almora. Ranikhet, Darjeeling and Kulu Manali are some of the hill stations in Himalayan mountains.

Question 6.
In which river the Gerosappa (jog) fall is found?
Answer:
Sharavathi River

India: Location, Relief and Drainage Additional Questions

I. Choose the correct answer.

Question 1.
India covers an area of ………. million sq.kms.
(a) 3.2
(b) 3.5
(c) 3.8
Answer:
(a) 3.2

Question 2.
India’s longest border is with:
(a) Bangladesh
(b) Srilanka
(c) Bhutan
(d) Afghanistan
Answer:
(a) Bangladesh

Question 3.
In India, there is a vast plain in ………
(a) north
(b) south
(c) east
Answer:
(a) north

Question 4.
Till 2024 the ………………… will be the capital for both the States of Andhra pradesh and Telangana.
(a) Hyderabad
(b) Secunderabad
(c) Nellore
(d) Amaravati
Answer:
(a) Hyderabad

Question 5.
India is situated into southern part of ……….
(a) Indo-Myanmar
(b) Asia
(c) Sri Lanka
Answer:
(b) Asia

Question 6.
The ………………… ranges of the Himalayas is the most continuous of all the ranges.
(a) Himachal
(b) Aravalli
(c) Siwaliks
(d) Himadri
Answer:
(d) Himadri

Question 7.
India occupies ………. % of the world’s land area.
(a) 3.5
(b) 2.4
(c) 7.5
Answer:
(b) 2.4

Question 8.
The deltaic region of lower Ganga, the uplands are known as:
(a) Chars
(b) Bhangar
(c) Terai
(d) Bils
Answer:
(a) Chars

Question 9.
India is the ……….. largest country regarding its area.
(a) seventh
(b) third
(c) fourth
Answer:
(a) seventh

Question 10.
The ………………… river system is the largest drainage system of India.
(a) Yamuna
(b) Godavari
(c) Ganga
(d) Kaveri
Answer:
(c) Ganga

Question 11.
India is ………….. times smaller than USA.
(a) ten
(b) seven
(c) three
Answer:
(c) three

Question 12.
India has a predominant position in the ……. realm.
(a) Bay of Bengal
(b) Indian Ocean
(c) Arabian Sea
Answer:
(b) Indian Ocean

Question 13.
No other country has such a large ……….. line.
(a) mountain
(b) plateau
(c) coastal
Answer:
(c) coastal

Question 14.
India is connected with China, Japan and Australia through ………..
(a) Malaccan Strait
(b) Suez canal
(c) Palk Strait
Answer:
(a) Malaccan Strait

Question 15.
The 0° Meridian passes through Greenwich in ……………
(a) New york
(b) England
(c) Los Angeles
Answer:
(b) England

Question 16.
Dehra Dun is the capital of ………..
(a) Uttar Pradesh
(b) Sikkim
(c) Uttranchal
Answer:
(c) Uttranchal

Question 17.
India is a ……….. country with total freedom of worship.
(a) republic
(b) democratic
(c) secular
Answer:
(c) secular

Question 18.
According to gaseous mass theory, the earth was separated from the ………….
(a) Universe
(b) Sun
(c) Milky way
Answer:
(b) Sun

Question 19.
The ……… force on Tethys sea gave rise to “Himalayas”.
(a) depressional
(b) compressed
(c) longitudinal
Answer:
(b) compressed

Question 20.
The Himalayas started growing up due to ………. of the agents of denudation.
(a) deposition
(b) transportation
(c) erosion
Answer:
(c) erosion

Question 21.
The Gangetic plain was formed due to ………..
(a) erosion
(b) transportation
(c) deposition
Answer:
(c) deposition

Question 22.
The major physiographical units of India are ………
(a) six
(b) five
(c) four
Answer:
(a) six

Question 23.
The Himalayas are made of ………… rocks.
(a) sedimentary
(b) igneous
(c) volcanic
Answer:
(a) sedimentary

Question 24.
Pamir knot is in the ………… part of India.
(a) north-east
(b) north-west
(c) south-east
Answer:
(b) north-west

Question 25.
Ladakh is in the north-west of ……….
(a) Himachal Pradesh
(b) Kashmir
(c) Kerala
Answer:
(b) Kashmir

Question 26.
The northernmost range of Himalayas is ………
(a) Himachal
(b) Siwaliks
(c) Himadri
Answer:
(c) Himadri

Question 27.
Himachal is highly ………. topography.
(a) smooth
(b) rugged
(c) narrow
Answer:
(b) rugged

Question 28.
Pirpanjal lies in ……… state.
(a) Kashmir
(b) Himachal Pradesh
(c) Arunachal Pradesh
Answer:
(a) Kashmir

Question 29.
Amamath, Kedranath and Badrinath are the …….. of Himachal range.
(a) valleys
(b) hill resorts
(c) pilgrimages
Answer:
(c) pilgrimages

Question 30.
Siwaliks is the ………. range, made up of mud and soft rocks.
(a) discontinuous
(b) broad
(c) continuous
Answer:
(a) discontinuous

Question 31.
An example of longitudinal valley of Siwaliks is ………..
(a) Nainital
(b) Mussouri
(c) Dehra Dun
Answer:
(c) Dehra Dun

Question 32.
……… supports the growth of thick forest in siwaliks.
(a) Pebbles and gravels
(b) Muds and soft rock
(c) fine silt
Answer:
(c) fine silt

Question 33.
The Purvachal mountains are of ……… height.
(a) medium
(b) short
(c) very tall
Answer:
(a) medium

Question 34.
The Himalayas act as the ……. barrier protecting from the foreign invasions.
(a) physical
(b) natural
(c) climate
Answer:
(a) physical

Question 35.
…… soil helps the cultivation of crops in great plain.
(a) block
(b) alluvium
(c) fertile
Answer:
(b) alluvium

Question 36.
There are dense forests on the ……….. of the Himalayan.
(a) peak
(b) valleys
(c) slopes
Answer:
(c) slopes

Question 37.
The ……… attract the tourists by its scenic beauty and pleasant climate.
(a) hill stations
(b) pilgrimages
(c) valleys
Answer:
(a) hill stations

Question 38.
The great Plain was formed due to …….. force.
(a) compression
(b) depression
(c) longitudinal
Answer:
(b) depression

Question 39.
The depression was caused in the plains out of ………..
(a) erosion
(b) transportation
(c) deposition
Answer:
(c) deposition

Question 40.
The rivers passed their ways by eroding their ……. when Himalayas lifted themselves high.
(a) passes
(b) mountains
(c) valleys
Answers:
(c) valley

II. Fill in the blanks.

1. The land in India gets abundance sunshine from the ………. sun.
2. India is the ……….. largest country with respect to area.
3. The shallow sea divided the Angara and Gondwana land was ………
4. The south eastern part of the Deccan plateau is known as ……….
5. The description of the physical relief features of a country is known as ………
6. The Pamir Knot, popularly known as the ……….
7. Mount Everest is the highest peak of the world with an altitude of ………..
8. The largest tributary of the Ganga is ……..
9. The largest plateau of the peninsular region is ………
10. The largest and longest of the Peninsular river is …………
11. The Cauvery rises in hills of …….. district in Karnataka.
12. The ………. river rises in Agasthiar hills.
13. India is connected with Europe through ………….
14. Himalayas and Karakoram provide a ……… boundary in the north.
15. Godwin Austin known as ………. is the highest peak of India.
16. The water body around the compact land mass was called …………
17. Karakoram stretches out eastern from ………..
18. …………. is the longest glacier.
19. The gaps providing natural routes across the mountains are called …………
20. The beautiful Kashmir valley is found in ……… ranges.
21. The narrow longitudinal valleys of siwaliks are called ………..
22. A flat low lying land of the Great plain is made up of ……….
23. The ……….. is a narrow porous zone along the foothills of Siwaliks.
24. The ………. is a zone of dampers and marshes covered with forests.
25. The Indus river flows into Pakistan through ………..
26. Ganga river reaches plain at …………..
27. Dams across the rivers help in generating ……….. power.
28. The Sub division of the peninsular plateau are the …….. and the ………..
29. At the apex of the Deccan plateau is ……….
30. The other name of the Deccan plateau is ………. plateau.
31. The Passes in the Western Ghats are ………., ………. and ………
32. The rivers of the Peninsular India originate on the ……… ghats.
33. ………, ……….. are the distributaries of the cauvery.
34. ……… and ……….. of the special markages of the fast coast plain.
35. River Ganga is known as the river ………. in Bangladesh.
Answers
1. tropical
2. seventh
3. tethys
4. Telengana
5. physiography
6. Roof of the World
7. 8,848 metres
8. Hooghly
9. Deccan plateau
10. Godavari
11.coorg
12. Thamiraparani
13. suez canal
14. natural
15. Mount K2
16. Panthalasa
17. Pamir Knot
18. Siachin
19. Passes
20. dhauladhar
21. duns
22. alluvium
23. Bhabar
24. Terail
25. Kashmir
26. Haridwar
27. hydroelectric
28. Central highland, Deccan plateau
29. Kanyakumari
30. lava
31. Thai ghat, Bohr ghat, Pal ghat
32. western
33. vennar, vettar
34. Lagoons, delta
35. Padma

III. Match the following.

Answers
1. (b)
2. (c)
3. (d)
4. (e)
5. (a)

Answers:
1. (d)
2. (c)
3. (e)
4. (b)
5. (a)

Answers:
1. (c)
2. (e)
3. (b)
4. (d)
5. (a)

Answers:
1. (e)
2. (c)
3. (b)
4. (a)
5. (d)

Answers:
1. (d)
2. (c)
3. (e)
4. (a)
5. (b)

Answers:
1. (c)
2. (b)
3. (d)
4. (e)
5. (a)

Answers:
1. (d)
2. (e)
3. (b)
4. (a)
5. (c)

Answers:
1. (d)
2. (a)
3. (e)
4. (b)
5. (c)

IV. Distinguish Between:

Question 1.
Konkan Coast and Malabar Coast.
Answer:

Question 2.
Andaman and Nicobar Islands and Lakshadweep Islands.
Answer:

Question 3.
The Bundelkhand and The Baghelkhand.
Answer:

Question 4.
GMT and IST.
Answer:

Question 5.
The Bhabar Plain and The Khadar Plain.

V. Answer in One Word.

Question 1.
The tropic of Cancer does not pass through which state?
Answer:
Odisha

Question 2.
What is the easternmost longitude of India?
Answer:
97° 25′ E

Question 3.
Which latitude divides India into two equal halves?
Answer:
Tropic of Cancer (23° 30′ N)

Question 4.
Name the island group which lies towards east of India.
Answer:
Andaman and Nicobar

Question 5.
What is the percentage of land area that India occupies with respect to the world?
Answer:
2.4%

Question 6.
What is the distance between east and weat extremity of India in Kilometres?
Answer:
2933 km

Question 7.
What is the time lag from Gujarat to Arunachal Pradesh?
Answer:
2 hrs

Question 8.
The Standard Meridian of India passes through Mirzapur, it is located on which state of India.
Answer:
Uttar Pradesh

Question 9.
How much distance has been reduced between India and Europe by the construction of the Suez Canal?
Answer:
7000 km distance

Question 10.
Which is the smallest state of India area wise?
Answer:
Goa

Question 11.
What is the southernmost part of India?
Answer:
Indira Point

Question 12.
What is southernmost part of Indian mainland?
Answer:
Kanyakumari

Question 13.
Which states of India shares their border with Myanmar?
Answer:
Tripura, Mizoram, Manipur, and Nagaland.

Question 14.
Name the Eastern Coastal States of India.
Answer:
Odisha, West Bengal, Tamil Nadu, Andhra Pradesh.

Question 15.
Name the landmass bounded by sea on three sides.
Answer:
Peninsular

Question 16.
Which is the most of the volcanoes and earthquakes in the world located?
Answer:
The Peninsular Plateau

Question 17.
Name the mountain ranges in the eastern part of India forming its boundary with Myanmar.
Answer:
Purvanchal

Question 18.
What is the name given to longitudinal valley that lies between lesser Himalayas and Shivaliks?
Answer:
Duns

Question 19.
Name the part of the Himalayas lying between the Kali and Tista rivers.
Answer:
Nepal Himalayas

Question 20.
Which river has the largest inhabited riverine islands in the world?
Answer:
Brahmaputra

Question 21.
What is the soil of Bangar region that contains calcareous deposits locally known as?
Answer:
Kankar

Question 22.
What is the swampy and marshy region of Northern plains called as?
Answer:
Terai

Question 23.
What is the highest peak in the Eastern Ghats?
Answer:
Mahendragiri (1502 mt)

Question 24.
Name the second highest peak of Western Ghats.
Answer:
Doda Betta

Question 25.
Which type of soil is found in Deccan Trap region?
Answer:
Black soil

Question 26.
Name the major west flowing rivers in the peninsular plateau.
Answer:
Narmada and Tapti

Question 27.
Name the largest river of the Indian Ocean.
Answer:
Luni

Question 28.
What is the name given to the southern part of eastern coastal plains?
Answer:
Coromandel

Question 29.
Which is the largest salt water lake of India?
Answer:
Chilika

Question 30.
Which physiographic division is the storehouse of minerals?
Answer:
Peninsular Plateau

VI. Answer in brief.

Question 1.
India is a Sub-continent.
Answer:
India possesses distinct continental characteristics in physiography, climate, natural vegetation, minerals, human resources etc., Hence India is known as “Subcontinent”.

Question 2.
Mention the tributaries and distributaries of Cauvery.
Answer:
Tributaries: Amaravathi, Noyyal and Bhavani.
Distributaries: Vennar, vettar and Kudamurutti

Question 3.
Write the latitudinal and longitudinal extent of India.
Answer:
India extends from 8°4’N to 37°6 ‘N latitudes and 68°7 ‘E to 97°25 ‘E longitude. It is located in the North-Eastern hemisphere. The Tropic of cancer (23°30’N) passes through the middle of the country dividing it into two halves.

Question 4.
When did the Suez Canal start functioning and how did it benefit India?
Answer:
The Suez Canal started functioning in 1869.
Benefits of India.

1. It reduced the distance between India and Europe by 7,000 km.
2. The Canal in the boom for trade as it had reduced the transportation cost and number of days involved.

Question 5.
Name the major physiographic divisions of India.
Answer:
Based on the geological structure, India is divided into 6 major physiographic divisions namely.

1. The Himalayan Mountains (Northern mountains)
2. The Great Northern plains
3. The Peninsular plateau
4. The Great Indian desert
5. The Coastal plains
6. The Islands

Question 6.
Which features modified the relief features of India?
Answer:
Besides geological functions, a lot of processes such as weathering erosion and deposition have created and modified the relief to its present form.

Question 7.
Name the prominent ranges of Trans-Himalayas.
Answer:
The prominent ranges of Trans-Himalayas are Karakoram, Zaskar, Ladak and Kailash.

Question 8.
What is the extent of Northern plains of India?
Answer:
It spreads over an area of 7 lakh sq.km. The plain being about 2400 km long and 240 to 320 km broad.

Question 9.
What are ‘Duns’?
Answer:
The longitudinal valley found between the Siwaliks and Himachal range are called Duns in the West and Duars in the East. These are the ideal sites for human settlements and agriculture.

Question 10.
What is doab?
Answer:
A tract of land between two rivers is called a doab. It made up of two words ‘do’ meaning ‘two’ and ab meaning ‘water’.

Question 11.
To which parts are Ganga plains extended?
Answer:
The Ganga plain extends between Ghagger and Teesta rivers. It is spread over Haryana, Delhi, Uttar Pradesh, Jharkhand and West Bengal.

Question 12.
How are riverine islands formed?
Answer:
Rivers moving from mountains carry alluvium with them and do the depositional work. In the lower course because of gentle slope, the river velocity decrease, and therefore islands are formed.

Question 13.
Where is -Deccan Trap located and what does it composed of?
Answer:
The black soil area in the peninsular plateau regions is known as Deccan trap. They extend from Gujarat to Delhi in a southwest – northeast direction. This is of volcanic origin hence the rocks are igneous.

Actually these rocks have descended over time and are responsible for the formation of black soil. The Aravalli hills lie on the western and northwestern margins of the peninsular plateau. These are highly eroded hills and are found as broken hills.

Question 14.
What are Barchans?
Answer:
They are crescent shaped dunes.

Question 15.
What are the climate characteristics of Indian desert?
Answer:
The region recevies very low rainfall below 150 mm per year. Streams appear during the rainy season. Soon after they disappear into the sand as they do not have enough water to reach the sea.

VII. Answer the following in paragraph.

Question 1.
What do you mean by drainage system? Write a note on it.
Answer:

1. A drainage system is an integrated system of tributaries and trunk stream which collects and drains surface water into the sea, lake or some other water body.
2. The total area drained by a river and its tributaries is known as a drainage basin.
3. The drainage pattern of an area is the result of the geological structure of the respective areas.
4. The river system provides irrigation, drinking water, power generation and livelihood for a large number of population.
5. The drainage system of India is broadly divided into two major groups on the basis of their location. They are
   • The Himalayan rivers
   • Peninsular rivers

Question 2.
“Unity in Diversity”. Explain.
Answer:
India is a vast country with a diversity of Physical characteristics.

Diversity in Natural Phenomena

1. India has unique landforms ranging from the highest peaks to the lowest plains. In the north India, Mount Godwin Austin, otherwise known as Mt. K2 is the highest peak of India and coastal plains are the lowest in the South India.
2. The Climate varies from the tropical to the temperate zone. Mawsynram in Meghalaya receives the highest amount of rainfall whereas the thar desert receives very low rainfall.
3. We have wet dense tropical forest on the western Ghats, Mangrove trees in the sundarbans of West Bengal and the shrubs and sparse vegetation in the Thar desert.
4. The diversity of the physical environment and climate has made India an ideal habitat for varieties of flora and fauna.

Diversity in Natural Phenomena
India is a secular country with total freedom of worship. People follow Hinduism, Christianity, Islam, Sikkism, Buddhism, Jainism and Zoroastrianism with cultural diversities.

Unity in Diversity
Inspite of its physical, religion and racial varities, the ‘Indian Culture’ unites all people. Hence India is known for her “Unity in diversity”.

People shed all their differences and stand together when is a crisis. The best examples are kargil invasions and natural calamities like floods and Tsunami. Even if we differ in many factors, we are all one in the fact we are Indians.

Question 3.
Give an account on Punjab-Haryana plains.
Answer:

1. Punjab Haryana plains are one among the subdivisions of the Northern plains.
2. They are formed by the deposition of the rivers Sutlej, Beas and Ravi, the tributaries of River Indus.
3. They lie to the North-east of the Great Indian Desert.
4. These plains are found over an area of about 1.75 lakh sq.km.
5. These plains act as water-divide (Doab) to the rivers Yamuna-Sutlej and Ganga-Yamuna.

Question 4.
Explain the origin of the Himalayas.
Answer:

1. Millions of years ago there was only one landmass on the surface of the earth. It was surrounded by ocean on all sides.
2. The landmass was called “ Pangea”. It was surrounded by a waterbody Known as “Panthalasa”.
3. This large landmass split up into two parts. The northern part was known as “Angaraland” and the southern part was known as “Gondwana land”.
4. The Sea seperating these two was called the Tethys Sea.
5. This Sea stretched along an east-west direction.
6. The rivers from Angara and Gondwana deposited the silts along the Tethys Sea.
7. After a long period, due to tectonic forces the deposits up lifted to forms fold mountains called the Himalayan range.
8. It is an young fold mountain.
9. The Himalayas is the home of several high peak. However, it holds the record of having the maximum number of highest peaks among any mountains range in world. Out of the heights peaks in this world, Himalayas holds nine.

Question 5.
Give an account of the four divisions of Himalayas from west to east along with Purvanchal hills respectively.
Answer:
Divisions of the Himalayas of the basis of regions from west to east. These division have been demarcated by river valleys.

(i) Punjab Himalayas:
The part of the Himalayas lying between Indus and Sutlej has been traditionally known as Punjab Himalayas but it is also known as regionally as Kashmir and Himachal Himalaya from west to east respectively.

(ii) Kumaon Himalayas:
The part of the Himalayas lying between Satluj and Kali rivers is known as Kumaun Himalayas.

(iii) Assam Himalayas:
The Kali and Tista rivers demarcate the Nepal Himalayas and the past lying between Tista and Dihang rivers is known as Assam Himalayas.

Purvachal:

1. The Brahmaputra marks the eastern most boundary of the Himalayas. Beyond the Dihang gorge, the Himalayas bend sharply to the south and spread along the eastern boundary of India. They are known as Purvanchal or the Eastern hills and mountains.
2. These hills running through the north-eastern states are mostly composed of strong sandstones which are sedimentary rocks.
3. It covered with dense forests, they mostly runs as parallel ranges and valleys.
4. The Purvanchal comprises the Patkai hills, the Naga hills, Manipur hills and the Mizo hills.

Question 6.
Give a detailed account on the Great Northern Plains.
Answer:
Formation:
The Northern Plain has been formed by the interplay of the three major river systems namely – The Indus, The Ganga, and The Brahmaputra along with their tributaries. This plain is formed of alluvial soil. The deposition of alluvium in a vast basis lying of the foothills of the Himalayas over millions of years formed this fertile plain.

Extension:

It spread over an area of 7 lakh sq.km. The plain being about 2400 km long and 240 to 320 km broad, is a densely populated physiographic division.
Importance:
With a rich soil caves combined with adequate water supply and favourable climate it is agriculturally a very production past of India.

Important Features:

1. In the lower course, due to gentle slope, the velocity of the river decreases which results in the formation of riverine islands. Majuli, in the Brahmaputra River is the largest inhabited riverine island in the world.
2. The rivers in their lower course split into numerous channels due to the deposition of silt. These channels are known as tributaries.
3. The Northern Plain is broadly divided into three sections. Punjab plains, Ganga plains and Brahmaputra plains.
4. According to the variations in relief features, the Northern plains can be divided into four regions – Bhabar, Terai, Bhangar and Khadar.

Punjab Plains:
The Western part of the Northern plain is referred to as the Punjab plains formed by the Indus and its tributaries. The larger part of this plain lies in Pakistan Indus and its tributaries – The Jhelum, the Chenab, the Ravi, the Beas and the Satluj originate in the Himalayas. This section of the plain is dominated by the doabs.

Ganga Plains:
They extend between Ghagger and Teesta rivers. It is spread over the states of North India like Haryana, Delhi, U.P., Bihar, Partly Jharkhand and West Bengal.

Brahmaputra plains:
In the east of Ganga plains lies the Brahmaputra plains. They cover the areas of Assam and Arunachal Pradesh.
According to the variations in relief features, the Northern plains can be divided into four regions.

Bhabar:
The rivers after descending from the mountains deposit pebbles in a narrow belt of about 8 to 16 km in width lying parallel to the slopes of the Shivaliks. It is known as Bhabar. All the streams disappear in this Bhabar belt.

Terai:
South of this belt, the streams and rivers re-emerge and create a wet swampy and marshy region known as ‘Terai’. This was a thickly forested region full of wildlife.

The Terai is wider in the eastern parts of the Great plains, especially in Brahmaputra valley due to heavy rainfall. In many states, the Terai forest have been cleaned for cultivation.

Khadar:
The newer younger deposits of the flood plains are called Khadar. They are renamed almost every year and so are fertile. Thus ideal for intensive agriculture. The Khadar tracts are enriched by fresh deposits of silt every year during rainy seasons. The Khadar land consists of sand, silt, clay and mud. It is highly fertile soil.

Delta Plains:
The deitaic plains is an extension of the Khadar land. It covers about 1.9 sq.km in the lower reaches of the Ganga River. It is an area of deposition as the river flows in this tract sluggishly. The deltaic plain consists mainly of old mud, new mud and marsh. In the delta region, the uplands are called “chars”. While the marshy areas are called ‘Bils’

Question 7.
Write in brief about Peninsular Plateau.
Answer:
Location:

1. The Peninsular plateau is located to the south of northern great plains.
2. It covers an area of about 16 lakh sq.km.
3. Boundaries of the Peninsular plateau.
   • North – Aravalli, Vindhya, Satpura and Rajmahal ranges
   • West – Western Ghats
   • East – Eastern Ghats
4. The average height varies between 600 – 900 mts above the mean sea level.
5. The general slope is from west to east broad divisions, namely the Central Highlands and the Deccan Plateau.

Central Highlands:

• The part of the Peninsular plateau lying to the north of the Narmada river covering a major area of the Malwa Plateau is known as the Central Highlands

Malwa Plateau:

• Malwa Plateau is bounded by the Aravalli range, the Vindhya range and Bundelthand.
• The Chambal river and its tributaries have created ravines in the northern part of the plateau.

The Bundelkhand:

• It is located towards the south of the Yamuna river.
• It is composed of igneous and metamorphic rocks.
• In the northern part, the Ganga and Yamuna rivers have deposited alluvium.
• The hilly areas are made up of sandstone and granite.
• Betwa and Ken rivers have carved out deep gorges.

The Baghelkhand:

• It lies to the east of “Maikala Range”.
• It is made up of sandstone and limestone in the west and granite in the east.
• The central part of the plateau acts as water divide between the Son and the Mahanadhi drainage system.

The Chotanagpur Plateau:

• It is located towards the northeast.
• It is drained by Damodar, Sabamarekha, Koel and Barakar rivers.
• The Damodar river flows from west to east through the middle of this region.
• This region has a series of plateau and hills.

Deccan Plateau:

1. It is made up of lava and is covered with black soil.
   Boundaries of Deccan Plateau.
   • North-east – Vindhya and Satpura mountains
   • North – Mahadev and Maikala range
   • West – Western Ghats
   • East – Eastern Ghats
2. It is a tableland composed of the old crystalline, igneous and metamorphic rocks.
3. It is formed due to the breaking and drifting of the Gondwana land.
4. It has broad and shallow valleys and rounded hills.
5. It is a triangular landmass that lies to the south of the river Narmada.
6. The Satpura range flanks its broad base in the north while the Mahadev, the Kaimur hills and the Maikal range form its eastern extensions.
7. The Deccan plateau is higher in the weat and slopes gently eastwards.
8. Rivers like Mahanadhi, Godavari, Krishna and Kaveri flow eastwards and join the Bay of Bengal.
9. It is separated by a fault from the Chotanagpur plateau.
10. It made up of lava rocks and has black regur soils.

Question 8.
Write about the Indian Desert.
Answer:

1. The Indian desert lies towards the western margins of the Aravalli Hills.
2. It is an undulating sandy plain covered with sand dunes.
3. This regions receives very low rainfall below 150 mm per year.
4. It has arid climate with low vegetation cover.
5. Streams appear during the rainy season, soon after they disappear into the sand as they do not have enough water to reach the sea. Luni is the only large rivers in this region.
6. Barchans (Crescent-shaped dunes) covers large areas more prominent in Jaisalmer but longitudinal dunes become more prominent near the Indo-Pakistan boundary.
7. It is the world 7th largest desert and world 9th largest sub tropical desert located in western part of the India.
8. There are two major division on the Thar desert. They are known as the Actual desert region (Marusthali) and the semi desert region (Bhangar).

Question 9.
Differences between Himalayan rivers and Peninsular rivers.
Answer:

We think the data given here clarify all your queries of Chapter 1 and make you feel confident to attempt all questions in the examination. So, practice more & more from Tamilnadu State Board solutions for 10th Social Science Geography Chapter 1 India: Location, Relief and Drainage Questions and Answers & score well. Need any information regarding this then ask us through comments & we’ll give the best possible answers very soon.

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.4

9th Maths Exercise 3.4 Question 1.
Expand the following :
(i) (2x + 3y + 4z)²
(ii) (-p + 2q + 3r)²
(iii) (2p + 3)(2p – 4)(2p – 5)
(iv) (3a + 1)(3a – 2)(3a + 4)
Solution:
(i) (2x + 3y + 4z)²
(a + b + c)² ≡ a² + b² + c² + 2ab + 2bc + 2ca
∴ (2x + 3y + 4z)² = (2x)² + (3y)² + (4z)² + 2 (2x) (3y) + 2 (3y) (4z) + 2 (4z) (2x)
= 4x² + 9y² + 16z² + 12xy + 24yz + 16zx

(ii) (-p + 2q + 3r)²
(a + b + c)² ≡ a² + b² + c² + 2ab + 2bc + 2ca
(-p + 2q + 3r)² = (-p)² + (2q)² + (3r)² + 2(-p) (2q) + 2(2q) (3r) + 2 (-p) (3r)
= p² + 4q² + 9r² – 4pq +12qr – 6rp

(iii) (2p + 3)(2p – 4)(2p – 5)
(x + a) (x + b) (x + c) ≡ x³ + (a + b + c) x² + (ab + be + ca) x + abc
(2p + 3)(2p – 4)(2p – 5) = (2p)³ + (3 – 4 – 5) (2p)² + [(3 × – 4)² + (-4 × -5) + (-5 × 3)] 2p + 3 × – 4 × – 5
= 8p³ + (-6) (4p²) + [-12 + 20 + (-15)] 2p + 60
= 8p³ – 24p² + (-7)2p + 60
(2p + 3)(2p – 4)(2p – 5) = 8p³ – 24p² – 14p + 60

(iv) (3a + 1)(3a – 2)(3a + 4)
(x + a) (x + b) (x + c) ≡ x³ + (a + b + c) x² + (ab + bc + ca) x + abc
(3a + 1)(3a – 2)(3a + 4) = (3a)³ + (1 – 2 + 4) (3a)² + [1 × (- 2) + (-2 × 4) + 4 × 1] (3a) + 1 × -2 × 4
= 27a³ + 3 (9a²) + (-2 – 8 + 4)3a – 8
= 27a³ + 27a² – 8a – 8

9th Maths Algebra Exercise 3.4 Question 2.
Using algebraic identity, find the coefficients of x2, x and constant term without actual expansion.
(i) (x + 5)(x + 6)(x + 7)
(ii) (2x + 3)(2x – 5)(2x – 6)
Solution:
(i) (x + 5)(x + 6)(x + 7)
(x + a) (x + b) (x + c) ≡ x³ + (a + b + c) x² + (ab + bc + ca) x + abc
Co-efficient of x² = a + b + c = 5 + 6 + 7 = 18
Co-efficient of x² = ab + bc + ca = (5 × 6) + (6 × 7) + (7 × 5)
= 30 + 42 + 35 = 107
Constant term = abc = 5 × 6 × 7
Co-efficient of constant term = 210

(ii) (2x + 3)(2x – 5)(2x — 6)
∴ Co-efficient of x² = 4 (a + b + c) = 4 (3 + (-5) + (-6))
= 4 × (-8) = -32
Co-efficient of x = 2 (ab + bc + ca)
= 2 [3 × (-5) + (-5) (-6) + (-6) (3)]
= 2[-15 + 30- 18] = 2 × (-3) = -6
Constant term = abc = 3 × (-5) × (-6) = 90

9th Maths 3.4 Question 3.
If (x + a)(x + b)(x + c) = x³ + 14x² + 59x + 70, find the value of
(i) a + b + c
(ii) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
(iii) a² + b² + c²
(iv) \(\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}\)
Solution:
(x + a)(x + b)(x + c) = x³ + 14x² + 59x + 70 …………….. (1)
(x + a) (x + b) (x + c) ≡ x³ + (a + b + c) x² + (ab + bc + ca) x + abc …………. (2)
(i) Comparing (1) & (2)
We get, a + b + c = 14
(ii) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{b c+a c+a b}{a b c}=\frac{59}{70}\)
(iii) (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
= 14² – 2 (59) = 196 – 118 = 78
(iv) \(\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}=\frac{a^{2}+b^{2}+c^{2}}{a b c}=\frac{78}{70}=\frac{39}{35}\)

Exercise 3.4 Class 9 Question 4.
Expand
(i) (3a – 4b)³
(ii) (x + \(\frac{1}{y}\))³
Solution:
(i) (3a – 4b)³ We know that
(x – y)³ = x³ – 3x²y + 3xy² – y³
(3a-4b)³ = (3a)³ – 3 (3a)² (4b) + 3 (3a) (4b)² – (4b)³
= 27a³ – 108a²b + 144 ab² – 64b³

(ii) (x + \(\frac{1}{y}\))³
(x + y)³ ≡ x³ + 3x²y + 3xy² + y³
\(\left(x+\frac{1}{y}\right)^{3}=x^{3}+\frac{3 x^{2}}{y}+\frac{3 x}{y^{2}}+\frac{1}{y^{3}}\)

9th Maths Exercise 3.4 In Tamil Question 5.
Evaluate the following by using identities:
(i) 98³
(ii) 1001³
Solution:
(i) 98² = (100 – 2)³
(a – b)³ ≡ a³ – 3a²b + 3ab² – b³
98³ = (100 – 2)³ = 100³ – 3 × 100² × + 3 × 100 × 2² – 2³
= 1000000 – 3 × 10000 × 2 + 300 × 4 – 8
= 1000000 – 60000 + 1200 – 8 = 1001200 – 60008 = 941192

(ii) 1001³ = (1000 + 1)³
(a + b)³ ≡ a³ + 3a²b + 3ab² + b³
(1000 + 1)³ = 1000³ + 3(1000²) × 1 + 3 × 1000 × 1² + 1³
= 1000,000,000 + 3,000,000 + 3000 + 1 = 1,003,003,001

Ex 3.4 Class 9 Question 6.
If (x + y + z) = 9 and (xy + yz + zx) = 26 then find the value of x² + y² + z².
Solution:
(x + y + z) = 9 and(xy + yz + zx) = 26
x² + y² + z² = (x + y + z)² – 2 (xy + yz + zx)
= 9² – 2 × 26 = 81 – 52 = 29

9th Standard Maths Exercise 3.4 Question 7.
Find 27a³ + 64b³, if 3a + 4b = 10 and ab = 2.
Solution:
3a + 4b = 10, ab = 2
(3a + 4b)³ = (3a)³ + 3 (3a)² (4b) + 3 (3a) (4b)² + (4b)³
(27a³ + 64b³) = (3a + 4b)³ – 3 (3a) (4b) (3a + 4b)
∵ x³ + y³ = (x + y)³ – 3xy – (x + y)
= 10³ – 36 ab (10)= 1000 – 36 × 2 × 10
= 1000 – 720 = 280

Exercise 3.4 Class 9 Maths Solution Question 8.
Find x³ – y³, if x – y = 5 and xy = 14.
Solution:
x – y = 5, xy = 14
x³ – y³ = (x – y)³ + 3xy (x – y) = 5³ + 3 × 14 × 5
= 125 + 210 = 335

Class 9 Maths Exercise 3.4 Solutions Question 9.
If a + \(\frac{1}{a}\) = 6, then find the value of a³ + \(\frac{1}{a^{3}}\)
Solution:
a³ + b³ = (a + b)³ – 3ab (a + b)

9th Math 3.4 Solution Question 10.
If x² + \(\frac{1}{x^{2}}\) = 23, then find the value of x + \(\frac{1}{x}\) and x³ + \(\frac{1}{x^{3}}\)
Solution:

Maths Exercise 3.4 Class 9 Question 11.
If (y – \(\frac{1}{y}\))³ = 27, then find the value of y³ – \(\frac{1}{y^{3}}\)
Solution:

9th Class Maths Exercise 3.4 Solution Question 12.
Simplify : (i) (2a + 36 + 4c)(4a² + 9b² + 16c² – 6ab – 12bc – 12bc – 8ca)
(ii) (x – 2y + 3z) (x² + 4y² + 9z² + 2xy + 6yz – 3xz)
Solution:
(i) (2a + 36 + 4c)(4a² + 9b² + 16c² – 6ab – 12bc – 12bc – 8ca)
We know that
(a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³ + c³ × 3 abc
∴ (2a + 36 + 4c) (4a² + 9b² + 16 c² – 6 ab – 12 bc – 8 ca)
= (2a)³ + (3b)³ + (4c)³ – 3 × 2a × 36 × 4c
= 8a³ + 27b³ + 64c³ – 72 abc

(ii) (x – 2,y + 3z) (x² + 4y² + 9z² + 2xy + 6yz – 3xz)
(a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³ + c³ – 3abc .
∴ (x – 2y + 3z) (x² + 4y² + 9z² + 2xy + 6yz – 3xz)
= x³ + (-2y)³ + (3z)³ – 3 × x × (-2y) (3z)
= x³ – 8y³ + 27z³ + 18 xyz

9th Maths Exercise 3.4 Samacheer Kalvi Question 13.
By using identity evaluate the following:
(i) 7³ – 10³ + 3³
(ii) 1 + \(\frac{1}{8}-\frac{27}{8}\)
Solution:
(i) 7³ – 10³ + 3³
(a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³ + c³ – 3 abc
If a + b + c = 0, then a³ + b³ + c³ = 3 abc
∴ 7 – 10 + 3 = 0
⇒ 7³ – 10³ + 3³ = 3 × 7 × – 10 × 3
= 9 × -70 = -630

9th Class Math Exercise 3.4 Solution Question 14.
If 2x – 3y – 4z = 0, then find 8x³ – 27y³ – 64z³.
Solution:
If 2x – 3y – 4z = 0 then 8x³ – 27y³ – 64z³ = ?
If x + y + z = 0 then x³ + y³ + z³ = 3xyz
8x³ – 21 y³ – 64z³ = (2x)³ + (-3y)³ + (-4z)³
= 3 × 2x × -3y × – 4z = 72 xyz

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Tamilnadu Samacheer Kalvi 10th Social Science History Solutions Chapter 6 Early Revolts against British Rule in Tamil Nadu

Do you feel scoring more marks in the 10th Social Science History Grammar sections and passage sections are so difficult? Then, you have the simplest way to understand the question from each concept & answer it in the examination. This can be only possible by reading the passages and topics involved in the 10th Social Science History Board solutions for Chapter 6 Early Revolts against British Rule in Tamil Nadu Questions and Answers. All the Solutions are covered as per the latest syllabus guidelines. Check out the links available here and download 10th Social Science History Chapter 6 textbook solutions for Tamilnadu State Board.

Early Revolts against British Rule in Tamil Nadu Textual Exercise

I. Choose the correct answer.

Early Revolts Against British Rule In Tamil Nadu Question 1.
Who was the first Palayakkarar to resist the East India Company’s policy of territorial aggrandizement?
(a) Marudhu brothers
(b) Puli Thevar
(c) Velunachiyar
(d) Veera Pandya Kattabomman
Answer:
(b) Puli Thevar

Early Revolts Against British Rule In Tamil Nadu Book Back Answers Question 2.
Who had borrowed money from the East India Company to meet the expenses he had incurred during the Carnatic wars?
(a) Velunachiyar
(b) Puli Thevar
(c) Nawab of Arcot
(d) Raja of Travancore
Answer:
(c) Nawab of Arcot

Early Revolts Against British Rule Question 3.
Who had established close relationship with the three agents of Chanda Sahib?
(a) Velunachiyar
(b) Kattabomman
(c) Puli Thevar
(d) Oomai thurai
Answer:
(c) Puli Thevar

Early Revolts Against British Rule In Tamilnadu Question 4.
Where was Sivasubramanianar executed?
(a) Kayathar
(b) Nagalapuram
(c) Virupachi
(d) Panchalamkurichi
Answer:
(b) Nagalapuram

Question 5.
Who issued the Tiruchirappalli proclamation of Independence?
(a) Marudhu brothers
(b) Puli Thevar
(c) Veera Pandya Kattabomman
(d) Gopala Nayak
Answer:
(a) Marudhu brothers

Question 6.
When did the Vellore Revolt breakout?
(a) 24 May 1805
(b) 10 July 1805
(c) 10 July 1806
(d) 10 September 1806
Answer:
(c) 10 July 1806

Question 7.
Who was the Commander-in-Chief responsible for the new military regulations in Vellore Fort?
(a) Col. Fancourt
(b) Major Armstrong
(c) Sir John Cradock
(d) Colonel Agnew
Answer:
(c) Sir John Cradock

Question 8.
Where were the sons of Tipu Sultan sent after the Vellore Revolt?
(a) Calcutta
(b) Mumbai
(c) Delhi
(d) Mysore
Answer:
(a) Calcutta

II. Fill in the blanks.

1. The Palayakkarar system was put in place in Tamil Nadu by ……………..
2. Except the Palayakkarars of ……………., all other western Palayakkarars supported Puli-Thevar.
3. Velunachiyar and her daughter were under the protection of …………… for eight years.
4. Bennerman deputed ……………… to convey his message, asking Kattabomman to surrender.
5. Kattabomman was hanged to death at ……………..
6. The Rebellion of Marudhu Brothers was categorized in the British records as the ……………….
7. …………. was declared the new Sultan by the rebels in Vellore Fort.
8. ……………. suppressed the revolt in Vellore Fort.
Answers:
1. Viswanatha Nayaka
2. Sivagiri
3. GopalaNayaker
4. Ramalinga Mudaliar
5. Kayathar
6. South Indian Rebellion
7. Fateh Hyder
8. Col. Gillespie

III. Choose the correct statement.

Question 1.
(i) The Palayakkarar system was in practice in the Kakatiya Kingdom.
(ii) Puli Thevar recaptured Nerkattumseval in 1764 after the death of Khan Sahib.
(iii) Yusuf Khan who was negotiating with the Palayakkarars, without informing the Company administration was charged with treachery and hanged in 1764.
(iv) Ondiveeran led one of the army units of Kattabomman.
(a) (i) (ii) and (iv) are correct
(b) (i) (ii) and (iii) are correct
(c) (iii) and (iv) are correct
(d) (i) and (iv) are correct
Answer:
(b) (i) (ii) and (iii) are correct

Question 2.
(i) Under Colonel Campbell, the English Army went along with Mafuzkhan’s army.
(ii) After Muthu Vadugar’s death in Kalaiyar Kovil battle, Marudhu Brothers assisted Velunachiyar in restoring the throne to her.
(iii) Gopala Nayak spearheaded the famous Dindigul League.
(iv) In May 1799 Cornwallis ordered the advance of Company armies to Tirunelveli.
(a) (i) and (ii) are correct
(b) (ii) and (iii) are correct
(c) (ii) (iii) and (iv) are correct
(d) (i) and (iv) are correct.
Answer:
(b) (ii) and (iii) are correct

Question 3.
Assertion (A): Puli Thevar tried to get the support of Hyder Ali and the French.
Reason (R): Hyder Ali could not help Puli Thevar as he was already in a serious conflict with the Marathas.
(a) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
(b) Both (A) and (R) are wrong
(c) Both (A) and (R) are correct and R is the correct explanation of (A)
(d) (A) is wrong and (R) is correct
Answer:
(d) (A) is wrong and (R) is correct

Question 4.
Assertion (A): Apart from the new military Regulations the most objectionable was the addition of a leather cockade in the turban.
Reason (R): The leather cockade was made of animal skin.
(a) (A) is wrong and (R) is correct
(b) Both. (A) and (R) are correct and (R) is the correct explanation of (A)
(c) Both (A) and (R) are wrong
(d) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
Answer:
(b) Both. (A) and (R) are correct and (R) is the correct explanation of (A)

IV. Match the following.

Answer:
1. (e)
2. (c)
3. (b)
4. (a)
5.(d)

V. Answer the questions briefly.

Question 1.
What were the duties of the Palayakkarars?
Answer:

1. The Palayakkarars were free to collect revenue.
2. Administer the territory.
3. Settle disputes and maintain law and order also.
4. Helped the Nayak rulers to restore the kingdom.

Question 2.
Identify the Palayams based on the division of east and west.
Answer;
Among the 72 Palayakkarars, there were two blocks namely the eastern and the western Palayams.

• The eastern Palayams were – Sattur, Nagalapuram, Ettayapuram and Panchalam Kurichi.
• The western Palayams were – Uttrumalai, Thalavankottai, Naduvakurichi, Singampatti and Seithur.

Question 3.
Why was Heron dismissed from service?
Answer:

1. Puli Thevar continued were defy the authority of the English East India Company.
2. Col. Heron was urged by the company to deal the issue of Puli Thevar.
3. For want of canon and of supplies and pay to soldiers, colonel Heron abandoned the plan. Hence he was dismissed from service.

Question 4.
What was the significance of the Battle of Kalakadu?
Answer:
In the Battle of Kalakadu, Mahfuzkhan’s troops were routed by the huge forces of Puli Thevar.

Question 5.
What was the bone of contention between the Company and Kattabomman?
Answer:

1. The company gained the right to collect taxes from Panchalamkurichi which was under Kattabomman.
2. The company appointed collectors to collect taxes from all the Palayams.
3. The collectors humiliated the Palayakkarars and adopted force to collect the taxes.
4. This was the bone of contention between the company and Kattabomman.

Question 6.
Highlight the essence of the Tiruchirappalli Proclamation of 1801.
Answer:
The Tiruchirappalli Proclamation of 1801 was the first call to the Indians to unite against the British, cutting across region, caste, creed and religion. The Proclamation was pasted on the walls of the Nawab’s palace in Tiruchirappalli Fort and on the walls of the Srirangam Temple. As a result, many Palayakkarars of Tamil country rallied together to fight against the English.

Question 7.
Point out the importance of the Treaty of 1801.
Answer:

1. The Treaty of 1801 was known as “Carnatic Treaty”. Under the terms of the camatic Treaty of 31st July 1801.
2. The British assumed the direct control over Tamilagam.
3. The Palayakkarar system came to an end with the demolition of all forts and disbandment of their army.

VI. Answer all the questions given under each caption.

Question 1.
Velunachiyar
(a) Who was the military chief of Velunachiyar?
Answer:
Gopala Nayaker

(b) What were the martial arts in which she was trained?
Answer:
The martial arts in which she was trained were valari, stick fighting and to wield weapons.

(c) Whom did she marry?
Answer:
She was married to Muthu Vadugar, the Raja of Sivagangai.

(d) What was the name of her daughter?
Answer:
Her Daughter’s name was Vellachinachiar.

Question 2.

Dheeran Chinnamalai

(a) When was Dheeran Chinnamalai bom?
Answer:
Dheeran Chinnamalai was bom in 1756.

(b) How did he earn the title “Chinnamalai”?
Answer:
The tax money collected by Tipu’s Diwan was confiscated by Theerthagiri (original name of Dheeran Chinnamalai) While returning to Mysore. He let Diwan to go by instructing him to tell his sultan that “Chinnamalai” who is between Sivamalai and Chennimalai was the one who took away taxes. Thus he gained the name “Dheeran Chinnamalai”.

(c) Name the Diwan of Tipu Sultan.
Answer:
Diwan of Tipu Sultan was Mohammed Ali.

(d) Why and where was he hanged to death?
Answer:
Dheeran Chinnamalai refused to accept the overlordship of the British. So he was captured and imprisoned. In 31st July 1805 he was hanged to death at the top of the Sankagiri Fort.

VII. Answer the following in detail.

Question 1.
Attempt an essay of the heroic fight Veerapandya Kattabomman conducted against the East India Company.
Answer:
(i) Veera Pandya Kattabomman became the Palayakkarar of Panchalamkurichi after the death of his father, Jagavira Pandya Kattabomman. The company administrators did not give him much importance. But soon several events led to the conflict between him and the East India Company.

(ii) The Company had gained the right to collect taxes from Panchalamkurichi. The Collectors adopted force to collect the taxes. This was the bone of contention between the English and Kattabomman.

(iii) When Kattabomman refused to pay the land revenue, he was asked to meet Jackson, the company official in Ramanathapuram where he had to stand for hours before the official. Kattabomman somehow escaped from there.

(iv) On his return to Panchalamkuruchi, Kattabomman represented to the Madras council about how he was ill-treated by the collector Jackson. The council asked Kattabomman to appear before a Committee with William Brown, William Oram and John Casamajor as members. The committee found Kattabomman innocent and Jackson was dismissed from service.

(v) Thereafter Kattabomman along with Marudhu Brothers confronted with the English. Kattabomman was asked to surrender. But his ‘evasive reply’ prompted Major Bannerman to attack his fort. The major soon seiged Panchalamkuruchi.

(vi) Kattabomman escaped to Pudukottai where he was captured. During the trail Kattabomman bravely admitted all the charges levelled against him. He was hanged from a tamarind tree in the old fort of Kayathar.

Question 2.
Highlight the tragic fall of Sivagangai and its outcome.
Answer:
Tragic fall of Sivagangai: In May 1801, the English attacked the rebles in Thanjaviir and Tiruchirapaili.

1. The rebels went to piranmalai and Kalaiyar kovil.
2. They were again defeated by the able commanders and superior military strength of the English company.
3. The rebellion failed and Sivagangai was annexed in 1801.
4. In 24th October 1801 the Marudhu brothers were executed in the Fort of Thirupathur near Ramanathapuram.

Outcomes:

1. The exploits and sacrifices of the Palayakkarars inspired later generations.
2. The Rebellion of Murudhu brothers called as “South Indian Rebellion is a landmark event in the history of Tamil Nadu.
3. The rebellion resulted in the liquidation of all the chieftains of Tamil Nadu.
4. The British assumed direct control over Tamilagam.
5. The Palayakkarar system came to an end with the demolition of all Forts and disbandment of their army.

Question 3.
Account for the outbreak of Vellore Revolt in 1806.
Answer:
After the suppression of resistance of Kattabomman and Marudhu Brothers in 1801, the British charged the Nawab of Arcot with disloyalty and forced a treaty on him. According to this treaty, the Nawab was forced to cede the districts of North Arcot, South Arcot, Tiruchirappalli, Madurai and Tirunelveli to the Company and transfer all the administrative powers to it.

But the resistance did not die down. The dispossessed little kings and feudal chieftains were continuously deliberating on the future course of action against the company. This finally resulted in the Vellore Revolt of 1806. The sepoys of the British Indian army nursed a strong sense of resentment over low salary and poor prospects of promotion. The English army officers gave little respect for the social and religious sentiments of the Indian sepoys also angered them.

The immediate cause of the revolt came in the form of a new military regulation according to which the Indian soldiers were asked not to wear caste marks or ear rings when in uniform. They were to be cleanly shaven on the chin and maintain uniformity how their moustache looked. The new turban added fuel to fire. The new turban had the leather cockade made of animal skin. The sepoys refused to wear it.

On 10 July 1806, in the early hours, guns began to boom. The Indian sepoys revolted against the company rule. However, the revolt was suppressed brutally.

VIII. Activity

Question 1.
Teacher can ask the students to prepare an album of patriotic leaders of early revolts against the British rule in Tamil Nadu. Using their imagination they can also draw pictures of different battles in which they attained martyrdom.
Answer:
Do it yourself.

Question 2.
Stage play visualizing the conversation between Jackson and Kattabomman be attempted by students with the help of teachers.
Answer:
Do it yourself.

Question 3.
A comparative study of Vellore Revolt and 1857 Revolt by students be tried enabling them to find out to what extent Vellore Revolt had all the forebodings of the latter.
Answer:
The Vellore Mutiny took place on July 10, 1806. This major act of defiance happened even before the famous Rebellion of 1857. Though lasting only for a day, the Vellore Mutiny marked the first ever large-scale and violent mutiny by Indian sepoys against the East India Company. It was triggered by the English disregard to the religious sensitivities of the Hindu and Muslim Indian sepoys. The Revolt of 1857 was a rather large revolt which went on for days. It was fed by resentments bom of diverse perceptions, including invasive British-style social reforms, harsh land taxes, as well as scepticism about the improvements brought about by British rule.
Students can make a comparative study under the guidance of their teacher.

Early Revolts against British Rule in Tamil Nadu Additional Questions

I. Choose the correct answer.

Question 1.
Palayakkarar system was in practice during the rule of ………………. of Warangal.
(a) Rajendra Chola
(b) Prataba Rudhra
(c) Ashoka
Answer:
(b) Prataba Rudhra

Question 2.
Puli Thevar’s three major ports came under the control of Yusuf khan on
(a) 18^th August 1798
(b) 16^th May 1761
(c) 19^th September 1798
(d) 16^th October 1799
Answer:
(b) 16^th May 1761

Question 3.
On many occasions the Palayakarars helped the ………….. rulers to restore the kingdom.
(a) Nayak
(b) Pallava
(c) Pandya
Answer:
(a) Nayak

Question 4.
Muthu vadagar died in the battle of :
(a) Kalakad
(b) Tiruchirapalli
(c) Kalaiyar kovil
(d) Palayamkottai
Answer:
(c) Kalaiyar kovil

Question 5.
The Proclamation of ………………. was the first call to the Indians to unite against the British.
(a) 1805
(b) 1809
(c) 1801
Answer:
(c) 1801

Question 6.
As per this treaty of 1801 Nawab of Arcot transferred all the administrative powers to the company:
(a) Treaty of Mangalore
(b) Treaty of Madras
(c) Treaty of Carnatic
(d) None of the above
Answer:
(c) Treaty of Carnatic

Question 7.
Hyder Ali could not help Puli Thevar because ……………..
(a) he was not well
(b) he was in a serious conflict with the Marathas
(c) he did not like Puli Thevar
(d) he was hostile to Puli Thevar
Answer:
(b) he was in a serious conflict with the Marathas

Question 8.
Puli Thevar was defeated by:
(a) Colonel Heron
(b) Captain campbell
(c) Major cootes
(d) Colonel Fancourt
Answer:
(b) Captain campbell

Question 9.
Dheeran was well trained in ……………..
(a) modem warfare
(b) archery
(c) horse riding
(d) all of the above
Answer:
(d) all of the above

Question 10.
………………….. ordered the release of Sivasubramanianar and the suspension of the collector Jackson.
(a) Lord Wellesley
(b) John casamajor
(c) Governor edward clive
(d) General Bannerman
Answer:
(c) Governor edward clive

II. Fill in the blanks :

1. Chinna Marudhu collected nearly ……………. men to challenge the English army.
2. ……………. tried to get the support of Hyder Ali and the French.
3. Puli Thevar wielded much influence over the ………………
4. …………… was the military chief of Velunachiyar.
5. The leather cockade was made of ……………..
6. The Marudhu brothers were executed in the Fort of Tirupathur near …………… on 24 October 1801.
7. ………….. policy of the English split the forces of the Palayakkarars.
8. ………….. wielded much influence over the western Palayakkarars.
9. Trained by the French Dheeran mobilised the …………….. youth in thousands and fought the British together with Tipu.
10. Tipu was killed at the end of the Anglo-Mysore war in …………….
Answers :
1. 1.20,000
2. Puli Thevar
3. Western Palayakkarars
4. Gopala Nayakar
5. animal skin
6. Ramanathapuram
7. Divide and Rule
8. Puli Thevar
9. Kongu 10.1799

III. Choose the correct statement.

Question 1.
(i) Kuyili was a faithful friend of Velunachiyar.
(ii) She led the unit of women soldiers named after Udaiyaal.
(iii) Udaiyaal was a timid girl who divulged information on Kuyili.
(iv) Kuyili is said to have walked into the British arsenal after setting herself on fire.
(a) (i), (ii) and (iii) are correct
(b) (i), (iii) and (iv) are correct
(c) (i), (ii) and (iv) are correct
(d) (ii),(iii) and (iv) are correct
Answer:
(c) (i), (ii) and (iv) are correct

Question 2.
(i) Velunachiyar was crowned as Queen with the help of Puli Thevar.
(ii) She was the first female ruler or queen to resist the British colonial power in India.
(iii) Gopala Nayak drew inspiration from Tipu Sultan.
(iv) The Carnatic Treaty took place in the year 1801.
(a) (i), (ii) and (iv) are correct
(b) (ii), (iii) and (iv) are correct
(c) (i) and (iii) are correct
(d) (i), (iii) and (iv) are correct
Answer:
(b) (ii), (iii) and (iv) are correct

IV. Match the following

Question 1.

Answer:
1. (d)
2. (e)
3. (a)
4. (b)
5. (c)

Question 2.

Answer:
1. (c)
2. (a)
3. (e)
4. (b)
5. (d)

V. Answer briefly:

Question 1.
Name the Palayakkarars who revolted against the British rule in Tamil Nadu.
Answer:
Puli Thevar, Velunachiyar, Dheeran chinnamalai, Marudhu brothers and Veerapandiya Kattabomman were some of the prominent Palayakkarars who revolted against the British rule in Tamil Nadu.

Question 2.
Explain about Ondiveeran’s bravery.
Answer:
Ondiveeran led one of the army units of Puli Thevar. Fighting by the side of Puli Thevar, he caused much damage to the company’s army. According to oral tradition, in one battle, Ondiveeran’s hand was chopped off and Puli Thevar was saddened. But Ondiveeran said it was a reward for his penetration into enemy’s fort causing many heads to roll.

Question 3.
Who supported the British and who did not join the confederacy of Puli Thevar?
Answer:
The English succeeded in getting the support of the Raj as of Ramanathapuram and Pudukkottai. The Palayakkarars of Ettayapuram, Panchalamkurichi and Sivagiri did not join the confederacy of Puli Thevar.

Question 4.
Who was Muthu Vadugar? How was he killed?
Answer:
Muthu Vadugar was the Raja of Sivagangai. He was married to Velunachiyar. In 1772, the Nawab of Arcot and the Company troops under the command of Lt. col. Bon Jour stormed the Kalaiyar Kovil Palace. In the ensuing battle Muthu Vadugar was killed.

Question 5.
Write a short note on Velunachiyar.
Answer:
Velunachiyar was bom in 1730 she was the only daughter of Raja Sellamuthu sethupathy of Ramanathapuram. She was trained in martial arts like valari, stick fighting and wield weapons. She was expert in horse riding and archery and had proficiency in English, French and Urdu.

Question 6.
What status did the Palayakkarars avail during the seventeenth and eighteenth centuries?
Answer:
During the seventeenth and eighteenth centuries the Palayakkarars dominated the politics of Tamil country. They functioned as independent sovereign authorities within their respective Palayams.

Question 7.
Write a brief note on Marudhu brothers.
Answer:

1. Periya Marudhu or Vella Marudhu and his younger brother Chinna Marudhu were the able generals of Muthuvadugar of Sivagangai.
2. After Muthuvadugar’s death in the Kalaiyur kovil battle, they assisted in restoring the throne to Velunachiyar.
3. In the last years of the eighteenth century Marudhu brothers organised resistance against the British.

Question 8.
Give an estimate of the Vellore Revolt of 1806.
Answer:
The Vellore Revolt failed because there was no immediate help from outside. According to recent studies, the organizing part of the revolt was done perfectly by Subedars Sheik Adam and Sheik Hamid and Jamedar Sheik Hussain of the 2nd Battalion of 23^rd regiment. Vellore revolt had all the forebodings of the Great Rebellion of 1857. The only difference was that there was no civil rebellion following the mutiny.

VI. Answer all the questions given under each caption:

Question 1.
Vellore Revolt:

(a) When did the outbreak of Vellore revolt occur?
Answer:
Out break of Vellore revolt occurred on 10th July 1806.

(b) Who commanded the garrison?
Answer:
Colonel Fancourt commanded the garrison.

(c) By whom was the organising part of the revolt done?
Answer:
The organising part of the revolt was done perfectly by Subedars Sheik Adam and Sheik Hamid and Jamedar Shiek Hussain of 2nd battalion of 23rd regiment and subedar Jamedar Sheik Kasim of the 1st battalion of 1st regiment.

(d) Name the places where the Vellore revolt of 1806 echoed?
Answer:
Vellore Revolt of 1806 had its echoes in Bellary, Walajabad, Hyderabad, Bengaluru, Nandydurg and Sankaridurg.

Question 2.
Appearance Before Madras Council

(a) Whom did the Madras council ask to appear before the committee?
Answer:
Kattabomman.

(b) Who were the committee members of Madras council?
Answer:
William Brown, William oram, John Casamajor.

(c) When did Kattabomman appear before the committee?
Answer:
Kattabomman appeared before the committee on 15th December 1798.

(d) What did Kattabomman report to the committee? What was the result?
Answer:
Kattabomman reported on what transpired in Ramanathapuram. The committee found Kattabomman was not guilty.

VII. Answer the following in detail.

Question 1.
Give an estimation of Revolt at Vellore (1806).
Answer:

1. General Gillespie from Arcot along with the captain young cavalry commander crushed the revolt. Nearly eight hundred soldiers were found dead. The organising part of the revolt was done perfectly by the subedars of 1st battalion of the 1st regiment and the 2nd battalion of 23rd regiment.
2. The Vellore revolt failed because there was no immediate help from outside.
3. It was the 1st open uprising of the Indian soldiers under British army.
4. This revolt was not confined to Vellore Fort alone but echoed outside regions also.
5. Vellore Revolt had all the forebodings of the Great Rebellion of 1857.
6. This was also called as Vellore mutiny as it arose only from the soldiers.

Question 2.
Give on assessment of Velunachiyar’s resistance to the British colonial power in India,
Answer:
Velunachiyar belonged to the royal family. She was brought up as a princess and was trained in marital arts. She was also adept in horse riding and archery. She was married to Muthu Vadugar, the Raja of Sivangangai. When the Raja was killed in the battle, Velunachiyar had to live under the protection of Gopala Nayakar at Virupachi near Dindigul for eight years. During her period in hiding, Velunachiyar organised an army and succeeded in securing an alliance with not only Gopala Nayakar but Hyder Ali as well. Dalavay (military chief) Thandavarayanar wrote a letter to Sultan Flyder Ali on behalf of Velunachiyar asking for 5000 infantry and 5000 cavalry to defeat the English. Velunachiyar explained in detail in Urdu all the problems she had with East India Company.

She conveyed her strong determination to fight the English. Impressed by her courage, Hyder Ali ordered his commandant Syed in Dindigul Fort to provide the required military assistance. Velunachiyar employed agents for gathering intelligence to find where the British had stored their ammunition. With military assistance from Gopala Nayakar and Hyder Ali, She recaptured Sivagangai. She was crowned as Queen with the help of Marudhu Brothers. ” She became the first female ruler to resist the British colonial power in India.

Impotant Events And Years:

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Tamilnadu Samacheer Kalvi 7th Tamil Solutions Term 2 Chapter 2.5 ஒரெழுத்து ஒருமொழி, பகுபதம், பகாப்பதம்

மதிப்பீடு

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
நன்னூலின்படி தமிழிலுள்ள ஓரெழுத்து ஒரு மொழிகளின் எண்ணிக்கை …………..
அ) 40
ஆ) 42
இ) 44
ஈ) 46
Answer:
ஆ) 42

Question 2.
எழுதினான்’ என்பது ………………..
அ) பெயர்ப் பகுபதம்
ஆ) வினைப் பகுபதம் இ) பெயர்ப் பகாப்பதம்
ஈ) வினைப் பகாப்பதம்
Answer:
ஆ) வினைப் பகுபதம்

Question 3:
பெயர்ப்ப குபதம் ……………… வகைப்படும்.
அ) நான்கு
ஆ) ஐந்து
இ) ஆறு
ஈ) ஏழு
Answer:
இ) ஆறு

Question 4.
காலத்தைக் காட்டும் பகுபத உறுப்பு ……………….
அ) பகுதி
ஆ) விகுதி
இ) இடைநிலை
ஈ) சந்தி
Answer:
இ) இடைநிலை

பொருத்துக

1. பெயர்ப் பகுபதம் – வாழ்ந்தான்
2. வினைப் பகுபதம் – மன்
3. இடைப் பகாப்பதம் – நனி
4. உரிப் பகாப்பதம் – பெரியார்
Answers:
1. பெயர்ப் பகுபதம் – பெரியார்
2. வினைப் பகுபதம் – வாழ்ந்தான்
3. இடைப் பகாப்பதம் – மன்
4. உரிப் பகாப்பதம் – நனி

சரியான பகுபத உறுப்பை எழுதுக

1. போவாள் – போ + வ் + ஆள்
போ – பகுதி
வ் – எதிர்கால இடைநிலை
ஆள் – படர்க்கைப் பெண்பால் வினைமுற்று விகுதி

2. நடக்கின்றான் – நட + க் + கின்று + ஆன்
நட – பகுதி
க் – சந்தி
கின்று – நிகழ்கால இடைநிலை
ஆன் – படர்க்கை ஆண்பால் வினைமுற்று விகுதி

பின்வரும் சொற்களைப் பிரித்துப் பகுபத உறுப்புகளை எழுதுக

1. பார்த்தான் – பார் + த் + த் + ஆன்
பார் – பகுதி
த் – சந்தி
த் – இறந்தகால இடைநிலை
ஆன்- படர்க்கை ஆண்பால் வினைமுற்று விகுதி

2. பாடுவார் – பாடு + வ் + ஆர்
பாடு – பகுதி
வ் – எதிர்கால இடைநிலை
ஆர் – படர்க்கைப் பலர்பால் வினைமுற்று விகுதி

குறுவினா

Question 1.
ஓரெழுத்து ஒருமொழி என்றால் என்ன?
Answer:
(i) ஓரெழுத்து தனித்து நின்று பொருள் தரும் சொல்லாக அமைவதே ஓரெழுத்து ஒரு மொழி ஆகும்.
(ii) எ.கா. (தீ, நீ, வா, போ).

Question 2.
பதத்தின் இரு வகைகள் யாவை?
Answer:
பதம் இரண்டு வகைப்படும். அவை, பகுபதம், பகாப்பதம்.

Question 3.
பகுபத உறுப்புகள் எத்தனை வகைப்படும்?
Answer:
அவை யாவை? பகுபத உறுப்புகள் ஆறு வகைப்படும். அவை, பகுதி, விகுதி, இடைநிலை, சந்தி, சாரியை, விகாரம்.

சிறுவினா

Question 1.
விகுதி எவற்றைக் காட்டும்?
Answer:
சொல்லின் இறுதியில் நிற்கும் உறுப்பே விகுதி ஆகும். இது திணை, பால், எண், இடம்,
முற்று, எச்சம் போன்றவற்றைக் காட்டும். (எ.கா.) படித்தான் = ஆன் – விகுதி

Question 2.
விகாரம் என்பது யாது? எடுத்துக்காட்டுடன் விளக்குக.
Answer:
பகுதி, விகுதி, சந்தி, இடைநிலை இவற்றில் ஏற்படும் மாற்றமே விகாரம் எனப்படும்.
(எ. கா.) வந்தான் = வா – பகுதி வா ‘வ’ எனக் குறுகியது விகாரம்.

Question 3.
பெயர்ப்பகுபதம் எத்தனை வகைப்படும்? அவை யாவை?
Answer:
பெயர்ப் பகுபதம் ஆறு வகைப்படும். அவை
(i) பொருள் – பொன்னன் (பொன் + அன்)
(ii) இடம் – நாடன் (நாடு + அன்)
(iii) காலம் சித்திரையான் (சித்திரை + ஆன்)
(iv) சினை கண்ண ன் (கண் + அன்)
(v) பண்பு இனியன் (இனிமை + அன்)
(vi) தொழில் – உழவன் (உழவு + அன்)

கற்பவை கற்றபின்

Question 1.
பாடப்பகுதியில் இடம்பெற்ற சொற்களில் பகுபதம், பகாப்பதம் ஆகியவற்றைக் கண்டறிந்து தனித்தனியே தொகுக்க.
Answer:
பகுபதம்
பெயர்ப்பகுபதம் :
பொருள் – பொன்னன் (பொன் + அன்)
இடம் – நாடன் (நாடு + அன்)
காலம் – சித்திரையான் (சித்திரை + ஆன்)
சினை – கண்ண ன் (கண் + அன் )
பண்பு – இனியன் (இனிமை + அன்)
தொழில் – உழவன் (உழவு + அன்)
வினைப்பகுபதம் : உண்கின்றான் – உண் + கின்று + ஆன்

பகாப்பதம் :
பெயர்ப் பகாப்பதம் – நிலம், நீர், நெருப்பு, காற்று
வினைப் பகாப்பதம் – நட, வா, படி, வாழ்.
இடைப் பகாப்பதம் – மன், கொல், தில், போல்
உரிப் பகாப்பதம் – உறு, தவ, நனி, கழி.

Question 2.
உங்கள் வகுப்பு மாணவ – மாணவிகளின் பெயர்களைப் பகுபதம், பகாப்பதம் என வகைப்படுத்துக.
Answer:
மாணவர்கள் தாங்களாகவே செய்ய வேண்டியவை.

கூடுதல் வினாக்கள்

நிரப்புக.

Question 1.
தே என்பதன் பொருள் ………….. எனப்படும்.
Answer:
கடவுள்

Question 2.
நன்னூல் என்னும் இலக்கண நூலை எழுதியர் ………….
Answer:
பவணந்தி முனிவர்

ஓரெழுத்து ஒரு மொழிகளும் அவற்றின் பொருளும்

மொழியை ஆள்வோம்

கேட்க.

Question 1.
சிறந்த கல்வியாளர்களின் சொற்பொழிவுகளை இணையத்தில் கேட்டு மகிழ்க.
Answer:
மாணவர்கள் தாங்களாகவே செய்ய வேண்டியவை

கீழ்க்காணும் தலைப்பில் இரண்டு நிமிடம் பேசுக

Question 1.
கல்வியின் சிறப்பு
Answer:
அனைவருக்கும் வணக்கம்!
நான் கல்வியின் சிறப்பு என்ற தலைப்பில் பேசவிருக்கிறேன். இன்றைய உலகின் தே இன்றியமையாத ஒன்றாகத் திகழ்வது யாதெனில் கல்வியே ஆகும். அனைவருக்கும் பகுத்தறிவு தேவைப்படுகிறது. அவர்களுக்குப் பகுத்தறியும் சக்தியைக் கொடுப்பதே கல்வியின் சிறப்பாகும்.

கல்வியின் சிறப்பு என்றாலே கல்வி கற்றவன் எங்குச் சென்றாலும் சிறப்பிக்கப்படுபவன் – என்பதே நினைவுக்கு வரும். அவனுக்கு எல்லா நாட்டினரும் உறவினர்கள் ஆவார்கள். எல்லா நாடும் சொந்த நாடாகும். கல்வி கற்கவில்லையெனில் வாழ்நாள் முழுவதும் அனைவராலும் அவமதிக்கப்படுவான். இதனையே வள்ளுவர்.

“யாதானும் நாடாமல் ஊர்ஆமால் என்னொருவன் சாந்துணையும் கல்லாத வாறு. என்று கூறுகிறார்.
அதுமட்டுமா? கற்றவரைக் கண்ணுடையார்’ என்றும் கல்லாதவரை முகத்தில் இரண்டு புண்ணுடையவர் என்றும் இடித்துரைக்கிறார் வள்ளுவர்.

கற்றோர்க்கு அணிகலன் கல்வியே; கற்றோரே கண்ணுடையவர்; கற்றாரே தேவர் எனப் போற்றப்படத்தக்கவர்; கற்றோரே மேலானவர் என்பதை அனைவரும் உணர வேண்டும்.
“கல்வி வந்தது எனில் கடைத்தேறிற்று உலகே!” என்று புரட்சிக்கவி கூறுகின்றார். கல்வியால் எல்லா வளங்களும் கிடைக்கும் என்பதே இதன் பொருளாகும்.

கல்வி உடையவர் எல்லா மக்களிடமும், நன்றாக பழகிக் கொள்வதோடு மட்டுமல்லாமல் அவர்களுடன் மகிழ்ச்சியாக சேர்ந்து வாழ்வதையே விரும்புவர்.

மனிதன் வாழ்நாள் முழுவதும் கற்றுக் கொண்டே இருக்க வேண்டும். கற்க மறுப்பவன் வாழ மறுப்பவன் ஆகின்றான். கல்வி என்னும் விளக்கால் வாழ்க்கையில் எதிர்ப்படும் இருள்களையெல்லாம் நீக்க முடியும். கல்வி போல மனப்பயத்தைப் போக்கும் மருந்து வேறொன்றுமில்லை. கல்வித் துணை வறுமையில் கை கொடுக்கும். கல்வியின் பயனே மனித வாழ்வின் பெரும்பேறாகும்.

கல்வி, தொழிலுக்கு வழி காட்டும். கல்வி என்பது வாழ்வதற்கு உதவும் கருவியாகும். வாழ்க்கையின் வெற்றிக்குக் கல்வி மிகவும் இன்றியமையாததாகிறது. வாழ்க்கையை நெறிப்படுத்தவும் மேம்படுத்தவும் கல்வி பயன்படுகிறது. கல்வி கற்ற பண்பு, நீதி, நேர்மை இவைகள் அனைத்தும் ஒருங்கே அமைந்து காணப்படும்.

கல்வியினால் மட்டுமே உலக அறிவினை வளர்த்துக் கொள்ள முடியும். உலகை முழுமையாகப் படிக்கவும் முடியும். கல்வி மனிதனுக்கு ஓர் உன்னதமான தேவையாகும்”
“கற்கை நன்றே! கற்றை நன்றே! பிச்சைப் புகினும் கற்கை நன்றே!” என்ற கூற்றினை மனதில் நிறுத்தி அள்ள அள்ளக் குறையாதக் கல்வியை அள்ளிப் பருகுவோம்.
கல்வி என்பது பலமே !
கற்றல் என்பது சுகமே!

Question 2.
குழந்தைத் தொழிலாளர் முறை ஒழிப்பு
Answer:
ஒரு குழந்தை கூலிக்காக வேலை பார்ப்பது மிகவும் தவறு. இளமைக் காலம் கல்வி கற்பதற்கே. குழந்தைத் தொழிலாளர்கள் இல்லாத நிலை உருவாக வேண்டும்.
பள்ளி செல்லாத குழந்தைகள், குழந்தைத் தொழிலாளர் ஆகிறார்கள். இது ஒரு பக்கம் இருக்க; வேலைவாய்ப்புக்காக இடம்பெயர்ந்து மாநிலம் விட்டு மாநிலம், மாவட்டம் விட்டு மாவட்டம் என்று செல்கின்றனர். இவ்விரண்டு நிலைகளில் ஒரு சிலர் தங்கள் பெற்றோருடன் சேர்ந்து வேலை செய்வார்கள் அல்லது தனியாகச் செல்வார்கள்.

மூன்றாவது நிலையில் உள்ளவர்கள் மிகவும் பரிதாபத்துக்குரியவர்கள். கொத்தடிமை முறையில் வேலைக்குச் செல்கிறார்கள். பெற்றோர் வாங்கிய கடனை ஈடுகட்டுவதற்காக பிள்ளைகள் வேலைக்குச் செல்கிறார்கள்.

குழந்தைத் தொழிலாளர்கள் உடல் ரீதியாகப் பாதிக்கப்படுகிறார்கள். உளவியல் ரீதியான பாதிப்பு, உணர்வு மற்றும் சமூக ரீதியான பாதிப்பு ஏற்படுகிறது.

கொடிய வறுமை, ஊட்டச்சத்துக் குறைவு, கல்வியறிவு பெற முடியாத நிலை, உடல் நலனைப் பாதிக்கக் கூடிய ஆபத்தான சூழல், காற்றோட்டம் இல்லாத குறுகிய அறை போன்றவை குழந்தைகளின் உடல் நலனைப் பெரிதும் பாதிப்பதால் ஆஸ்துமா, காசநோய் போன்ற நோய்கள் தாக்குகிறது. இதனைத் தடுக்க வேண்டும்.

குழந்தைகளுக்குத் தொடக்கக் கல்வி கட்டாயம் ஆக்கப்பட வேண்டும். பள்ளி மற்றும் கல்லூரிகளில் குழந்தைத் தொழிலாளர் ஒழிப்பு பற்றிய விழிப்புணர்வு முகாம்களை நடத்தலாம்.

சொல்லக் கேட்டு எழுதுக

1. இளமைப் பருவத்திலேயே கல்வி கற்க வேண்டும்.
2. கல்வியே அழியாத செல்வம்.
3. கல்வி இல்லாத நாடு விளக்கு இல்லாத வீடு.
4. பள்ளித்தலம் அனைத்தும் கோயில் செய்குவோம்.
5. நூல்களை ஆராய்ந்து ஆழ்ந்து படிக்க வேண்டும்.

கீழ்க்காணும் சொற்களை அறுவகைப் பெயர்களாக வகைப்படுத்துக

நல்லூர், வடை, கேட்டல், முகம், அன்னம், செம்மை, காலை, வருதல், தோகை, பாரதிதாசன், பள்ளி, இறக்கை, பெரியது, சோலை, ஐந்து மணி, விளையாட்டு, புதன்

Answer:

அறிந்து பயன்படுத்துவோம்

மூவிடம் :

இடம் மூன்று வகைப்படும். அவை 1. தன்மை, 2. முன்னிலை, 3. படர்க்கை
தன்னைக் குறிப்பது தன்மை.
(எ.கா.) நான், நாம், நாங்கள், என், எம், எங்கள்.
முன்னால் இருப்பவரைக் குறிப்பது முன்னிலை.
(எ.கா.) நீ , நீங்கள், நீர், நீவிர், உன், உங்கள்.
தன்னையும், முன்னால் இருப்பவரையும் அல்லாமல் மூன்றாமவரைக் குறிப்பது – படர்க்கை .
(எ.கா.) அவன், அவள், அவர், அவர்கள், அது, அவை, இவன், இவள், இவை.

சரியான சொல்லைக் கொண்டு நிரப்புக

(அது, நீ, அவர்கள், அவைகள், அவை, நாம், உன்)

Question 1.
……… பெயர் என்ன ?
Answer:
உன்

Question 2.
ஏழாம் வகுப்பு மாணவர்கள்.
Answer:
நாம்

Question 3.
………….. எப்படி ஓடும்?
Answer:
அது

Question 4.
………………என்ன செய்து கொண்டிருக்கிறாய்?
Answer:
நீ

Question 5.
…. வந்து கொண்டு இருக்கிறார்கள்.
Answer:
அவர்கள்

பின்வரும் தொடர்களில் மூவிடப் பெயர்களை அடிக்கோடிடுக. அவற்றை வகைப்படுத்துக.

1. எங்கள் வீட்டு நாய்க்குட்டி ஓடியது.
2. இவர்தான் உங்கள் ஆசிரியர்.
3. நீர் கூறுவது எனக்குப் புரியவில்லை
4. எனக்கு, அது வந்ததா என்று தெரியவில்லை , நீயே கூறு.
5. உங்களோடு நானும் உணவு உண்ணலாமா?

Answer:

கடிதம் எழுதுக

உங்கள் பகுதியில் நூலகம் ஒன்று அமைத்துத் தர வேண்டி நூலக ஆணையருக்குக் கடிதம் எழுதுக.
Answer:
அனுப்புநர் :
ஊர்ப் பொதுமக்கள்,
மறைமலை நகர்,
காஞ்சிபுரம் மாவட்டம்

பெறுநர் :
நூலக ஆணையர்,
பொதுநூலகத் துறை,
சென்னை – 600 002.

மதிப்பிற்குரிய ஐயா,
பொருள் : நூலகம் அமைத்துத் தர வேண்டுதல் தொடர்பாக

எங்கள் ஊர் மறைமலைநகர். இங்கு இரண்டாயிரம் பேருக்கு மேல் வாழ்கிறோம். பள்ளி மாணவர்கள், கல்லூரி மாணவர்கள், போட்டித் தேர்வுக்குப் படிக்கும் இளைஞர்கள், அன்றாடச் செய்தியை அறிந்து கொள்ளும் ஆர்வலர், பணி ஓய்வு பெற்றவர்கள் எனப் பலரும் உள்ளனர்.

அவரவர்களுக்குத் தேவையான நூல்கள், செய்தித்தாள்கள், இதழ்கள் போன்றவை இங்குக் கிடைப்பதற்கரிதாக உள்ளது. இவர்கள் அனைவரும் பயன்பெறும் வகையில் நூலகம் ஒன்றை எங்கள் ஊரில் அமைத்துத்தர ஆவன செய்யுமாறு கேட்டுக் கொள்கிறோம்.

நன்றி!
இடம் : மறைமலை நகர்,
தேதி : 5-2-2020

இப்படிக்கு ,
தங்கள் உண்மையுள்ள,
ஊர்ப் பொதுமக்கள்

மொழியோடு விளையாடு

கீழே உள்ள குறிப்புகளைப் பயன்படுத்திக் கட்டத்தில் எழுத்துகளை நிரப்புக.
1. காலையில் பள்ளி மணி……………………
2. திரைப்படங்களில் விலங்குகள் ……………………… காட்சி குழந்ை தகளுக்குப் பிடிக்கும்.
3. கதிரவன் காலையில் கிழக்கே ………………
4. நாள்தோறும் செய்தித்தாள் ………………. வழக்கம் இருக்க வேண்டும்.

Answer:

1. காலையில் பள்ளி மணி அடிக்கும்.
2. திரைப்படங்களில் விலங்குகள் நடிக்கும் காட்சி குழந்ை தகளுக்குப் பிடிக்கும்.
3. கதிரவன் காலையில் கிழக்கே உதிக்கும்.
4. நாள்தோறும் செய்தித்தாள் படிக்கும் வழக்கம் இருக்க வேண்டும்.

ஓர் எழுத்துச் சொற்களால் நிரப்புக

Question 1.
…………………… புல்லை மேயும்.
Answer:
ஆ

Question 2.
……………………… சுடும்.
Answer:
தீ

Question 3.
……………….. பேசும்.
Answer:
நா

Question 4.
…………………… பறக்கும்.
Answer:
ஈ

Question 5.
…………….. மணம் வீசும்
Answer:
பூ

பின்வரும் எழுத்துகளுக்குப் பொருள் எழுதுக

(எ.கா.) தா – கொடு

Answer:

1. தீ – நெருப்பு
2. பா – பாடல்
3. தை – தை மாதம்
4. வை – புல், வைக்கோல்
5. மை – அஞ்சனம்

பின்வரும் சொற்களை இருபொருள் தருமாறு தொடரில் அமைத்து எழுதுக

(ஆறு, விளக்கு, படி, சொல், கல், மாலை, இடி)
(எ.கா.) ஆறு – ஈ ஆறு கால்களை உடையது.
தஞ்சாவூரில் காவிரி ஆறு பாய்கிறது.

விளக்கு : இலக்கணப் பாடத்தை விளக்கிக் கூறு.
அறியாமை என்னும் இருளைப் போக்குவது கல்வி என்னும் விளக்கு.

படி : காலையில் தினமும் படி.
மாடிப்படி ஏறி வா.

சொல் : சொற்கள் சேர்ந்தால் பாமாலை.
பெரியோர் சொல் கேட்டு சிறியோர் நடக்க வேண்டும்.

கல் : கற்களால் ஆனது கோபுரம்.
இளமையில் கல்.

மாலை : நேற்று மாலை பூங்காவிற்குச் சென்றேன்.
பூ மாலை நல்ல மணம் வீசியது.

இடி இடிக்கும் சப்தம் கேட்டது.
தவறுகளைக் கண்டால் இடித்துரைத்தல் வேண்டும்.

நிற்க அதற்குத் தக

என் பொறுப்புகள்

1. பாடப்புத்தகங்கள் மட்டுமன்றிப் பிற புத்தகங்களையும் படிப்பேன்.
2. பெற்றோர், ஆசிரியர், மூத்தோர் இவர்களை எப்போதும் மதித்து நடப்பேன்.

கலைச்சொல் அறிவோம்

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

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Samacheer Kalvi 11th Chemistry Chapter 3 Periodic Classification of Elements Textual Evaluation Solved

Choose The Correct Answer from The Following

11th Chemistry Chapter 3 Book Back Answers Question 1.
What would be the IUPAC name for an element with atomic number 222?
(a) bibibiium
(b) bididium
(c) didibium
(d) bibibium
Answer:
(d) bibibium

11th Chemistry Lesson 3 Book Back Answers Question 2.
The electronic configuration of the elements A and B are 1s², 2s², 2p⁶, 3s² and 1s², 2s², 2p⁵, respectively. The formula of the ionic compound that can be formed between these elements is ……….
(a) AB
(b) AB₂
(c) A₂B
(d) none of the above.
Answer:
(a) AB₂

11th Chemistry Unit 3 Book Back Answers Question 3.
The group of elements in which the differentiating electron enters the anti-penultimate shell of atoms are called –
(a) p-block elements
(b) d-block elements
(c) s-block elements
(d) f-block elements
Answer:
(d) f-block elements

11th Chemistry 3rd Lesson Answers Question 4.
In which of the following options the order of arrangement does not agree with the variation of property indicated against it? (NEET 2016 Phase 1)
(a) I < Br < Cl < F (increasing electron gain enthalpy)
(b) Li < Na < K < Rb (increasing metallic radius)
(c) Al³⁺< Mg²⁺< Na⁺ < F^– (increasing ionic size)
(d)  B < C < O < N (increasing first ionization enthalpy)
Answer:
(a) I < Br < Cl < F (increasing electron gain enthalpy)

11th Chemistry 3rd Lesson Book Back Answers Question 5.
Which of the following elements will have the highest electro negativity?
(a) Chlorine
(b) Nitrogen
(c) Cesium
(d) Fluorine
Answer:
(d) Fluorine

Question 6.
Various successive ionization enthalpies (in kJ mol^-1) of an element are given below. The element is ………….

(a) phosphorus
(b) sodium
(c) aluminium
(d) silicon table
Answer:
(c) aluminium

11th Chemistry 3rd Lesson Question 7.
In the third period, the first ionization potential is of the order …………..
(a) Na > Al > Mg > Si > P
(b) Na < Al < Mg < Si < P
(c) Mg > Na > Si > P > Al
(d) Na< Al < Mg < Si < P
Answer:
(b) Na < Al < Mg < Si < P

11th Chemistry Chapter 3 Question 8.
Identify the wrong statement ……………..
(a) Among st the iso electronic species, smaller the positive charge on cation, smaller is the ionic radius
(b) Among-st iso electric species greater the negative charge on the anion, larger is the ionic radius
(c) Atomic radius of the elements increases as one moves down the first group of the periodic table
(d) Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table.
Answer:
(a) Among-st the iso electronic species, smaller the positive charge on cation, smaller is the ionic radius

Samacheer Kalvi Guru 11th Chemistry Question 9.
Which one of the following arrangements represent the correct order of least negative to most negative electron gain enthalpy?
(a) Al< O<C< Ca< F
(b) Al < Ca<O< C< F
(c) C < F < O < Al < Ca
(d) Ca < Al < C < O < F
Answer:
(d) Ca < Al < C < O < F

Samacheer Kalvi Class 11 Chemistry Solutions Question 10.
The correct order of electron gain enthalpy with negative sign of F, Cl, Br and I having atomic number 9, 17, 35 and 53, respectively is ………..
(a) J > Br > Cl >F
(b) F > Cl > Br >I
(c) Cl > F > Br >I
(d) Br > I > Cl > F
Answer:
(c) Cl > F > Br > I

Samacheer Kalvi 11 Chemistry Solutions Question 11.
Which one of the following is the least electro negative element?
(a) Bromine
(b) Chlorine
(c) Iodine
(d) Hydrogen
Answer:
(d) Hydrogen.
Solution:
Hydrogen is the least electro negative element. Since electro negativity increases across the period from left to right. Hydrogen is the first element and it has less electro negativity and down the group electro negativity decreases.

Samacheer Kalvi 11th Chemistry Chapter 1 Solutions Question 12.
The element with positive electron gain enthalpy is ……….
(a) hydrogen
(b) sodium
(c) argon
(d) fluorine
Answer:
(c) argon
Solution:
Argon has completely filled configuration. So addition of the electron is not possible and has positive electron gain enthalpy.

11th Chemistry 3rd Chapter Question 13.
The correct order of decreasing electro negativity values among the elements X, Y, Z and A with atomic numbers 4, 8, 7 and 12 respectively –
(a) Y > Z > X > A
(b) Z > A > Y > X
(c) X > Y > Z > A
(d) X > Y >A >Z
Answer:
(a) Y > Z > X > A

Periodic Classification Of Elements Class 11 Question 14.
Assertion : Helium has the highest value of ionization energy among all the elements known Reason: Helium has the highest value of electron affinity among all the elements known –
(a) Both assertion and reason are true and reason is correct explanation for the assertion
(b) Both assertion and reason are true but the reason is not the correct explanation for the assertion
(c) Assertion is true and the reason is false
(d) Both assertion and the reason are false
Answer:
(c) Assertion is true and the reason is false

Periodic Classification Of Elements Class 11 Notes Pdf Question 15.
The electronic configuration of the atom having maximum difference in first and second ionization energies is ……….
(a) 1s², 2s², 2p⁶, 3s¹
(b) 1s², 2s², 2p⁶, 3s²
(c) 1s², 2s², 2p⁶, 3s², 3s², 3p⁶, 4s¹
(d) 1s², 2s², 2p⁶, 3s², 3p¹
Answer:
(a) 1s², 2s², 2p⁶, 3s¹

Chemistry Chapter 3 Answers Question 16.
Which of the following is second most electro negative element?
(a) Chlorine
(b) Fluorine
(c) Oxygen
(d) Sulphur
Answer:
(a) Chlorine

Periodic Classification Of Elements Class 11 Questions And Answers Question 17.
IE₁  and IE₂ of Mg are 179 and 348 k cal mol^-1 respectively. The energy required for the reaction
Mg → Mg²⁺ + 2e^– is ……..
(a) +169 kcal mol^-1
(b) -169 kcal mol^-1
(c) +527 kcal mol^-1
(d) -527 kcal mol^-1
Answer:
(c) +527 kcal mol^-1

Class 11 Chemistry Chapter 3 Important Questions With Answers Question 18.
In a given shell the order of screening effect is …………..
(a) s > p > d > f
(b) s > p > f > d
(c) f > d > p > s
(d) f > p > s > d
Answer:
(a) s > p > d > f

Samacheerkalvi.Guru 11th Chemistry Question 19.
Which of the following orders of ionic radii is correct?
(a) H^– > H⁺ > H
(b) Na⁺ > F“ > O^–
(c) F > O^2- > Na⁺
(d) None of these
Answer:
(d) None of these

Samacheer Kalvi.Guru 11th Chemistry Question 20.
The first ionization potential of Na, Mg and Si are 496, 737 and 786 kJ mol^-1 respectively. The ionization potential of Al will be closer to
(a) 760 kJ mol^-1
(b) 575 kJ mol^-1
(c) 801 kJ mol^-1
(d) 419 kJ mol^-1
Answer:
(b) 575 kJ mol^-1

Class 11 Chemistry Samacheer Solutions Question 21.
Which one of the following is true about metallic character when we move from left to right in a period and top to bottom in a group?
(a) Decreases in a period and increases along the group
(b) Increases in a period and decreases in a group
(c) Increases both in the period and the group
(d) Decreases both in the period and in the group
Answer:
(a) Decreases in a period and increases along the group

11th Chemistry Solutions Samacheer Kalvi Question 22.
How does electron affinity change when we move from left to right in a period in the periodic table?
(a) Generally increases
(b) Generally decreases
(c) Remains unchanged
(d) First increases and then decreases
Answer:
(a) Generally increases.

Samacheer Kalvi Guru 11 Chemistry Question 23.
Which of the following pairs of elements exhibit diagonal relationship?
(a) Be and Mg
(b) Li and Mg
(c) Be and B
(d) Be and Al
Answer:
(d) Be and Al

II. Write brief answer to the following questions

Samacheer Kalvi 11th Chemistry Solution Question 24.
Define modern periodic law.
Answer:
The modem periodic law states that, “The physical and chemical properties of the elements are periodic function of their atomic numbers.”

Question 25.
What are isoelectronic ions? Give examples.
Answer:
There are some ions of different elements having the same number of electrons are called isoelectronic ions.
Example:
Na⁺, Mg²⁺, Al³⁺, F^– , O^2- and N^3-

Question 26.
What is effective nuclear charge?
Answer:
The net nuclear charge experienced by valence electrons in the outermost shell is called the effective nuclear charge.
Z_eff = Z – S
Where,
Z = Atomic number
S = Screening constant calculated by using Slater’s rules.

Question 27.
Is the definition given below for ionization enthalpy is correct?
“Ionization enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom”
Answer:
No. It is not correct. The accurate and absolute definition is as follows:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 28.
Magnesium loses electrons successively to form Mg⁺, Mg²⁺ and Mg³⁺ ions. Which step will have the highest ionization energy and why?
Answer:

• The third step will have the highest ionization energy. I.E₃>I.E₂>I.E₁
• Because from a neutral gaseous atom, the electron removal is easy and less amount of energy is required. But from a di positive cation, there will be more number of protons than the electrons and there is more forces of attraction between the nucleus and electron. So the removal of electron in a di positive cation, becomes highly difficult and more energy is required.

Question 29.
Define electro negativity.
Answer:
Electro negativity is the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

Question 30.
How would you explain the fact that the second ionization potential is always higher than first ionization potential?
Answer:

• Second ionization potential is always higher than first ionization potential.
• Removal of one electron from the valence orbit of a neutral gaseous atom is easy so first ionization energy is less. But from a uni positive ion, removal of one more electron becomes difficult due to the more forces of attraction between the excess of protons and less number of electrons.
• Due to greater nuclear attraction, second ionization energy is higher than first ionization energy.

Question 31.
Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10^-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of kJ mol^-1.
Answer:
Energy of an electron in the ground state of the hydrogen atom = -2.18 x 10^-18 J
H → H⁺ + e^–
Energy required to ionize 1 mole of hydrogen atoms, we multiply by the Avogadro constant.
E = 2.18 x 10^-18 x 6.023 x 10²³
= 13.123 x 10⁵ J mol^-1
I.E = +1312 K J mol^-1

Question 32.
The electronic configuration of an atom is one of the important factor which affects the value of ionization potential and electron gain enthalpy. Explain.
Answer:

• Electronic configuration of an atom affects the value of ionization potential and electron gain enthalpy.
• Half filled valence shell electronic configuration and completely filled valence shell electronic configuration are more stable than partially filled electronic configuration.
• For e.g. Beryllium (Z = 4) 1s² 2s² (completely filled electronic configuration)
  Nitrogen (Z = 7) 1s²  2s²  2p_x¹  2p_y¹ 2p_z¹ (half filled electronic configuration) Both beryllium and nitrogen have high ionization energy due to more stable nature.
• In the case of beryllium (1s² 2s²), nitrogen (1s² 2s² 2p³) the addition of extra electron will disturb their stable electronic configuration and they have almost zero electron affinity.
• Noble gases have stable ns² np⁶ configuration and the addition of further electron is unfavorable and they have zero electron affinity.

Question 33.
In what period and group will an element with Z = 118 will be present?
Answer:
The element with atomic number Z = 118 is present in 7^th period and 18^th group.

Question 34.
Justify that the fifth period of the periodic table should have 18 elements on the basis of quantum numbers.
Answer:
Fifth period of the periodic table have 18 elements. 5^th period starts from Rb to Xe (18 elements). 5^th period starts with principal quantum number n = 5 and 1 = 0, 1,2,3 and 4. When n = 5, the number of orbitals = 9.
1 for 5s
5 for 4d
3 for 5p
Total number of orbitals = 9.
Total number of electrons that can be accommodated in 9
orbitals = 9 x 2 = 18. Hence the number of elements in 5th period is 18.

Question 35.
Elements a, b, c and d have the following electronic configurations:
a : 1s², 2s², 2p⁶
b : 1s², 2s², 2p⁶, 3s², 3p¹
c : 1s², 2s², 2p⁶ 3s²,3p⁶
d : 1s², 2s², 2p¹
Which elements among these will belong to the same group of periodic table?
Answer:

1. 
2. In the above elements, Ne and Ar belong to same group (Noble gases – 18th group).
3. Al and B belong to the same group (13^th group).

Question 36.
Give the general electronic configuration of lanthanides and actinides?
Answer:

• The electronic configuration of lanthanides is 4f^1-14 5d^0-16s².
• The electronic configuration of actinides is 5f^1-14 6d^0-1 7s².

Question 37.
Why halogens act as oxidizing agents?
Answer:
Halogens act as oxidizing agents. Their electronic configuration is ns² np⁵. So all the halogens are ready to gain one electron to attain the nearest inert gas configuration. An oxidizing agent is the one which is ready to gain an electron. So all the halogens act as oxidizing agents. Also halogens are highly electro negative with low dissociation energy and high negative electron gain enthalpies. Therefore, the halogens have a high tendency to gain an electron. Hence they act as oxidizing agents.

Question 38.
Mention any two anomalous properties of second period elements.
Answer:

• In the 1^st group, lithium forms compounds with more covalent character while the other elements of this group form only ionic compounds.
• In the 2^nd group, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.

Question 39.
Explain the Pauling’s method for the determination of ionic radius.
Answer:
1. Ionic radius is defined as the distance from the center of the nucleus of the ion up-to which it exerts its influence on the electron cloud of the ion.
2. Ionic radius of uni-univalent crystal can be calculated from the inter-ionic distance between the nuclei of the cation and anion.
3. Pauling assumed that ions present in a crystal lattice are perfect spheres and they are in contact with each other, therefore
d = r_C⁺ + r_A^– ………(1)
Where, d = distance between the center of the nucleus of cation C+ and the anion A-
r_C⁺ = radius of cation
r_A^– = radius of anion.
4. Pauling assumed that the radius of the ion having noble gas configuration (Na⁺ and F^– having 1s², 25², 2p⁶ configuration) is inversely proportional to the effective nuclear charge felt at the periphery of the ion.

Where Z_eff is the effective nuclear charge
Z_eff = Z – S
5. Dividing the equation (2) by (3)

On solving equation (1) and (4), the values of r_C⁺ and r_A^– can be obtained.

Question 40.
Explain the periodic trend of ionization potential.
Answer:
(a) The energy required to remove the most loosely held electron from an isolated gaseous atom is called as ionization energy.
(b) Variation in a period:
Ionization energy is a periodic property. On moving across a period from left to right, the ionization enthalpy value increases. This is due to the following reasons.

• Increase of nuclear charge in a period
• Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus. Therefore, ionization enthalpy increases.

(c) Variation in a group:
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons.

• A gradual increase in atomic size
• Increase of screening effect on the outermost electrons due to the increase of number of inner electrons.

Hence, ionization enthalpy is a periodic property.

Question 41.
Explain the diagonal relationship.
Answer:

• On moving diagonally across the periodic table, the second and the third period elements show certain similarities.
• Even though the similarity is not same as we see in a group, it is quite pronounced in the following pair of elements.
• 
• The similarity in properties existing between the diagonally placed elements is called “diagonal relationship”.

Question 42.
Why the first ionization enthalpy of sodium is lower than that of magnesium while its second ionization enthalpy is higher than that of magnesium?
Answer:
The 1st ionization enthalpy of magnesium is higher than that of Na due to higher nuclear charge and slightly smaller atomic radius of Mg than Na. After the loss of first electron, Na⁺ formed has the electronic configuration of neon (2,8). The higher stability of the completely filled noble gas configuration leads to very high second ionization enthalpy for sodium. On the other hand, Mg⁺ formed after losing first electron still has one more electron in its outermost (3 s) orbital. As a result, the second ionization enthalpy of magnesium is much smaller than that of sodium.

Question 43.
By using Pauling’s method calculate the ionic radii of K⁺ and Cl^– ions in the potassium chloride crystal. Given that \(\mathrm{d}_{\mathrm{K}}+_{-} \mathrm{cl}^{-}\) = 3.14 Å
Answer:
Given

We know that,

(Z_eff)_Cl^– = Z – S
= 17 – [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 17 – 11.25 = 5.75
(Z_eff)_K⁺ = Z – S
= 19 – [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 19 – 11.25 = 7.75

r_(K⁺) = 0.74 r_(Cl^–)
Substitute the value of r_(K⁺) in equation (1)
0.74 r_(Cl^–) + r_(Cl^–) = 3.14 Å
1.74 r_(Cl^–) = \(\frac {3.14 Å}{1.74}\) = 1.81 Å.

Question 44.
Explain the following, give appropriate reasons.

1. Ionization potential of N is greater than that of O
2. First ionization potential of C-atom is greater than that of B-atom, where as the reverse is true for second ionization potential.
3. The electron affinity values of Be, Mg and noble gases are zero and those of N (0.02 eV) and P (0.80 eV) are very low
4. The formation of F^– (g) from F(g) is exothermic while that of O^2-(g) from O (g) is endothermic.

Answer:
1. N (Z = 7) 1s² 2s² 2p_x¹ 12p_y¹ 2p_z¹. It has exactly half filled electronic configuration and it is more stable. Due to stability, ionization energy of nitrogen is high.
O (Z = 8) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹. It has incomplete electronic configuration and it requires less ionization energy.
I.E₁ N > I.E₁O

2. C (Z = 6) 1s² 2s² 2p_x¹ 2p_y¹. The electron removal from p orbital is very difficult. So carbon has highest first ionization potential. B (Z = 5) 1s² 2s² 2p¹. In boron nuclear charge is less than that of carbon, so boron has lowest first ionization potential.
I.E₁ C > I.E₁ B
But it is reverse in the case of second ionization energy. Because in case of B+ the electronic configuration is 1s² 2s², which is completely filled and it has high ionization energy. But in C+ the electronic configuration is 1s² 2s² 2p¹, one electron removal is easy so it has low ionization energy.
I.E₂ B > I.E₂ C

3. Be (Z = 4) 1s² 2s²
Mg (Z = 12) 1s² 2s² 2p⁶ 3s²
Noble gases has the electronic configuration of ns² np⁶. All these are completely filled and are more stable. For all these elements Be, Mg and noble gases, addition of electron is unfavorable and so they have zero electron affinity.

Nitrogen (Z = 7) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹. It has half filled electronic configuration. So addition of electron is unfavorable and it has very low electron affinity value of 0.02 eV. Phosphorus (Z = 15) 1s² 2s² 2p⁶ 3s² 3p_x¹ 3p_y¹ 3p_z¹. It also has half filled electronic configuration. Due to the symmetry and more stability, it has very low electron affinity value of 0.80 eV.

4. F_(g) + e^– → F_(g)^– exothermic
F (Z = 9) 1s² 2s² 2p⁵. It is ready to gain one electron to attain the nearest inert gas configuration. By gaining one electron, energy is released, so it is an exothermic reaction.
O_(g) + 2e^– → O^2-_(g) endothermic
O (Z = 8) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹. It is the small atom with high electron density. The first electron affinity is negative because energy is released in the process of adding one electron to the neutral oxygen atom. Second electron affinity is always endothermic (positive) because the electron is added to an ion which is already negative, therefore it must overcome the repulsion.

Question 45.
What is screening effect? Briefly give the basis for Pauling’s scale of electro negativity. Screening effect:
Answer:
The repulsive force between inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect . is called shielding effect (or) screening effect.

Pauling’s scale:

• Electro negativity is the relative tendency of an element present in a covalently bonded molecule to attract the shared pair of electrons towards itself.
• Pauling assigned arbitrary value of electronegativities for hydrogen and fluorine as 2.2 and 4, respectively.
  • Based on this the electronegativity values for other elements can be calculated using the following expression.
    (X_A-X_B) = 0.182 √E_AB – (E_AA E_BB)
    Where E_AB , E_AA and E_BB are the bond dissociation energies of AB, A₂ and B₂ molecules respectively.
    X_A and X_B are electronegativity values of A and B.

Question 46.
State the trends in the variation of electro negativity in period and group.
Answer:
Variation of electron negativity in a period:
The electro negativity increases across a period from left to right. Since the atomic radius decreases in a period, the attraction between the valence electron and the nucleus increases. Hence the tendency to attract shared pair of electrons increases. Therefore, electro negativity increases in a period.

Variation of electro negativity in a group:
The electro negativity decreases down a group. As we move down a group, the atomic radius increases and the nuclear attractive force on the valence electron decreases. Hence electro negativity decreases in a group.

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements  In-Text Question – Evaluate your self

Question 1.
What is the basic difference in approach between Mendeleev’s periodic table and modern periodic table?
Answer:
The main basic difference between Mendeleev’s periodic table and modem periodic table is that first one is constructed on the basis of atomic weight and the later is constructed on the basis of atomic number.

Question 2.
The element with atomic number 120 has not been discovered so far. What would be the IUPAC name and the symbol for this element? Predict the possible electronic configuration of this element.
Answer:
Atomic number : 120
IUPAC temporary symbol : Unbinilium
IUPAC temporary symbol : Ubn
Possible electronic configuration : [Og] 8s²

Question 3.
Predict the position of the element in periodic table satisfying the electronic configuration (n – 1 )d² ns² where n = 5?
Answer:
Electronic Configuration : (n – 1 )d² ns²
for n = 5, the electronic configuration is,
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d² 5s²
Atomic number : 40
4th group 5th period (d block element) = Zirconium

Question 4.
Using Slater’s rule calculate the effective nuclear charge on a 3p electron in aluminium and chlorine. Explain how these results relate to the atomic radii of the two atoms.
Answer:
Electronic Configuration of Aluminium

Effective nuclear charge = Z – S = 13 – 9.5
(Z_eff)_Al = 3.5
Electronic Configuration of chlorine

Effective nuclear charge = Z- S = 17 – 10.9
(Z_eff)_Cl = 6.1
(Z_eff)_Cl > (Z_eff)_Cl and hence r_Cl< r_Al

Question 5.
A student reported the ionic radii of iso electronic species X³⁺ , Y²⁺ and Z^– as 136 pm, 64 pm and 49 pm respectively. Is that order correct? Comment.
Answer:
X³⁺, Y²⁺, Z^– are iso electronic.
∴ Effective nuclear charge is in the order
(Z_eff)_Cl^– < (Z_eff)Y_Y²⁺ < (Z_eff)_X³⁺ and hcnce, ionic radii should be in the order r_Z^– > r_Y²⁺ > r_X³⁺
∴ The correct values are:

Question 6.
The first ionisation energy (IE₁) and second ionisation energy (IE₂) of elements X, Y and Z are given below.

Which one of the above elements is the most reactive metal, the least reactive metal and a noble gas?
Answer:
Noble gases:
Ioniation energy ranging from 2372 KJmol^-1 to 1037 kJ mol^-1. For element X, the IE₁ value is in the range of noble gas, moreover for this element both IE₁ and IE₂ are higher and hence X is the noble gas. For Y, the first ionisation energy is low and second ionisation energy is very high and hence Y is most reactive metal.
For Z, both IE₁ and IE₂ are higher and hence it is least reactive.

Question 7.
The electron gain enthalpy of chlorine is 348 kJ mol^-1. How much energy in kJ is released when 17.5 g of chlorine is completely converted into Cl^– ions in the gaseous state?
Cl_(g)+ e^– → Cl^–_(g)
∆H = 348 kJ mol^-1
For one mole (35.5g) 348 kJ is released.
∴ For 17.5g chlorine,  energy leased.
∴ The amount of energy released = \(\frac {348}{2}\) = 174 kJ.

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements  Additional Questions

Question 1.
The chemical symbol of carbon and cobalt are
(a) Ca and CO
(b) Ca and Cl
(c) C and CO
(d) Cr and Cb
Answer:
(c) C and CO

Question 2.
Consider the following statements.
(i) The chemical symbol of nickel is N.
(ii) An element is a material made up of different kind of atoms.
(iii) The physical state of bromine is liquid.
Which of the above statement is/are not correct?
(a) (i) and (iiii)
(b) (iii) only
(c) (ii) and (iii)
(d) (i) and (ii)
Answer:
(d) (i) and (ii)

Question 3.
Match the list-I and list-II using the correct code given below the list.
List – I
A. Jewels
B. Bolts and cot
C. Table salt
D. Utensils

List – II
1. Sodium chloride
2. Copper
3. Gold
4. Iron

Answer:

Question 4.
The law of triads is not obeyed by
(a) Ca, Sr, Ba
(b) Cl, Br, I
(c) Li, Na, K
(d) Be, B, C
Answer:
(d) Be, B, C

Question 5.
The law of triads is obeyed by
(a) Fe, CO, Ni
(b) C, N, O
(c) He, Ne, Ar
(d) Al, Si, P
Answer:
(a) Fe, CO, Ni

Question 6.
Match the list-I and list-II using the code given below the list.
List-I
A. Law of triads
B. Law of octaves
C. First periodic law
D. Modem periodic law

List-II
1. Chancourtois
2. Henry Moseley
3. Newland
4. Johann Dobereiner

Answer:

Question 7.
Consider the following statements.
(i) In Chancourtois classification, elements differed from each other in atomic weight by 16 or multiples of 16 fell very nearly on the same vertical line.
(ii) Mendeleev’s periodic law is based on atomic weight.
(iii) Mendeleev listed the 117 elements known at that time and are arranged in the order of atomic numbers.
Which of the following statement is/are not correct?
(a) (i) only
(b) (ii) and (iii)
(c) (iii) only
(d) (i),(ii), (iii)
Answer:
(c) (iii) only

Question 8.
Which of the following elements were unknown at that time of Mendeleev?
(a) Na, Mg
(b) Fe, CO
(c) K, Cu
(d) Ga, Ge
Answer:
(d) Ga, Ge

Question 9.
Consider the following statements.
(i) Position of hydrogen could not be made clear.
(ii) Isotopes find correct place in Mendeleev’s periodic table.
(iii) Mendeleev’s periodic table could not explain the variable valencies of elements.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) and (iii)
(c) (ii) only
(d) (i), (ii), (iii)
Answer:
(c) (ii) only

Question 10.
According to modem periodic law, the physical and chemical properties of the elements are periodic functions of their
(a) atomic volume
(b) atomic numbers
(c) atomic weights
(d) valency
Answer:
(B) atomic numbers

Question 11.
Which period contain 32 elements?
(a) Period 1
(b) Period 4
(c) Period 5
(d) Period 6
Answer:
(d) Period 6

Question 12.
There are horizontal rows of the periodic table known as
(a) groups
(b) periods
(c) families
(d) chalcogens
Answer:
(b) periods

Question 13.
The shortest period contains elements.
(a) H, He
(b) Li, Be
(c) B, C
Answer:
(a) H, He

Question 14.
The longest form of periodic table was constructed by
(a) Dmitri Mendeleev
(b) Henry Moseley
(c) Lothar Meyer
(d) New lands
Answer:
(b) Henry Moseley

Question 15.
Match the list-I and list-II using the correct code given below the list.
List – I
Z = 100
Z = 101
Z = 102
Z = 103

List – II
1. Mendelevium
2. Lawrencium
3. Fermium
4. Nobelium

Answer:

Question 16.
Which one of the following is the first transition series?
(a) Sc
(b) Zn
(c) Ti
(d) Cu
Answer:
(a) Sc

Question 17.
Which period mostly include man made radioactive elements?
(a) 4^th period
(b) 7^th period
(c) 6^th period
(d) 31 period
Answer:
(b) 7^th period

Question 18.
Which one of the following is called halogen family?
(a) Group 17
(b) Group 16
(c) Group 1
(d) Group 2
Answer:
(a) Group 17

Question 19.
Group 16 constitutes family.
(a) halogen
(b) nobel gas
(c) chalcogen
(d) alkali metals
Answer:
(c) chalcogen

Question 20.
Consider the following statements.
(i) The valency of the elements increases from left to right in a period.
(ii) Valency decreases from 7 to I with respect to oxygen.
(iii) The metallic character of the elements decreases across a period.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) only
(c) (i) and (iii)
(d) (i) (ii) and (iii)
Answer:
(b) (ii) only

Question 21
Match the list – I and list -II using the correct code given below the list.
List – I
A. Li
B. Na
C. K
D. Cs

List – II
1. 2,8,8,1
2. 2,1
3. 2,8, 18, 18, 8, 1
4. 2,8, 1

Answer:

Question 22.
What will be the change in valency down the group in the periodic table?
(a) increases
(b) decreases
(c) remains same
(d) zero
Answer:
(c) remains same

Question 23.
Which one of the following is a metalloid?
(a) N
(b) P
(c) Bi
(d) Sb
Answer:
(d) Sb

Question 24.
Which one of the following is a metal?
(a) N
(b) Br
(c) Bi
(d) As
Answer:
(c) Bi

Question 25.
Match the list – I and list – II using the correct code given below the list.
List – I
A. Alkali metal
B. Alkaline earth metals
C. d-block elements
D. p-block elements

List – II
1. ns² np^1-6
2. ns¹
3. ns²
4. (n – 1)d^1-10 ns^0-2

Answer:

Question 26.
Consider the following statements.
(i) Oxidation character increases from left to right in a period.
(ii) Reducing character increases from left to right in a period.
(iii) Metallic character increases from left to right in a period.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (i), (ii), (iii)
Answer:
(c) (ii) and (iii)

Question 27.
The general electronic configuration of d-block elements is
(a) ns² nd^1-10 10
(b) (n-1)d^1-10 ns^0-2
(c) (n-2)d^1-10 (n – 1)^0-2
(d) ns²nd⁵
Answer:
(b) (n-1)d^1-10 ns^0-2

Question 28.
Consider the flowing statements.
(i) d-block elements show variable oxidate states.
(ii) Mostly d-block elements form colourless compounds.
(iii) Mostly d-block elements are diamagnetic due to paired electrons.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) only
(c) (i) and (ii)
(d) (ii) and (iii)
Answer:
(d) (ii) and (iii)

Question 29.
All the s – block and p-block elements excluding l8 group are called elements.
(a) representative
(b) transition
(c) inner – transition
(d) trans uranium
Answer:
(a) representative

Question 30.
Which of the following is the correct electronic configuration of noble gases?
(a) ns² np⁶ nd¹⁰
(b) ns² np⁵
(c) ns² np⁶
(d) ns² np³
Answer:
(c) ns² np⁶

Question 31.
Group numbers 13 to 12 in the periodic table are called …………..
(a) inner transition elements
(b) Representative elements
(c) synthetic elements
(d) transition elements
Answer:
(d) transition elements

Question 32.
Which one of the following is in solid state at room temperature?
(a) Bromine
(b) Mercury
(c) Bismuth
(d) Gallium
Answer:
(c) Bismuth

Question 33.
Which of the following is not a metalloid (or) semi-metal?
(a) Silicon
(b) Arsenic
(c) Germanium
(d) Sodium
Answer:
(d) Sodium

Question 34.
Which of the following metal is not in liquid state?
(a) Gallium
(b) Aluminium
(c) Mercury
(d) Calsium
Answer:
(b) Aluminium

Question 35.
Which of the following is not a periodic property?
(a) Atomic radius
(b) Ionization enthaphy
(c) Electron affinity
(d) Oxidation number
Answer:
(d)Oxidation number

Question 36.
Which of the following property increases as we go down the group in the periodic property?
(a) ionization energy
(b) Electro negativity
(c) Atomic radius
(d) Electron affinity
Answer:
(c) Atomic radius

Question 37.
The metallic radius of copper is ………………
(a) 0.99 Å
(b) 1.28 Å
(c) 1.98 Å
(d) 2.56 Å
Answer:
(5) 1.28 Å

Question 38.
Consider the following statements………………
(i) Atomic radius of elements increases with increase in atomic number as we go down the group.
(ii) Atomic radius of elements increases with increase in atomic number as we go across the period.
(iii) Atomic radius of elements decreases as we go from left to right in a period.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) and (ii)
(c) (ii) only
(d) (i) and (iii)
Answer:
(c) (ii) only

Question 39.
Which one of the following is not an iso electronic ion?
(a) Na⁺
(b) Mg²⁺
(c) Cl^–
(d) O^2-
Answer:
(c) Cl^–

Question 40.
Which one of the following is not an isoelectronic ion?
(a) Al³⁺
(b) N^3-
(c) Mg²⁺
(d) K⁺
Answer:
(d) K⁺

Question 41.
Which of the following possess almost same properties due to lanthanide contraction?
(a) Zr, HF
(b) Na, K
(c) Zn, Cd
(d) Ag. Au
Answer:
(a) Zr, HF

Question 42.
Consider the following statements.
(i) Ionization is always an exothermic process.
(ii) Ionization energies always increase in the order I.E₁> IE₂>I.E₃.
(iii) Ionization energy measurements are carried out with atoms in the solid state.
Which of the above statement is/are not correct?
(a) (i) only
(b) (i) and (ii)
(c) (iii) only
(d) (i), (ii) and (iii)
Answer:
(d) (i), (ii) and (iii)

Question 43.
Statement-I : Ionization enthalpy of Be is greater than that of 13.
Statement-II : The nuclear charge of B is greater than that of Be.
(a) Statement-I and II are correct and statement-II is the correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not the correct explanation of statement-I.
(c) Statement-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(b) Statement-I and II are correct but statement-II is not the correct explanation of statement-I.

Question 44.
Statement-I: Ionization enthalpy of nitrogen is greater than that of oxygen.
Statement-lI: Nitrogen has exactly half filled electronic configuration which is more stable than electronic configuration of oxygen.
(a) Statement-I is wrong but statement-II is correct.
(b) Statement-I is correct but statement-II is wrong.
(c) Statement-I and II are correct and statement-TI is the correct explanation of statement-I.
(d) Statement-I and II are correct but statement-II is not the correct explanation of statement-I.
Answer:
(c) Statement-I and II are correct and statement-II is the correct explanation of statement-I.

Question 45.
Which of the following does not have zero electron gain enthalpy?
(a) Be
(b) Cl
(c) Mg
(d) N
Answer:
(b) Cl

Question 46.
Which of the following have zero electron gain enthalpy?
(a) Halogens
(b) Noble gases
(c) Chalcogens
(d) Gold
Answer:
(b) Noble gases

Question 47.
Which of the following have the highest value of electronegativity?
(a) Halogens
(b) Alkali metals
(c) Alkaline earth metals
(d) Transition metals
Answer:
(a) Halogens

Question 48.
Among all the elements which one has the highest value of electronegativity?
(a) Chlorine
(b) Bromine
(c) Fluorine
(d) Iodine
Answer:
(c) Fluorine

Question 49.
Among the alkali metals which one form compounds with more covalent character?
(a) Sodium
(b) Potassium
(c) Rubidium
(d) Lithium
Answer:
(d) Lithium

Question 50.
Which of the following pair is not diagonally related?
(a) Li, Mg
(b) Li, Na
(c) Be, Al
(d) B, Si
Answer:
(b) Li, Na

Question 51.
In the modern periodic table, the period indicates the value of
(a) atomic number
(b) mass number
(c) principal quantum number
(d) azimuthal quantum number
Answer:
(c) principal quantum number
Hint:
In the modern periodic table, each period begins with the filling of a new shell. Therefore, the period indicates the value of principal quantum number. Thus, option (c) is correct.

Question 52.
Which of the following statements related to the modern periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitais in a p-subshell.
(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitais in a d-subsheli.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (1) for the last subshell that received electrons in building up the electronic configuration.
Answer:
(b)The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitais in a d-subshell.

Question 53.
The size of isoelectronic species- F^–, Ne and Na⁺ is affected by
(a) nuclear charge (Z)
(b) valence principal quantum number (n)
(c) electron-electron interaction in the outer orbitais
(d) none of the factors because their size is the same
Answer:
(a) nuclear charge (Z).

Question 54.
Which of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electrons from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitais bearing lower n value is easier than from orbital having high n value.
Answer:
(d)Removal of electron from orbitais bearing lower n value Is easier than from orbital having high n value.

Question 55.
Considering the elements B, Al, Mg and K, the correct order of their metallic character is:
(a) B > Al >Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Answer:
(d) K > Mg > Al> B
Hint:
In a period, metallic character decreases as we move from left to right. Therefore, metallic character of I< Mg and Al decreases in the order: K> Mg > Al. However, within a group, the metallic character, increases from top to bottom. Thus, Al is more metallic than B.Therefore, the correct sequence of decreasing metallic character is K> Mg >Al > B, i.e,
option (d) is correct.

Question 56.
Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is
(a) B>C>Si>N>F
(b) Si>C>B>N>F
(c) F>N>C>B>Si
(d) F>N>C>Si>B
Answer:
(c) F>N>C>B> Si
Hint:
In a period, the non-metallic character increases from left to right. Thus, among B, C, N and F, non-metallic character decreases in the order: F > N > C> B. However, within a group, non-metallic character decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correct sequence of decreasing non-metallic character is: F> N > C > B > Si, i.e., option (c) is correct.

Question 57.
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is
(a) F>Cl>O>N
(b) F>O>Cl>N
(c) Cl>F>O>N
(d) O>F>N>Cl
Answer:
(b) F>O>Cl>N.
Hint:
Within a period, the oxidizing character increases from left to right. Therefore, among F, O and N,oxidizing power decreases in the order: F> O> N. However, within a group, oxidizing power decreases from top to bottom. Thus, F is a stronger oxidizing agent than Cl. Further because O is more electronegative than Cl, therefore, O is a stronger oxidizing agent than Cl. Thus, overall decreasing order of oxidizing power is: F > O > Cl > N, i.e.,
option (b) is correct.

Question 58.
The highest ionization energy is exhibited by ………………
(a) halogens
(b) alkaline earth metals
(c) transition metals
(d) noble gases
Answer:
(b) alkaline earth metals

Question 59.
Which of the following is arranged in order of increasing radius?
(a) K⁺_(aq) < Na⁺_(aq) <Li⁺_(aq)
(b) K⁺_(aq)> Na⁺_(aq)> Zn²⁺_(aq)
(c) K⁺_(aq)> Li⁺_(aq) > Na⁺_(aq)
(d) Li⁺_(aq)< Na⁺_(aq) < K⁺_(aq)
Answer:
(d) Li⁺_(aq)< Na⁺_(aq) < K⁺_(aq)

Question 60.
Among the following elements, which has the least electron affinity?
(a) Phosphorous
(b) Oxygen
(c) Sulphur
(d) Nitrogen
Answer:
(d) Nitrogen

Question 61.
Which one of the following is isoelectronic with Ne?
(a) N^3-
(b) Mg²⁺
(c) Al³⁺
(d) All the above
Answer:
(d) All the above

Question 62.
Which clement has smallest size?
(a) B
(b) N
(c) Al
(d) P
Answer:
(b) N

Question 63.
In halogens, which of the following decreases from iodine to fluorine?
(a) Bond length
(b) Electronegativity
(c) ionization energy
(d) Oxidizing power
Answer:
(a) Bond length

Question 64.
What is the electronic configuration of the elements of group 14?
(a) ns² np⁴
(b) ns² np⁶
(c) ns² np²
(d) ns²
Answer:
(c) ns² np²

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements  2-Mark Questions

Write brief answer to the following questions:

Question 1.
State Johann Dobereiner’s law of triads.
Answer:
Johann Dobereiner noted that elements with similar properties occur in groups of three which he called triads. It was seen that invariably, the atomic weight of the middle number of the triad was nearly equal to the arithmetic mean of the weights of the other two numbers of the triad.
For e.g.,

Question 2.
Write a note about Chancourtois classification.
Answer:
in this system, elements that differed from each other in atomic weight by 16 or multiples of 16 fell very nearly on the same vertical line. Elements lying directly under each other showed a definite similarity. This was the first periodic law.

Question 3.
State the New land’s law of octaves.
The Law of octaves states that, “when elements are arranged in the order of increasing atomic weights, the properties of the eighth element are a repetition of the properties of the first element.

Question 4.
State Mendeleev’s periodic law.
Answer:
This law states that “The physical and chemical properties of elements are a periodic function of their atomic weights.”

Question 5.
Explain about the relationship between the atomic number of an element and frequency of the X-ray emitted from the elements.
Answer:
Henry Moseley studied the X-ray spectra of several elements and determined their atomic numbers (Z). He noticed that the frequencies of X-ray emitted from the elements concerned could be correlated by the equation
\(\sqrt{υ}\) = a(Z – b)
Where, υ Frequency of the X-ray emitted by the element.
a and b = Constants and have same values for all the elements.
Z = Atomic number of the element.

Question 6.
State modern periodic law.
Answer:
The modem periodic law states that, “The physical and chemical properties of the elements are periodic function of their atomic numbers.”

Question 7.
What are the anomalies of the long form of periodic table?
Answer:
The long form of periodic table need clarification about the following:

• Position of hydrogen is not defined till now.
• Lanthanides and actinides still find place in the bottom of the table.

Question 8.
Mention the names of the elements with atomic number 101, 102, 109 and 110.
Z = 101  IUPAC  name : Mendelevium
Z = 102  IUPAC  name : Nobelium
Z = 109  IUPAC  name : Meitnerium
Z = 110  IUPAC  name : Darmstadtium

Question 9.
Write a note about the electronic configuration of elements in groups.
Answer:
A vertical column of the periodic table is called a group. A group consists of a series of elements having similar configuration of the outermost shell. There are 18 groups in the periodic table. it may be noted that the elements belonging to the same group are said to constitute a family. For example, elements of group 17 are called halogen family.

Question 10.
Give the name and electronic configuration of elements of group and 2 group.
Answer:

• Elements of 1^st group are called alkali metals. Their electronic configuration is ns¹.
• Elements of 2^nd group are called alkaline earth metals. Their electronic configuration is ns².

Question 11.
Write any two characteristic properties of alkali metals.
Answer:

• Alkali metals readily lose their outermost electron to form +1 ion.
• Alkali metals are soft metals with low melting and boiling points.

Question 12.
Write any two characteristic properties of alkaline earth metals.
Answer:

• Alkaline earth metals readily lose their outermost electrons to form +2 ion.
• As we go down the group. their metallic character and reactivity are increased.

Question 13.
Groups from 13 to 1 in the periodic table are called p-block elements. Give reason.
Answer:

• The elements whose last electron enters into the p-orbital of the outermost shell are having similar properties and thus form a group. The ‘np’ orbital of these elements is being progressively tilled. Hence, these elements are named as p-block elements.
• The groups of 3^th to 18^th in the periodic table belongs to p-block.

Question 14.
Why noble gases do not show much of chemical reactivity?
Answer:
Noble gases having closed valence shell configuration as ns² np⁶. The valence shell orbitais of noble gases are completely filled by electrons and it is very difficult to alter this stable arrangement by the addition or removal of electrons. Because of these reasons noble gases do not show much of chemical reactivity.

Question 15.
Halogens and chalcogens have highly negative electron gain enthalpies. Why?
Answer:

• Group 16 (chaicogens) and Group 17 (halogens) are interested to add two or one electrons respectively to attain stable noble gas configuration.
• Because of this interest these elements have highly negative electron gain enthalpies.

Question 16.
What are d-biock elements? Why are they called so?
Answer:

• The elements whose last electron enters into the d-orbitalof the penultimate shell (n-1) are having similar properties and called as d-block elements.
• The groups of 3 to 12 in the center of the periodic table belongs to d-block.

Question 17.
Elements Zn, Cd and Hg with electronic configuration (n-1)d¹⁰ ns² do not show most of transition elements properties. Give reason.
Answer:

• Zn, Cd and Hg are having completely filled d-orbitais (d¹⁰ electronic configuration).
• They do not have partially filled d-orbitais Like other transition elements. So they do not show much of the transition elements properties.

Question 18.
Why Zn, Cd and Hg are considered as soft metals?
Answer:

• Zinc, cadmium and mercury are metals with low melting points. This is because they have an especially stable electronic configuration.
• Mercury is so poor at forming metallic bonds that it is liquid at room temperature.
• Zinc and cadmium arc soft metals that oxidize to the +2 oxidation states.

Question 19.
Why d-block elements are called as transition elements?
Answer:
d-block elements form a bridge between the chemically active metals of s-block elements and the less active elements of groups of 13^th and 14^th and thus take their familiar name transition elements.

Question 20.
What are f-block elements? how many series are there? Why they are called f-block elements?
Answer:

• The elements, whose last electron enters into the f-orbital of the ante-penultimate shell (n-2) are having similar properties are called f-block elements. In these elernents (n-2)f orbitais are being filled progressively.
• The two rows of elements placed at the bottom of the periodic table. They are lanthanides and actinides.

Question 21.
Write the electronic configuration of lanthanides and actinides?
Answer:

• The electronic configuration of lanthanides is 4f^1-144 5d^0-11 6s².
• The electronic configuration of actinides is 5f^1-14 6d^0-17s².

Question 22.
What are lanthanides and actinides? ,
Answer:

• In 4f senes, 4f’orbitals arc being progressively filled with electrons, 4f^1-14 5d^0-1 6s². These elements lie in 6^th period and are called rare earths or lanthanides or lanthanones.
• In 5f series, 5f orbitais are being progressively filled with electrons, 5f¹6d^0-1 7s². These elements lie in 7^th period and are called actinides or actonones.

Question 23.
What are semi-metals? Give example.
Answer:

• Some elements in the periodic table show properties that are characteristic of both metals and non-metals. These elements are called semi-metals or metalloids.
• Example: Silicon. germanium, arsenic, antimony and tellurium.

Question 24.
What are periodic properties? Give example.
Answer:
The term periodicity of properties indicates that the elements with similar properties reappear at certain regular intervals of atomic number in the periodic table.
Example:

• Atomic radii
• Ionisation energy
• Electron affinity
• Electronegativity.

Question 25.
Define ionic radius.
Answer:
The ionic radius of an ion is the distance between the center of the ion and the outermost point of its electron cloud.

Question 26.
Cationic radius is smaller than its corresponding neutral atom. Justify this statement.
Answer:

• When an neutral atom lose one or more electrons it forms cation.
  Na → Na⁺ + e^–
• The radius of this cation (r_Na⁺)is decreased than its parent atom (r_Na).
• When an atom is charged to cation, the number of nuclear charge becomes greater than the number of orbital electrons. Hence the remaining electrons are more strongly attracted by the nucleus. Hence the cationic radius is smaller than its corresponding neutral atom.

Question 27.
Anionic radius is higher than the corresponding neutral atom. Give reason.
Answer:
When an atom gain one or more electrons it forms anion. During the formation of anion, the number of orbital electrons become greater than the nuclear charge. Hence, the electrons are not strongly attracted by the lesser number of nuclear charges. Hence anionic radius is higher than the corresponding neutral atom.

Question 28.
What are iso electronic ions? Give example.
Answer:
There are some ions of different elements having the same number of electrons are called isoelectronic ions.
Example: Na⁺ Mg²⁺, Al³⁺, F^–, O^2-, N^3-

Question 29.
Define ionization energy. Give its unit.
Answer:
The energy required to remove the most loosely held electron from an isolated gaseous atom is called ioniiation energy.
M_(g) + energy M⁺_(g) + electron
The unit of ionization energy is KJ mole^-1.

Question 30.
Ionization energy of beryllium is greater than the ionization energy of boron. Why?
Answer:
Be (Z= 4) 1s² 2s². it has completely filled valence electrons, which requires high IE₁.
B (Z =5) 1s² 2s² 2p¹. It has incompletely filled valence electrons, which requires comparatively
less IE₁ Hence I.E₁ Be > I.E₁ B.

Question 31.
Ionization energy of nitrogen is greater than the ionization energy of oxygen. Give reason.
Answer:
₇N 1s² 2s¹ 2p³ (or) . Nitrogen has exactly half filled valence electrons, which requires high I.E₁.
₈O 1s² 2s² 2p⁴ (or). Oxygen has incomplete valence shell electrons, which requires comparatively less I.E₁.
I.E₁N>I.E₁O

Question 32.
Define electron gain enthalpy or electron affinity. Give its unit.
Answer:
The electron gain enthalpy of an element is the amount of energy released when an electron is added to the neutral gaseous atom.
A + electron → A^– + energy (E.A)
Unit of electron affinity is KJ mole^–.

Question 33.
Electron gain enthalpy of F ¡s less negative than Cl. Why?
Answer:
When an electron is added to F, the added electron goes to the L shell (n = 2). As the ‘L’ shell possess smaller region of space, the added electron feels significant repulsion from the other electrons present in this level.
E.A of F = -328 KJ mole^-1
E.A of Cl = -349 KJ mole^-1

Question 34.
Electron affinity of oxygen is less negative than sulphur. Justify this statement.
Answer:
When an electron is added to oxygen, the added electron goes to the ‘L’ shell (n 2). As the ‘L’ shell possess smaller region of space, the added electron feels significant repulsion from the other electrons present in this level.
E.A of O = – 141 KJ mole^–.
E.A of S = – 200 KJ mole^–.

Question 35.
Explain about the factors that affect electro negativity.
Answer:

• Effective nuclear charge:
  As the nuclear charge increases, electro negativity also increases along the periods.
• Atomic radius:
  The atoms in smaller size will have larger electronegativity.

Question 36.
Explain about periodic variation of electro negativity across a period.
Answer:
As we move from left to right in a period, electro negativity increases. This is due to the following reasons:

• Nuclear charge increases in a period
• Atomic size decrease in a period

Halogens have the highest value of electro negativity in their respective periods.

Question 37.
Explain about the period variation of electro negativity along a group.
Answer:
As we move down from top to bottom in a group, electro negativity decreases due to increased atomic radius. Fluorine has the highest value of clectro negativity among all the elements.

Question 38.
Define valency. How is it determined?
Answer:
The valency of an element may be defined as the combining capacities of elements. The electrons present in the outermost shell are called valence electrons and these electrons determine the valency of the atom.

Question 39.
What is the basic difference in approach between Mendeleev’s periodic law and the modern periodic law?
Answer:
The basic difference in approach between Mendeleev’s periodic law and modern periodic law is the change in basis of classification of elements from atomic weight to atomic number.

Question 40.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer:
The sixth period corresponds to sixth shell. The orbitais present in this shell are 6s, 4f. 5p and 6d. The maximum number of electrons which can be present in these sub-shell is 2 + 14 + 6 + 10 = 32. Since the number of elements in a period corresponds to the number of electrons in the shells, the sixth period should have a maximum of 32 elements.

Question 41.
Why do elements in the same group have similar physical and chemical properties?
Answer:
The elements in a group have same valence shell electronic configuration and hence have similar physical and chemical properties.

Question 42.
How do atomic radius vary in a period and in a group? How do you explain the variation.
Answer:
Within a group atomic radius increases down the group Reason :
This is due to continuous increases in the number of electronic shells or orbit numbers in the structure of atoms of the elements down a group.

Variation across period:
Atomic radii:
From left to right across a period atomic radii generally decreases due to increase in effective nuclear charge from left to right across a period.

Question 43.
Explain why cation are smaller and anions are larger ¡n radii than their parent atoms?
Answer:
A cation is smaller than the parent atom because it has fewer electrons while its nuclear charge remains the same. The size of anion will be larger than that of parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

Question 44.
What is basic difference between the terms electron gain enthalpy and electronegativity?
Answer:
Electron gain enthalpy refers to tendency of an isolated gaseous atom to accept an additional electron to form a negative ion. Whereas electronegativity refers to tendency of the atom of an clement to attract shared pair of electrons towards it in a covalent bond.

Question 45.
Would you expect the first ionization enthalpies of two isotopes of the same element to be the same or different? Justify your answer.
Answer:
Ionization enthalpy, among other things, depends upon the electronic configuration (number of electrons) and nuclear charge (number of protons). Since isotopes of an element have the same electronic configuration and same nuclear charge, they have same ionization enthalpy.

Question 46.
Write the general electronic configuration of s-, p-, d-, and f-block elements?
Answer:

• s-block elements : ns^1-2 where n 2 – 7.
• p-block elements : ns² np^1-6 where n = 2 – 6.
• d-block elements : (n – 1) d^1-0 ns^0-2 where n = 4 – 7.
• f-block elements : (n – 2) f^0-14 (n – 1) d^0-1 ns² where n = 6 – 7.

Samacheer Kalvi 11th Chemistry Periodic Classification of Elements 3-Mark Questions

Question 1.
Why there is a need for classification of elements?
Answer:

• Classification is a fundamental and essential process in our day-to-day life for the effective utilization of resources, daily events and materials.
• In such a way, for the effective utilization of discovered elements becomes fundamentally essential process.
• The periodic classification of the elements is one of the outstanding contributions to the progress of chemistry.

Question 2.
Prove that the halogens, chlorine, bromine and iodine follow the law of triads.
Answer:
When the halogens, chlorine, bromine and iodine are placed on below the others, they had similar properties. The atomic weight of bromine was close to the average of the atomic weights of chlorine and iodine.

Question 3.
What are the salient features of New land’s law of octaves?
Answer:

1. This law is quite well for lighter elements but not supported to heavier elements.
2. Elements were arranged in increasing atomic masses without taking an account on the properties of elements.
3. This law was seemed to be applicable only for elements upto calcium.

Question 4.
How the properties of Eka – silicon was related to germanium?
Answer:

Question 5.
Compare the properties of Eka – aluminium and gallium.
Answer:

Question 6.
How Moseley determined the atomic number of an element using X-rays?
Answer:

• Henry Moseley studied the X-ray spectra of several elements and determined their atomic numbers (Z).
• He discovered a correlation between atomic number and the frequency of X-rays generated by bombarding an clement with high energy of electrons.
• Moseley correlated the frequency of the X-ray emitted by an equation as,
  \(\sqrt{v}\) = a (Z – b)
  Where υ = Frequency of the X-rays emitted by the elements.
  a and b = Constants.
• From the square root of the measured frequency of the X-rays emitted, he determined the atomic number of the element.

Question 7.
What are the reasons behind the Moseley’s attempt in finding atomic number?
Answer:

• The number of electrons increases by the same number as the increase in the atomic number.
• As the number of electrons increases, the electronic structure of the atom changes.
• Electrons in the out cannost shell of an atom (valence shell electrons) determine the chemical properties of the elements.

Question 8.
Draw a simplified form of periods and elements present in modern period table.
Answer:

Question 9.
Write the electronic configuration of alkali metals ₂Li,₁₁Na, ₁₉K, ₃₇Rb, ₅₅Cs and ₈₇Fr.
Answer:

Question 10.
Explain about the classification of elements based on electronic configuration.
Answer:

• The distribution of electrons into orbitais, s, p, d and f of an atom is called its electronic configuration. The electronic configuration of an atom is characterized by a set of four quantum numbers, n, l, m and s. of these the principal quantum number (n) defines the main energy level known as shells.
• The position of an element in the periodic table is related to the configuration of that element and thus reflects the quantum numbers of the last orbital filled.
• The electronic configuration of elements in the periodic table can be studied along the periods and groups separately for the best classification of elements.
• Elements placed in a horizontal row of a periodic table is called a period. There are seven periods.
• A vertical column of the periodic table is called a group. A group consists of a series of elements having similar configuration of the outermost shell. There are 18 groups in periodic table.

Question 11.
Write about the electronic configuration of 1^st and 2^nd period.
Answer:
Electronic configuration of t period:
In 1 period only two elements are present. This period starts with the filling of electrons in first energy level, n¹. This level has only one orbital as is. Therefore it can accommodate two electrons maximum.

Electronic configuration of 2nd period:
In the 2’ period 8 elements are present. This period starts with filling of electrons in the second energy level, n = 2. In this level four orbitais (one 2s and three 2p) are present. Hence the second energy level can accommodate 8 electrons. Thus, second period has eight elements.

Question 12.
How many elements are there in 4^th period? Prove it.
Answer:
In fourth period, 18 elements are present. In this period electrons are entering into fourth energy level, i.e., n = 4. it starts with the filling of 4s-orbitals. However, after the 4s, but before the 4p orbitais, there are five 3d-orbitais also to be filled. Thus, nine orbitais (one 4s, five 3d and three 4p) have to be filled. These nine orbitais can accommodate (9 x 2 = 18)18 electrons. Hence, period contain 18 elements in it.

Question 13.
How many elements are there in 6th period? Prove it.
Answer:
In sixth period, 32 elements are present. This period starts with the filling of 6th energy shell, n = 6. There are sixteen orbitais (one 6s, seven 4f, five 4d and three 6p) to be filled. These sixteen orbitais can accommodate 32 (16 × 2 = 32) electrons. Hence, 32 elements are present in sixth period.

Question 14.
What are the two exceptions of block division in the periodic table?
Answer:
1. Helium has two electrons. Its electronic configuration is 1s². As per the configuration, it is supposed to be placed in ‘s’ block, but actually placed in s group which belongs to ‘p’ block. Because it has a completely filled valence shell as the other elements present in 18^th group. It also resembles with 18^th group elements in other properties. Hence helium is placed with other noble gases.

2. The other exception is hydrogen. it has only one s-electron and hence can be placed in group 1. It can also gain an electron to achieve a noble gas arrangement and hence it can behave as halogens (17th group elements). Because of these assumptions, position of hydrogen becomes a special case. Finally, it is placed separately at the top of the periodic table.

Question 15.
Explain about the salient features of metals.
Answer:

• Metals comprise more than 78% of all known elements. They are present on the left side of the periodic table.
• They are usually solids at room temperature. [Mercury is an exception (Hg-liquid), gallium (303 K) and cesium (302 K) also have very low melting points].
• Metals usually have high melting and boiling points.
• They are good conductors of heat and electricity.
• They are malleable and ductile, and also can be flattened into thin sheets by hammering and drawn into thin wires.

Question 16.
Explain about the characteristics of non-metals.
Answer:

• Non-metals are located at the top right hand side of the periodic table.
• In a period, as we move from left to right the non-metallic character increases while the metallic character increases as we go down a group.
• Non-metals are usually solids or liquids or gases at room temperature with low melting and low boiling points (Exceptions : boron and carbon).
• They are poor conductors of heat and electricity.
• Most of the non-metallic solids are brittle and are neither malleable nor ductile.

Question 17.
Periodic change in electronic configuration is responsible for the physical and chemical properties of element. Justify this statement.
Answer:

• The electronic configuration of the elements changes periodically in a period and group as well.
• We could find a pattern in the physical and chemical properties as we go down in a group or move across a period.
• For example : The chemical reactivity is high at the beginning, lower at the middle and increases to a maximum at group 17 in a period.
• The reactivity increases on moving down the group of alkali metals. But the reactivity decreases on moving down the group of halogens.
• Atomic radii increases down the group and decreases across the period.

Question 18.
What is covalent radius? How would you determine the covalent radius of chlorine atom?
Answer:
The distance between the nuclei of two covalent bonded atoms is known as covaLent distance or inter-nuclear distance. The one – half of this inter-nuclear distance is called covalent radius. The covalent distance (Cl – Cl) of Cl₂ molecule is experimentally found as 198 pm (1.98 A). Its covalent radius is 99 pm (0.99 Å).
Cl – Cl Inter nuclear distance = 1.98 Å
∴ r_Cl = 1.98 / 2=0.99 Å.

Question 19.
Write a note about metallic radius.
Answer:

• It is defined as one half of the distance between the centers of nuclei of the two adjacent atoms in the metallic crystal.
• The metallic radius is always larger than its covalent radius.
• The distance between two adjacent copper atoms in solid copper is 2.56 A. Hence, the metallic radius of copper is 1.28 A.

Question 20.
Arrange Na⁺, Mg²⁺ and Al³⁺ in the increasing order of ionic radii. Give reason.
Answer:
Na⁺, Mg²⁺ and Al³⁺ are iso electronic cations.

The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is,
r_Na⁺ > r_Mg²⁺ > r_Al³⁺

Question 21.
Arrange the ions F^–, O^2- and N^3- ¡n the increasing order of their ionic radii. Give reason.
Answer:
F^–, O^2- and N^3- are isoelectronic species.

The anion with the greater negative charge will have a larger radius because of the lesser attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is,
r_Na^3– > r_O^2– > r_F^–

Question 22.
Mention some characteristics of ionization energy.
Answer:

• Ionization is always an endothermic process. It absorbs energy.
• Ionization energies always increase in the order, I.E₁< I.E₂<IE₃.
• Ionization energy measurements are carried out with atoms in the gaseous state.

Question 23.
Why ionization energy and electron affinity are calculated in gaseous state?
Answer:

• Inter molecular force can affect the value of ionization energy and electron affinity.
• In gaseous state, there is little inter molecular force in a substance and it can be considered negligible In some cases. So the value of I.E and E.A are almost unaffected if they are calculated for gaseous atoms.
• When we arc talking about ionization energy and electron affinity, we need to consider
  atoms and we can find free atoms only when the substance is in gaseous state.

Question 24.
How is screening effect of inner electrons affect ionization energy?
Answer:

• The electrons of inner shells form a cloud of negative charge. This cloud shields the outer electron from the nucleus and also acts as a screen between the nucleus and the outer electrons.
• This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. Therefore, the outer electrons do not feel the full effect of the positive charge of the nucleus.
• If screening effect increases, ionization energy decreases.

Question 24.
How is screening effect of inner electrons affect ionization energy?
Answer:

• The electrons of inner shells form a cloud of negative charge. This cloud shields the outer electron from the nucleus and also acts as a screen between the nucleus and the outer electrons.
• This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. Therefore, the outer electrons do not feel the full effect of the positive charge of the nucleus.
• If screening effect increases, ionization energy decreases.

Question 25.
Ionization energy of Mg is greater than that of Al. Why?
Answer:
Mg (Z = 12) 1s² 2s² 2p⁶ 3s².
Al (Z = 13) 1s² 2s² 2p⁶ 3s² 3p¹.
Although the nuclear charge of aluminium is greater than that of magnesium, I.E of Mg is greater than that of Al. It is because Mg atom has more stable configuration than Al atom. IE₁ of Mg > IE₁ of Al.

Question 26.
What are all the factors that influences electron gain enthalpy?
Answer:
1. Size of the atom:
The new electron which was added experiences stronger attraction to its nucleus if the atoms are smaller in size.
Atomic size α \(\frac {1}{ Electron gain enthalpy }\)

2. Nuclear charge:
The new electron which was added experiences stronger attraction to its nucleus if the atom possess greater nuclear charge. Nuclear charge α Electron gain enthalpy

3. Electronic configuration:
An atom with stable electronic configuration has no tendency to gain an electron. Such atoms have zero or almost zero electron gain enthalpy.

Question 27.
Explain about the periodic variation of electron gain enthalpy in a period and in a group.
Answer:
1. The electron gain enthalpy increases as we move from left to right in a period due to the increase of nuclear charge. However, Be, Mg, N and noble gases have almost zero value of electron gain enthalpy due to extra stability of completely and half filled orbitais.

2. When we move in a group of periodic table, the size and nuclear charge increase. But the effect of increase in atomic size is much more pronounced than that of nuclear charge and thus the additional electron feels less attraction by the large atom. Consequently, electron gain enthalpy decreases.

Question 28.
Explain about the electro negativity and non-metallic character across the period and down the group.
Answer:
Eletro negativity α Non-metallic character:

• As the electro negativity is directly proportional to the non-metallic character, thus across the period, with an increase in electro negativity. the non-metallic character also increases.
• As we move down the group. decrease in electro negativity is accompanied by a decrease in non-metallic character.

Question 29.
Prove that valency is a periodic property.
Answer:
Variation in period:
The number of valence electrons increases from I to 8 on moving across a period. The valency of the elements with respect to hydrogen and chlorine increases from 1 to 4 and then decreases from 4 to zero.

Variation in group:
On moving down a group, the number of valence electrons remains same. All the elements in a group exhibit same valency. For example, all the elements of group I have valency equal to 1. Hence. valency is a periodic property.

Question 30.
Write a note about periodic trends and chemical reactivity.
Answer:

• The group 1 elements are extremely reactive because these elements can lose one electron to form cation. Their ionization enthalpy is also least.
• The high reactivity of halogens is due to the ease with which these elements can gain an electron to form anion. Their electron gain enthalpy is most negative.
• The elements at the extreme left (alkali) exhibit strong reducing behavior, whereas the elements at the extreme right (halogens) exhibit strong oxidizing behaviour.
• The reactivity of elements at the center of the periodic table becomes low when compared with extreme right and left.

Question 31.
How would you explain the fact that the first ionization enthalpy of sodium ¡s lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer:
Electronic configuration of Na and Mg are
Na = 1s² 2s² 2p⁶ 3s¹
Mg = 1s² 2s² 2p⁶ 3s²
First electron in both cases has to be removed from 3s-orbital but the nuclear charge of Na (+11) is lower than that of Mg (+12) therefore first ionization energy of sodium is lower than that of magnesium.

After the loss of first electron, the electronic configuration of
Na = 1s² 2s² 2p⁶
Mg = 1s²2s²2p⁶3s¹
Here electron is to be removed from inert (neon) gas configuration which is very stable and hence removal of second electron requires more energy when compared to Mg. Therefore, second ionization enthalpy of sodium is higher than that of magnesium.

Question 32.
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down the group?
Answer:
Atomic size:
With the increase in atomic size, the number of electron shells increase. Therefore, the force that binds the electrons with the nucleus decreases. The ionization enthalpy thus decreases with the increase in atomic size.

Screening or shielding effect of inner shell electron:
With the addition of new shells, the number of inner electron shells which shield the valence electrons increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionization enthalpy decreases.

Question 33.
Which of the following pairs of elements would have more negative electron gain enthalpy?

1. O or F
2. F or Cl.

Answer:
1. O or F. Both O and F lie in 2e” period. As we move from O to F the atomic size decreases. Due to smaller size off nuclear charge increases. Further, gain of one electron by
F → F^–
F^– ion has inert gas configuration, While the gain of one electron by
O → O^–
gives O ion which does not have stable inert gas configuration. Consequently, the energy released is much higher in going from
F → F^–
than going from O → O^–. In other words electron gain enthalpy off is much more negative than that of oxygen. –

2. The negative electron gain enthalpy of Cl (e.g. ∆H = – 349 mol^-1) is more than that of F (e.g. ∆H = -328 U mol^-1).

The reason for the deviation is due to the smaller size off. Due to its small size, the electron repulsion in the relatively compact 2p-subshell are comparatively large and hence the attraction for incoming electron is less as in the case of Cl.

Question 34.
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
For oxygen atom:
O_(g) + e^– → O^-1_(g) (e.g. ∆H = -141 Id mor^-1)
O^-1_(g) + e^– → O^2-_(g) (e.g. ∆H = + 780 kJ mol^-1)
The first electron gain enthalpy of oxygen is negative because energy is released when a gaseous atom accepts an electron to form monovalent anion. The second electron gain enthalpy is positive because energy is needed to overcome the force of repulsion between monovalent anion and second incoming electron.

Question 35.
What are major differences between metals and non-metals?
Metals:

• Have a strong tendency to lose electrons to form cations.
• Metals are strong reducing agents.
• Metals have low ionization enthalpies.
• Metals form basic oxides and ionic compounds.

Non-Metals:

• Non-metals have a strong tendency to accept electrons to form anions.
• Non-metals are strong oxidizing agents.
• Non-metals have high ionization enthalpies.
• Non-metals form acidic oxides and covalent compounds.

Question 36.
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb Cl> Br>. Explain.
Answer:
The elements of group 1 have only one electron in their respective valence shells and thus have a strong tendency to lose this electron. The tendency to lose electrons in turn, depends upon the ionization enthalpy. Since the ionization enthalpy decreases down the group therefore, the reactivity of group 1 elements increases in the same order Li

Question 37.
Arrange the following as stated:

1. N₂ O₂, F₂, Cl₂ (Increasing order of bond dissociation energy)
2. F, Cl, Br, I (Increasing order of electron gain enthalpy)
3. F₂, N₂, Cl₂, O₂ (Ipcreasing order of bond length).

Answer:

1. F₂, N₂, Cl₂, O₂
2. I< Br < F < Cl
3. N₂ O₂, F₂, Cl₂