Category: Class 11

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Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

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Samacheer Kalvi 11th Physics Chapter 3 Laws of Motion Textual Questions Solved

Samacheer Kalvi 11th Physics Laws of Motion Multiple Choice Questions

11th Physics Chapter 3 Book Back Answers Question 1.
When a car takes a sudden left turn in the curved road, passengers are pushed towards the right due to
(a) inertia of direction
(b) inertia of motion
(c) inertia of rest
(d) absence of inertia
Answer:
(a) inertia of direction

Samacheer Kalvi 11th Physics Solution Chapter 3 Question 2.
An object of mass m held against a vertical wall by applying horizontal force F as shown in the figure. The minimum value of the force F is [IIT JEE 1994]
(a) Less than mg
(b) Equal to mg
(c) Greater than mg
(d) Cannot determine
Answer:
(c) Greater than mg

11th Physics Lesson 3 Book Back Answers Question 3.
A vehicle is moving along the positive x direction, if sudden brake is applied, then
(a) frictional force acting on the vehicle is along negative x direction
(b) frictional force acting on the vehicle is along positive x direction
(c) no frictional force acts on the vehicle
(d) frictional force acts in downward direction
Answer:
(a) frictional force acting on the vehicle is along negative x direction

11th Physics 3rd Chapter Book Back Answers Question 4.
A book is at rest on the table which exerts a normal force on the book. If this force is considered as reaction force, what is the action force according to Newton’s third law?
(a) Gravitational force exerted by Earth on the book
(b) Gravitational force exerted by the book on Earth
(c) Normal force exerted by the book on the table
(d) None of the above
Answer:
(c) Normal force exerted by the book on the table

Laws Of Motion Class 11 State Board Question 5.
Two masses m₁ and m₂ are experiencing the same force where m₁ < m₂ The ratio of their acceleration \(\frac{a_{1}}{a_{2}}\) is –
(a) 1
(b) less than 1
(c) greater than 1
(d) all the three cases
Answer:
(c) greater than 1

Samacheer Kalvi Guru 11th Physics Question 6.
Choose appropriate free body diagram for the particle experiencing net acceleration along negative y direction. (Each arrow mark represents the force acting on the system).

Answer:

Class 11 Physics Solutions Samacheer Kalvi Question 7.
A particle of mass m sliding on the smooth double inclined plane (shown in figure) will experience –
(a) greater acceleration along the path AB
(b) greater acceleration along the path AC
(c) same acceleration in both the paths
(d) no acceleration in both the paths

Answer:
(a) greater acceleration along the path AC

Samacheer Kalvi 11th Physics Question 8.
Two blocks of masses m and 2m are placed on a smooth horizontal surface as shown. In the first case only a force F₁  is applied from the left. Later only a force F₂  is applied from the right. If the force acting at the interface of the two blocks in the two cases is same, then F₁ : F₂ is [Physics Olympiad 2016]

(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer:
(c) 2 : 1

11th Physics 3rd Lesson Book Back Answers Question 9.
Force acting on the particle moving with constant speed is –
(a) always zero
(b) need not be zero
(c) always non zero
(d) cannot be concluded
Answer:
(b) need not be zero

11th Physics 3rd Chapter Exercise Question 10.
An object of mass m begins to move on the plane inclined at an angle 0. The coefficient of static friction of inclined surface is lay. The maximum static friction experienced by the mass is –
(a) mg
(b) µ_s mg
(c) µ_s mg sin θ
(d) µ_s mg cos θ
Answer:
(d) µ_s mg cos θ

Class 11 Samacheer Physics Solutions Question 11.
When the object is moving at constant velocity on the rough surface –
(a) net force on the object is zero
(b) no force acts on the object
(c) only external force acts on the object
(d) only kinetic friction acts on the object
Answer:
(a) net force on the object is zero

Samacheer Kalvi 11th Physics Book Back Answers Question 12.
When an object is at rest on the inclined rough surface –
(a) static and kinetic frictions acting on the object is zero
(b) static friction is zero but kinetic friction is not zero
(c) static friction is not zero and kinetic friction is zero
(d) static and kinetic frictions are not zero
Answer:
(c) static friction is not zero and kinetic friction is zero

Class 11 Physics Samacheer Kalvi Question 13.
The centrifugal force appears to exist –
(a) only in inertial frames
(b) only in rotating frames
(c) in any accelerated frame
(d) both in inertial and non-inertial frames
Answer:
(b) only in rotating frames

11th Physics Unit 3 Book Back Answers Question 14.
Choose the correct statement from the following –
(a) Centrifugal and centripetal forces are action reaction pairs
(b) Centripetal forces is a natural force
(c) Centrifugal force arises from gravitational force
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion
Answer:
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion

Laws Of Motion Class 11 Numericals With Solutions Pdf Question 15.
If a person moving from pole to equator, the centrifugal force acting on him –
(a) increases
(b) decreases
(c) remains the same
(d) increases and then decreases
Answer:
(a) increases

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions

Question 1.
Explain the concept of inertia. Write two examples each for inertia of motion, inertia of rest and inertia of direction.
Answer:
The inability of objects to move on its own or change its state of motion is called inertia. Inertia means resistance to change its state. There are three types of inertia:
1. Inertia of rest:
The inability of an object to change its state of rest is called inertia of rest.
Example:

• When a stationary bus starts to move, the passengers experience a sudden backward push.
• A book lying on the table will remain at rest until it is moved by some external agencies.

2. Inertia of motion:
The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.
Example:

• When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front seat.
• An athlete running is a race will continue to run even after reaching the finishing point.

3. Inertia of direction:
The inability of an object to change its direction of motion on its own is called inertia of direction.
Example:

• When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle.
• When a bus moving along a straight line takes a turn to the right. The passengers are thrown towards left.

Question 2.
State Newton’s second law.
Answer:
The force acting on an object is equal to the rate of change of its momentum –
\(\overline{\mathrm{F}}\) = \(\frac{d \bar{p}}{d t}\)

Question 3.
Define one newton.
Answer:
One newton is defined as the force which acts on 1 kg of mass to give an acceleration 1 ms^-2 in the direction of the force.

Question 4.
Show that impulse is the change of momentum.
Answer:
According to Newton’s Second Law
F = \(\frac {dp}{dt}\) i.e. dp = Fdt
Integrate it over a time interval from t_i  to t_f

P_i → initial momentum of the object at t_i
P_f → Final momentum of the object at t_f
P_f – P_i = ∆p = change in momentum during the time interval ∆t.
\(\int_{t_{i}}^{t_{f}} \mathrm{F} \cdot d t=\mathrm{J}\) is called the impulse.
If the force is constant over the time interval ∆t, then

Hence the proof.

Question 5.
Using free body diagram, show that it is easy to pull an object than to push it.
Answer:
When a body is pushed at an arbitrary angle θ [0 to \(\frac {π}{2}\)], the applied force F can be resolved into two components as F sin 0 parallel to the surface and F cos 0 perpendicular to the surface as shown in figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is no acceleration along the vertical direction the normal force N is equal to
N_push = mg + F cos θ …………(1)
As a result the maximal static friction also increases and is equal to
\(f_{S}^{\max }\) = \(\mu_{r} \mathrm{N}_{\mathrm{push}}\) = µ_s(mg + F cos θ) ……(2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.

When an object is pulled at an angle θ, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is –
N_pull = mg – F cos θ ………….(3)

Equation (3) shows that the normal force is less than – N_push. From equations (1) and (3), it is easier to pull an object than to push to make it move.

Question 6.
Explain various types of friction. Suggest a few methods to reduce friction.
Answer:
There are two types of Friction:
(1) Static Friction:
Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force f_s lies between
\(0 \leq f_{s} \leq \mu_{s} \mathrm{N}\)
where, µ_s – coefficient of static friction
N – Normal force

(2) Kinetic friction:
The frictional force exerted by the surface when an object slides is called as kinetic friction. Also called as sliding friction or dynamic friction,
f_k – µ_kN
where µ_k – the coefficient of kinetic friction
N – Normal force exerted by the surface on the object

Methods to reduce friction:
Friction can be reduced

• By using lubricants
• By using Ball bearings
• By polishing
• By streamlining

Question 7.
What is the meaning by ‘pseudo force’?
Answer:
Pseudo force is an apparent force which has no origin. It arises due to the non-inertial nature of the frame considered.

Question 8.
State the empirical laws of static and kinetic friction.
Answer:
The empirical laws of friction are:

• Friction is independent of surface of contact.
• Coefficient of kinetic friction is less than coefficient of static friction.
• The direction of frictional force is always opposite to the motion of one body over the other.
• Frictional force always acts on the object parallel to the surface on which the objet is placed,
• The magnitude of frictional force between any two bodies in contact is directly proportional to the normal reaction between them.

Question 9.
State Newton’s third law.
Answer:
Newton’s third law states that for every action there is an equal and opposite reaction.

Question 10.
What are inertial frames?
Answer:
Inertial frame is the one in which if there is no force on the object, the object moves at constant velocity.

Question 11.
Under what condition will a car skid on a leveled circular road?
Answer:
On a leveled circular road, if the static friction is not able to provide enough centripetal ’force to turn, the vehicle will start to skid
\(\mu_{s}<\frac{v^{2}}{r g}\)

Samacheer Kalvi 11th Physics Laws of Motion Long Answer Questions

Question 1.
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it.
Answer:
In nature, conservation laws play a very important role. The dynamics of motion of bodies can be analysed very effectively using conservation laws. There are three conservation laws in mechanics. Conservation of total energy, conservation of total linear momentum, and conservation of angular momentum. By combining Newton’s second and third laws, we can derive the law of conservation of total linear momentum. When two particles interact with each other, they exert equal and opposite forces on each other.

The particle 1 exerts force \(\overrightarrow{\mathrm{F}}_{12}\) on particle 2 and particle 2 exerts an exactly equal and opposite force \(\overrightarrow{\mathrm{F}}_{12}\) on particle 1 according to Newton’s third law.
\(\overrightarrow{\mathrm{F}}_{12}\) = –\(\overrightarrow{\mathrm{F}}_{12}\) ……..(1)
In terms of momentum of particles, the force on each particle (Newton’s second law) can be written as –
\(\overrightarrow{\mathrm{F}}_{12}\) = \(\frac{d \vec{p}_{1}}{d t}\) and \(\overrightarrow{\mathrm{F}}_{21}\) = \(\frac{d \vec{p}_{2}}{d t}\) ………(2)
Here \(\vec{p}_{1}\) is the momentum of particle 1 which changes due to the force \(\overrightarrow{\mathrm{F}}_{12}\) exerted by particle 2. Further Po is the momentum of particle \(\vec{p}_{2}\) This changes due to \(\overrightarrow{\mathrm{F}}_{21}\) exerted by particle 1.
Substitute equation (2) in equation (1)
\(\frac{d \vec{p}_{1}}{d t}\) = – \(\frac{d \vec{p}_{2}}{d t}\) …………(3)
\(\frac{d \vec{p}_{1}}{d t}\) + \(\frac{d \vec{p}_{2}}{d t}\) = 0 ………(4)
\(\frac {d}{dt}\)(\(\vec{p}_{1}\) + \(\vec{p}_{2}\)) = 0
It implies that \(\vec{p}_{1}\) + \(\vec{p}_{2}\) = constant vector (always).
\(\vec{p}_{1}\) + \(\vec{p}_{2}\) is the total linear momentum of the two particles (\(\vec{p}_{tot}\) = \(\vec{p}_{1}\) + \(\vec{p}_{2}\)).It is also called as total linear momentum of the system. Here, the two particles constitute the system. From this result, the law of conservation of linear momentum can be stated as follows.

If there are no external forces acting on the system, then the total linear momentum of the system (\(\vec{p}_{tot}\)) is always a constant vector. In other words, the total linear momentum of the system is conserved in time. Here the word ‘conserve’ means that \(\vec{p}_{1}\) and \(\vec{p}_{2}\) can vary, in such a way that \(\vec{p}_{1}\) + \(\vec{p}_{2}\) is a constant vector.

The forces \(\overrightarrow{\mathrm{F}}_{12}\) and \(\overrightarrow{\mathrm{F}}_{21}\) are called the internal forces of the system, because they act only between the two particles. There is no external force acting on the two particles from outside. In such a case the total linear momentum of the system is a constant vector or is conserved.

Meaning of law of conservation of momentum:
1. The Law of conservation of linear momentum is a vector law. It implies that both the magnitude and direction of total linear momentum are constant. In some cases, this total momentum can also be zero.

2. To analyse the motion of a particle, we can either use Newton’s second law or the law of conservation of linear momentum. Newton’s second law requires us to specify the forces involved in the process. This is difficult to specify in real situations. But conservation of linear momentum does not require any force involved in the process. It is convenient and hence important.

For example, when two particles collide, the forces exerted by these two particles on each other is difficult to specify. But it is easier to apply conservation of linear momentum during the collision process.

Examples:
Consider the firing of a gun. Here the system is Gun+bullet. Initially the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let \(\vec{p}_{1}\) be the momentum of the bullet and \(\vec{p}_{2}\) the momentum of the gun before firing. Since initially both are at rest,

Total momentum before firing the gun is zero, \(\vec{p}_{1}\) + \(\vec{p}_{2}\) = 0.
According to the law of conservation of linear momentum, total linear momentum has to be zero after the firing also.

When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from \(\vec{p}_{1}\) to \(\vec{p}_{1}\) To conserve the total linear momentum of the system, the momentum of the gun must also change from \(\vec{p}_{2}\) to \(\vec{p}_{2}^{\prime}\). Due to the conservation of linear momentum, \(\vec{p}_{1}\) + \(\vec{p}_{2}^{\prime}\) = 0.

It implies that \(\vec{p}_{1}^{\prime}\)= \(\vec{p}_{2}^{\prime}\), the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet. This is the reason after firing, the gun suddenly moves backward with the momentum (-\(\vec{p}_{2}^{\prime}\)). It is called ‘recoil momentum’. Th is is an example of conservation of total linear momentum.

Question 2.
What are concurrent forces? State Lami’s theorem.
Answer:
Concurrent force:
A collection of forces is said to be concurrent, if the lines of forces act at a common point.

Lami’s Theorem:
If a system of three concurrent and coplanar forces is in equilibrium, then Lami’s theorem states that the magnitude of each force of the system is proportional to sine of the angle between the other two forces.
i.e. |\(\overrightarrow{\mathrm{F}}_{1}\)|∝ sin α, |\(\overrightarrow{\mathrm{F}}_{2}\)| ∝ sin β, |\(\overrightarrow{\mathrm{F}}_{2}\)| ∝ sin γ,

Question 3.
Explain the motion of blocks connected by a string in

1. Vertical motion
2. Horizontal motion.

Answer:
When objects are connected by strings and When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Case 1:
Vertical motion:
Consider two blocks of masses m₁ and m₂ (m₁> m₂) connected by a light and in-extensible string that passes over a pulley as shown in Figure.

Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving, m₂ vertically upward and m_k, downward with same acceleration a. The gravitational force m₁g on mass m₁ is used in lifting the mass m₂. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.

Applying Newton’s second law for mass m₂ T \(\hat{j}\) – m₂ g \(\hat{j}\) = m₂ a \(\hat{j}\) The left hand side of the above equation is the total force that acts on m₂ and the right hand side is the product of mass and acceleration of m₂ in y direction.
By comparing the components on both sides, we get
T = m₂ g = m₂ a ……….(1)
Similarly, applying Newton’s second law for mass m₂
T \(\hat{j}\) – m₁ g\(\hat{j}\) = m₁a\(\hat{j}\)
As mass m₁ moves downward (-\(\hat{j}\)), its acceleration is along (-\(\hat{j}\))
By comparing the components on both sides, we get
T = m₁g = -m₁a
m₁g – T = m₁a ………..(2)
Adding equations (1) and (2), we get
m₁g – m₂g = m₁a + m₂a
(m₁ – m₂)g = (m₁ + m₂)a …………(3)
From equation (3), the acceleration of both the masses is –
a = (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ………..(4)
If both the masses are equal (m₁ = m₂), from equation (4)
a = 0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
T = m₂g = m₂(\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))
T = m₂g + m₂ (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ……….(5)
By taking m₂g common in the RHS of equation (5)

Equation (4) gives only magnitude of acceleration.
For mass m₁, the acceleration vector is given by \(\vec{a}\) = –\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\)\(\hat{j}\)
For mass m₂, the acceleration vector is given by \(\vec{a}\) = \(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\) \(\hat{j}\)

Case 2:
Horizontal motion:
In this case, mass m₂ is kept on a horizontal table and mass m₁, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface

As both the blocks are connected to the un stretchable string, if m₁ moves with an acceleration a downward then m₂ also moves with the same acceleration a horizontally.
The forces acting on mass m₂ are

• Downward gravitational force (m₂g)
• Upward normal force (N) exerted by the surface
• Horizontal tension (T) exerted by the string

The forces acting on mass m₁ are

• Downward gravitational force (m₁g)
• Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure.

Applying Newton’s second law for m₁
T\(\hat{i}\) – m₁g\(\hat{j}\) = -m₁a\(\hat{j}\) (alongy direction)
By comparing the components on both sides of the above equation,
T – m₁g = -m₁a …………(1)
Applying Newton’s second law for m₂
Ti = m₁ai (along x direction)
By comparing the components on both sides of above equation,
T = m₂a ………….(2)
There is no acceleration along y direction for m₂.
N\(\hat{j}\) – m₂g\(\hat{j}\) = 0
By comparing the components on both sides of the above equation
N – m₂g = 0
N = m₂g ……….(3)
By substituting equation (2) in equation (1), we can find the tension T
m₂a – m₁g = -m₁a
m₂a + m₁a = m₁g
a = \(\frac{m_{1}}{m_{1}+m_{2}}\)g …………(4)
Tension in the string can be obtained by substituting equation (4) in equation (2)
T = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\)g ………..(5)

Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

Question 4.
Briefly explain the origin of friction. Show that in an inclined plane, angle of friction is pqual to angle of repose.
Answer:
If a very gentle force in the horizontal direction is given to an object at rest on the table it does not move. It is because of the opposing force exerted by the surface on the object which resists its motion. This force is called the frictional force. During the time of Newton and Galileo, frictional force was considered as one of the natural forces like gravitational force. But in the twentieth century, the understanding on atoms, electron and protons has changed the perspective.

The frictional force is actually the electromagnetic force between the atoms on the two surfaces. Even well polished surfaces have irregularities on the surface at the microscopic level. The component of force parallel to the inclined plane (mg sin θ) tries to move the object down. The component of force perpendicular to the inclined plane (mg cos θ) is balanced by the Normal force (N).
N = mg cos θ ………(1)

When the object just begins to move, the static friction attains its maximum value
f_s = \(f_{s}^{\max }\) = µ_s N
This friction also satisfies the relation
\(f_{s}^{\max }\) = µ_s mg sin θ ……….(2)
Equating the right hand side of equations (1) and (2),
(\(f_{s}^{\max }\))/N = sin θ / cos θ
From the definition of angle of friction, we also know that
tan θ = µ_s ………..(3)
in which θ is the angle of friction.
Thus the angle of repose is the same as angle of friction. But the difference is that the angle of repose refers to inclined surfaces and the angle of friction is applicable to any type of surface.

Question 5.
State Newton’s three laws and discuss their significance.
Answer:
First Law:
Every object continues to be in the state of rest or of uniform motion (constant velocity) unless there is external force acting on it.

Second Law:
The force acting on an object is equal to the rate of change of its momentum

Third Law:
For every action there is an equal and opposite reaction.

Significance of Newton’s Laws:
1. Newton’s laws are vector laws. The equation \(\overline{\mathrm{F}}\) = m\(\overline{\mathrm{a}}\) is a vector equation and essentially it is equal to three scalar equations. In Cartesian coordinates, this equation can be written as F_x\(\hat{i}\) + F_y\(\hat{j}\) + F_z\(\hat{j}\) = ma_x\(\hat{i}\) + ma_y\(\hat{j}\) + ma_z\(\hat{j}\)
By comparing both sides, the three scalar equations are

F_x = ma_x The acceleration along the x-direction depends only on the component of force acting along the x – direction.
F_y = ma_y The acceleration along the y direction depends only on the component of force acting along the y – direction.
F_z = ma_z The acceleration along the z direction depends only on the component of force acting along the z – direction.
From the above equations, we can infer that the force acting along y direction cannot alter the acceleration along x direction. In the same way, F_z cannot affect a_y and a_x. This understanding is essential for solving problems.

2. The acceleration experienced by the body at time t depends on the force which acts on the body at that instant of time. It does not depend on the force which acted on the body before the time t. This can be expressed as
\(\overline{\mathrm{F}}\)(t) = m\(\overline{\mathrm{a}}\)(t)
Acceleration of the object does not depend on the previous history of the force. For example, when a spin bowler or a fast bowler throws the ball to the batsman, once the ball leaves the hand of the bowler, it experiences only gravitational force and air frictional force. The acceleration of the ball is independent of how the ball was bowled (with a lower or a higher speed).

3. In general, the direction of a force may be different from the direction of motion. Though in some cases, the object may move in the same direction as the direction of the force, it is not always true. A few examples are given below.

Case 1:
Force and motion in the same direction:
When an apple falls towards the Earth, the direction of motion (direction of velocity) of the apple and that of force are in the same downward direction as shown in the Figure.

Case 2:
Force and motion not in the same direction:
The Moon experiences a force towards the Earth. But it actually moves in elliptical orbit. In this case, the direction of the force is different from the direction of motion as shown in Figure.

Case 3:
Force and motion in opposite direction:
If an object is thrown vertically upward, the direction of motion is upward, but gravitational force is downward as shown in the Figure.

Case 4:
Zero net force, but there is motion:
When a raindrop gets detached from the cloud it experiences both downward gravitational force and upward air drag force. As it descends towards the Earth, the upward air drag force increases and after a certain time, the upward air drag force cancels the downward gravity. From then on the raindrop moves at constant velocity till it touches the surface of the Earth. Hence the raindrop comes with zero net force, therefore with zero acceleration but with non-zero terminal velocity. It is shown in the Figure

4. If multiple forces \(\overrightarrow{\mathrm{F}}_{1}\), \(\overrightarrow{\mathrm{F}}_{2}\), \(\overrightarrow{\mathrm{F}}_{3}\),……. \(\overrightarrow{\mathrm{F}}_{n}\) act on the same body, then the total force (\(\overrightarrow{\mathrm{F}}_{net}\)) is equivalent to the vectorial sum of the individual forces. Their net force provides the acceleration.
\(\overrightarrow{\mathrm{F}}_{net}\) = \(\overrightarrow{\mathrm{F}}_{1}\) + \(\overrightarrow{\mathrm{F}}_{2}\) + \(\overrightarrow{\mathrm{F}}_{3}\) + ……… + \(\overrightarrow{\mathrm{F}}_{n}\)

Newton’s second law for this case is –
\(\overrightarrow{\mathrm{F}}_{net}\) = m\(\overline{\mathrm{a}}\)
In this case the direction of acceleration is in the direction of net force.
Example:
Bow and arrow

5. Newton’s second law can also be written in the following form.
Since the acceleration is the second derivative of position vector of the body \(\left(\vec{a}=\frac{d^{2} \vec{r}}{d t^{2}}\right)\)
the force on the body is –
\(\overline{\mathrm{F}}\) = m\(\frac{d^{2} \vec{r}}{d t^{2}}\)
From this expression, we can infer that Newton’s second law is basically a second order ordinary differential equation and whenever the second derivative of position vector is not zero, there must be a force acting on the body.

6. If no force acts on the body then Newton’s second law, m = \(\frac{d \vec{v}}{d t}\) = 0
It implies that \(\overline{\mathrm{v}}\) = constant. It is essentially Newton’s first law. It implies that the second law is consistent with the first law. However, it should not be thought of as the reduction of second law to the first when no force acts on the object. Newton’s first and second laws are independent laws. They can internally be consistent with each other but cannot be derived from each other.

7. Newton’s second law is cause and effect relation. Force is the cause and acceleration is the effect. Conventionally, the effect should be written on the left and cause on the right hand side of the equation. So the correct way of writing Newton’s second law is –
m\(\overline{\mathrm{a}}\) = \(\overline{\mathrm{F}}\) or \(\frac{d \vec{p}}{d t}\) = \(\overline{\mathrm{F}}\)

Question 6.
Explain the similarities and differences of centripetal and centrifugal forces.
Answer:
Salient features of centripetal and centrifugal forces.
Centripetal Force:

• It is a real force which is exerted on the body by the external agencies like gravitational force, tension in the string, normal force etc.
• Acts in both inertial and non-inertial frames
• It acts towards the axis of rotation or center of the circle in circular motion
  \(\left|\overrightarrow{\mathrm{F}}_{\mathrm{C}_{\mathrm{P}}}\right|\) = mω²r = \(\frac{m v^{2}}{r}\)
• Real force and has real effects
• Origin of centripetal force is interaction between two objects.
• In inertial frames centripetal force has to be included when free body diagrams are drawn.

Centrifugal Force:

• It is a pseudo force or fictitious force which cannot arise from gravitational force, tension force, normal force etc.
• Acts only in rotating frames (non-inertial frame)
• It acts outwards from the axis of rotation or radially outwards from the center of the circular motion
  \(\left|\overrightarrow{\mathrm{I}}_{\mathrm{C}_{\mathrm{f}}}\right|\) = mω²r = \(\frac{m v^{2}}{r}\)
• Pseudo force but has real effects
• Origin of centrifugal force is inertia. It does not arise from interaction. In an inertial frame the object’s inertial motion appears as centrifugal force in the rotating frame.
• In inertial frames there is no centrifugal force. In rotating frames, both centripetal and centrifugal force have to be included when free body diagrams are drawn.

Question 7.
Briefly explain ‘centrifugal force’ with suitable examples.
Answer:
To use Newton’s first and second laws in the rotational frame of reference, we need to include a Pseudo force called centrifugal force. This centrifugal force appears to act on the object with respect to rotating frames.

Circular motion can be analysed from two different frames of reference. One is the inertial frame (which is either at rest or in uniform motion) where Newton’s laws are obeyed. The other is the rotating frame of reference which is a non – inertial frame of reference as it is accelerating.

When we examine the circular motion from these frames of reference the situations are entirely different. To use Newton’s first and second laws in the rotational frame of reference, we need to include a pseudo force called ‘centrifugal force’. This ‘centrifugal force’ appears to act on the object with respect to rotating frames. To understand the concept of centrifugal force, we can take a specific case and discuss as done below.

Free body diagram of a particle including the centrifugal force Consider the case of a whirling motion of a stone tied to a string. Assume that the stone has angular velocity ω in the inertial frame (at rest). If the motion of the stone is observed from a frame which is also rotating along with the stone with same angular velocity ω then, the stone appears to be at rest.

This implies that in addition to the inward centripetal force – mω²r there must be an equal and opposite force that acts on the stone outward with value +mω²r . So the total force acting on the stone in a rotating frame is equal to zero (-mω²r + mω² r = 0). This outward force +mω²r is called the centrifugal force. The word ‘centrifugal’ means ‘flee from center’.

Note that the ‘centrifugal force’ appears to act on the particle, only when we analyse the motion from a rotating frame. With respect to an inertial frame there is only centripetal force which is given by the tension in the rstring. For this reason centrifugal force is called as a ‘pseudo force’. A pseudo force has no origin. It arises due to the non inertial nature of the frame considered. When circular motion problems are solved from a rotating frame of reference, while drawing free body diagram of a particle, the centrifugal force should necessarily be included as shown in the figure.

Question 8.
Briefly explain ‘rolling friction’.
Answer:
The invention of the wheel plays a crucial role in human civilization. One of the important applications is suitcases with rolling on coasters. Rolling wheels makes it easier than carrying luggage. When an object moves on a surface, essentially it is sliding on it. But wheels move on the surface through rolling motion. In rolling motion when a wheel moves on a surface, the point of contact with surface is always at rest. Since Rolling and kinetic friction the point of contact is at rest, there is no relative motion between the wheel and surface.

Hence the frictional fore is very less. At the same time if an object moves without a wheel, there is a relative motion between the object and the surface. As a result frictional force is larger. This makes it difficult to move the object. The figure shows the difference between rolling and kinetic friction. Ideally in pure rolling, motion of the point of contact with the surface should be at rest, but in practice it is not so.

Due to the elastic nature of the surface at the point of contact there will be some deformation on the object at this point on the wheel or surface as shown in figure. Due to this deformation, there will be minimal friction between wheel and surface. It is called ‘rolling friction. In fact, rolling friction’ is much smaller than kinetic friction.

Question 9.
Describe the method of measuring angle of repose.
Answer:
When objects are connected by strings and When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Case 1:
Vertical motion:
Consider two blocks of masses m₁ and m₂ (m₁> m₂) connected by a light and in extensible string that passes over a pulley as shown in Figure.

Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving, m₂ vertically upward and m_k, downward with same acceleration a. The gravitational force m₁g on mass m₁ is used in lifting the mass m₂. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.

Applying Newton’s second law for mass m₂
T \(\hat{j}\) – m₂g\(\hat{j}\) = m₂a\(\hat{j}\) The left hand side of the above equation is the total force that acts on m2 and the right hand side is the product of mass and acceleration of m₂ in y direction.
By comparing the components on both sides, we get
T = m₂g = m₂a ……….(1)

Similarly, applying Newton’s second law for mass m₂
T \(\hat{j}\) – m₁g\(\hat{j}\) = m₁a\(\hat{j}\)
As mass mj moves downward (-\(\hat{j}\)), its acceleration is along (-\(\hat{j}\))
By comparing the components on both sides, we get
T = m₁g = -m₁a
m₁g – T = m₁a ………..(2)

Adding equations (1) and (2), we get
m₁g – m₂g = m₁a + m₂a
(m₁ – m₂)g = (m₁ + m₂)a …………(3)
From equation (3), the acceleration of both the masses is –
a = (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ………..(4)
If both the masses are equal (m₁ = m₂), from equation (4)
a = 0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
T = m₂g = m₂(\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))
T = m₂g + m₂ (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ……….(5)
By taking m2g common in the RHS of equation (5)

Equation (4) gives only magnitude of acceleration
For mass m₁, the acceleration vector is given by \(\vec{a}\) = –\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\)\(\hat{j}\)
For mass m₂, the acceleration vector is given by \(\vec{a}\) = \(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\) \(\hat{j}\)

Case 2:
Horizontal motion:
In this case, mass m₂ is kept on a horizontal table and mass m₁, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface

As both the blocks are connected to the un stretchable string, if m₁ moves with an acceleration a downward then m₂ also moves with the same acceleration a horizontally.
The forces acting on mass m₂ are

1. Downward gravitational force (m₂g)
2. Upward normal force (N) exerted by the surface
3. Horizontal tension (T) exerted by the string

The forces acting on mass m₁ are

1. Downward gravitational force (m₁g)
2. Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure 2.

Applying Newton’s second law for m₁
T\(\hat{i}\) – m₁g\(\hat{j}\) = -m₁a\(\hat{j}\) (alongy direction)
By comparing the components on both sides of the above equation,
T – m₁g = -m₁a …………(1)
Applying Newton’s second law for m₂
T\(\hat{i}\) = m₁a\(\hat{i}\) (along x direction)
By comparing the components on both sides of above equation,
T = m₂a ………….(2)
There is no acceleration along y direction for m₂.
N\(\hat{j}\) – m₂g\(\hat{j}\) = 0
By comparing the components on both sides of the above equation
N – m₂g = 0
N = m₂g ……….(3)
By substituting equation (2) in equation (1), we can find the tension T
m₂a – m₁g = -m₁a
m₂a + m₁a = m₁g
a = \(\frac{m_{1}}{m_{1}+m_{2}}\)g …………(4)
Tension in the string can be obtained by substituting equation (4) in equation (2)
T = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\)g ………..(5)

Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

Question 10.
Explain the need for banking of tracks.
Answer:
In a leveled circular road, skidding mainly depends on the coefficient of static friction py. The coefficient of static friction depends on the nature of the surface which has a maximum limiting value. To avoid this problem, usually the outer edge of the road is slightly raised compared to inner edge. This is called banking of roads or tracks. This introduces an inclination, and the angle is called banking angle. “Let the surface of the road make angle 9 with horizontal surface. Then the normal force makes the same angle 9 with the vertical. When the car takes a turn.

there are two forces acting on the car:
(a) Gravitational force mg (downwards)
(b) Normal force N (perpendicular to surface)
We can resolve the normal force into two components N cos θ and N sin θ. The component balances the downward gravitational force ‘mg’ and component will provide the necessary centripetal acceleration. By using Newton second law.

The banking angle 0 and radius of curvature of the road or track determines the safe speed of the car at the turning. If the speed of car exceeds this safe speed, then it starts to skid outward but frictional force comes into effect and provides an additional centripetal force to prevent the outward skidding.

At the same time, if the speed of the car is little lesser than safe speed, it starts to skid inward and frictional force comes into effect, which reduces centripetal force to prevent inward skidding. However if the speed of the vehicle is sufficiently greater than the correct speed, then frictional force cannot stop the car from skidding.

Question 11.
Calculate the centripetal acceleration of Moon towards the Earth.
Answer:
The centripetal acceleration is given by a = \(\frac{v^{2}}{r}\) This expression explicitly depends on Moon’s speed which is nontrivial. We can work with the formula
ω²R_m = a_m
a_m is centripetal acceleration of the Moon due to Earth’s gravity, ω is angular velocity
R_m is the distance between Earth and the Moon, which is 60 times the radius of the Earth.
R_m = 60R = 60 x 6.4 x 10⁶ = 384 x 10⁶ m
As we know the angular velocity ω = \(\frac { 2π}{ T }\) and T = 27.3 days = 27.3 x 24 x 60 x 60 second = 2.358 x 10⁶ sec.
By substituting these values in the formula for acceleration
a₆ = \(\frac{\left(4 \pi^{2}\right)\left(384 \times 10^{6}\right)}{\left(2.358 \times 10^{8}\right)^{2}}\) = 0.00272 ms^-2

Samacheer Kalvi 11th Physics Laws of Motion Conceptual Questions

Question 1.
Why it is not possible to push a car from inside?
Answer:
While trying to push a car from outside, he pushes the ground backwards at an angle. The ground offers an equal reaction in the opposite direction, so car can be moved. But the person sits inside means car and the person becomes a single system, and the force given will be a internal force. According to Newton’s third law, total internal force acting on the system is zero and it cannot accelerate the system.

Question 2.
There is a limit beyond which the polishing of a surface increases frictional resistance rather than decreasing it why?
Answer:
Friction is a contact force. Friction is directly proportional to area of contact. In the normal surfaces there are bumps to interlock the surfaces in contact. But the surfaces are polished beyond certain limit. The area of contact will be increased and the molecules come closer to each other. It increases electrostatic force between the molecules. As a result it increases friction.

Question 3.
Can a single isolated force exist in nature? Explain your answer.
Answer:
No. According to Newton’s third law, for every action, there is an equal and opposite reaction. So, whatever case we consider, if there is an action there is always a reaction. So it is impossible.

Question 4.
Why does a parachute descend slowly?
Answer:
A parachute descends slowly because the surface area of parachute is large so that air gives more resistance when it descends down.

Question 5.
When walking on ice one should take short steps. Why?
Answer:
Let R represent the reaction offered by the ground. The vertical component R cos θ will balance the weight of the person and the horizontal component R sin θ will help the person to walk forward.
Now, normal reaction = R cos θ
Friction force = R sin θ
Coefficient of friction, µ = \(\frac {R sin θ}{R cos θ }\) = tan θ
In a long step, θ is more. So tan θ is more. But p has a fixed value. So, there is danger of slipping in a long step.

Question 6.
When a person walks on a surface, the frictional force exerted by the surface on the person is opposite to the direction of motion. True or false?
Answer:
False. In frictional force exerted by the surface on the person is in the direction of his motion. Frictional force acts as an external force to move the person. When the person trying to move, he gives a push to ground on the backward direction and by Newton’s third law he is pushed by the ground in the forward direction. Hence frictional force acts along the direction of motion.

Question 7.
Can the coefficient of friction be more than one?
Answer:
Yes. The coefficient of friction can be more than one in some cases such as silicone rubber. Coefficient of friction is the ratio of frictional force to normal force, i.e. F = μR. If p is greater than one means frictional force is greater than normal force. But in general case the value is usually between 0 and 1.

Question 8.
Can we predict the direction of motion of a body from the direction of force on it?
Answer:
Yes. The direction of motion is always opposite to the force of kinetic friction. By using the principle of equilibrium, the direction of force of static friction can be determined. When the object is in equilibrium, the frictional force must point in the direction which results as a net force is zero.

Question 9.
The momentum of a system of particles is always conserved. True or false?
Answer:
True. The total momentum of a system of particles is always constant i.e. conserved. When no external force acts on it.

Samacheer Kalvi 11th Physics Laws of Motion Numerical Problems

Question 1.
A force of 50 N act on the object of mass 20 kg. shown in the figure. Calculate the acceleration of the object in x and y directions.
Answer:
Given F = 50 N and m = 20 kg
(1) component of force along x – direction
F_x = F cos θ
= 50 x cos 30° = 43.30 N
a_x = \(\frac{F_{x}}{m}\) = \(\frac {43.30}{20}\) =2.165 ms^-2

(2) Component of force along y – direction
F_y = F sin θ = 50 sin 30° = 25 N
a_y = \(\frac{F_{y}}{m}\) = \(\frac {25}{20}\) = 1.25 ms^-2

Question 2.
A spider of mass 50 g is hanging on a string of a cob web as shown in the figure. What is the tension in the string?
Answer:
Given m = 50 g, T = ?
Tension in the string T = mg
= 50 x 10^-2 x 9.8 = 0.49 N

Question 3.
What is the reading shown in spring balance?

Answer:
When a spring balance hung on a rigid support and load is attached at its other end, the weight of the load exerts a force on the rigid support in turn support exerts equal and opposite force on that load, so that balance will be stretched. This is the principle of spring balance. Flence the answer is 4 kg.
Given: m = 2 kg, 0 = 30°.
Resolve the weight into its component as mg sin θ and mg cos θ.
Here mg sin θ acts parallel to the surface
∴ W = mg sin θ
= 2 x 9.8 x sin 30° = 2 x 9.8 x \(\frac {1}{2}\) = 9.8 N

Question 4.
The physics books are stacked on each other in the sequence: +1 volumes 1 and 2; +2 volumes 1 and 2 on a table
(a) Identify the forces acting on each book and draw the free body diagram.
(b) Identify the forces exerted by each book on the other.
Answer:
Let m₁, m₂, m₃, m_4, are the masses of +1 volume I and II and +2 volumes I & II

(a) Force on book m₄

• Downward gravitational force acting downward (m₃g)
• Upward normal force (N₃) exerted by book of mass m₃
  |

(b) Force on book m₃

• Downward gravitational force (m₃g)
• Downward force exerted by m₄ (N₄)
• Upward force exerted by m₂ (N₂)
  

(c) Force on book m₂

• Downward gravitational force (m₂g)
• Downward force exerted by m₃ (N₃)
• Upward force exerted by m₁ (CN₁)
  

(d) Force on book m₁

• Downward gravitational force exerted by earth (m₁g)
• Downward force exerted by m₂ (N₂)
• Upward force exerted by the table (N_table)
  

Question 5.
A bob attached to the string oscillates back and forth. Resolve the forces acting on the bob in to components. What is the acceleration experienced by the bob at an angle θ.
Answer:
The gravitational force (mg) acting downward can be resolved into two components as mg cos θ and mg sin θ
T – tension exerted by the string.
Tangential force F_T = ma_T = mg sin θ
∴ Tangential acceleration a_T = g sin θ
Centripetal force Fc = ma_c = T – mg cos θ
a_c = \(\frac { T – mg cos θ }{ m }\)

Question 6.
Two masses m₁ and m₂ are connected with a string passing over a friction-less pulley fixed at the comer of the table as shown in the figure. The coefficient of static friction of mass m₁ with the table is µ_s Calculate the minimum mass m₃ that may be placed on m₁to prevent it from sliding. Check if m₁ = 15 kg, m₂ = 10 kg, m₃ = 25 and µ_s = 0.2.

Solution:
Let m₃ is the mass added on m₁
Maximal static friction
\(f_{s}^{\max }\) = µ_sN = µ_s (m₁ + m₃ )g
Here
N = (m₁ + m₃ )g
Tension acting on string = T = m₂ g
Equate (1) and (2)
µ_s(m₁ + m₃) = m₂g
µ_sm₁ + µ_sm₃ = m₂
m₃ = \(f_{s}^{\max }\) – m₁

(ii) Given,
m₁ = 15 kg, m₂ = 10 kg : m₃ = 25 kg and µ_s = 0.2
m₃ = \(f_{s}^{\max }\) – m₁
m₃ = \(\frac {10}{ 0.2 }\) – 15 = 50 – 15 = 35 kg
The minimum mass m₃ = 35 kg has to be placed on ml to prevent it from sliding. But here m₃ = 25 kg only.
The combined masses (m₁ + m₃) will slide.

Question 7.
Calculate the acceleration of the bicycle of mass 25 kg as shown in Figures 1 and 2.

Answer:
Given:
Mass of bicycle m = 25 kg
Fig. I:
Net force acting in the forward direction, F = 500 – 400 = 100 N
acceleration a = \(\frac { F}{ m }\) = \(\frac { 100 }{25}\) = 4 ms^-2

Fig. II:
Net force acting on bicycle F = 400 – 400 = 0
∴ acceleration a = \(\frac { F}{ m }\) = \(\frac { 0}{25}\) = 0

Question 8.
Apply Lami’s theorem on sling shot and calculate the tension in each string?

Answer:
Given F = 50 N, θ = 30°
Here T is resolved into its components as T sin θ and T cos θ as shown.
According to Lami’s theorem,
\(\frac { F}{sin θ}\) = \(\frac { T}{sin (180 – θ)}\) = \(\frac { T}{sin (180 – θ)}\)
\(\frac { F}{sin θ}\) = \(\frac { T}{sin θ}\)
\(\frac { F}{2 sin θ cos θ}\) = \(\frac { T}{sin θ}\) [T = \(\frac { T}{2 cos θ}\) ]
T = \(\frac {T}{2 cos θ}\) = \(\frac { 50}{ 2 cos 30}\) = 28.868 N

Question 9.
A football player kicks a 0.8 kg ball and imparts it a velocity 12 ms^-1. The contact between the foot and ball is only for one – sixtieth of a second. Find the average kicking force.
Answer:
Given,
Mass of the ball = 0.8 kg
Final velocity (V) =12 ms^-1 and time t =\(\frac {1}{60}\) s
Initial velocity = 0
We know the average kicking force
F = ma = \(\frac {m(v – u)}{t}\) = \(\frac{0.8(12-0)}{\left(\frac{1}{60}\right)}\)
F = 576 N

Question 10.
A stone of mass 2 kg is attached to a string of length 1 meter. The string can withstand maximum tension 200 N. What is the maximum speed that stone can have during the whirling motion?
Solution:
Given,
Mass of a stone = 2 kg,
length of a string = 1 m
Maximum tension = 200 N
The force acting on a stone in the whirling motion is centripetal force. Which is provided by tension of the string.
T_max = F_max = \(\frac{m \mathrm{V}_{\mathrm{max}}^{2}}{r}\)
200 = \(v_{\max }^{2}\) = 100
\(v_{\max }^{2}\) = 10 ms^-1

Question 11.
Imagine that the gravitational force between Earth and Moon is provided by an invisible string that exists between the Moon and Earth. What is the tension that exists in this invisible string due’ to Earth’s centripetal force? (Mass of the Moon = 7.34 x 1022 kg, Distance between Moon and Earth = 3.84 x 108 m).

Solution:
Given,
Mass of the moon = 7.34 x 1022 kg
Distance between moon and earth = 3.84 x 108 m
Centripetal force = F = \(\frac{m \mathrm{V}^{2}}{r}\) = \(\frac{7.34 \times 10^{22} \times\left(1.023 \times 10^{3}\right)^{2}}{3.84 \times 10^{8}}\) = 2 x 10²⁰

Question 12.
Two bodies of masses 15 kg and 10 kg are connected with light string kept on a smooth surface. A horizontal force F = 500 N is applied to a 15 kg as shown in the figure. Calculate the tension acting in the string.

Answer:
Given,
m₁  = 15 kg, m₂  = 10 kg, F = 500 N
Tension acting in the string T = \(\frac{m_{2}}{m_{1}+m_{2}}\) F
T = \(\frac {10}{25}\) x 500 = 200 N

Question 13.
People often say “For every action there is an equivalent opposite reaction”. Here they meant ‘action of a human’. Is it correct to apply Newton’s third law to human actions? What is meant by ‘action’ in Newton third law? Give your arguments based on Newton’s laws.
Answer:
Newton’s third law is applicable to only human’s physical actions which involves physical force. Third law is not applicable to human’s psychological actions or thoughts.

Question 14.
A car takes a turn with velocity 50 ms^-1 on the circular road of radius of curvature To m. Calculate the centrifugal force experienced by a person of mass 60 kg inside the car?
Answer:
Given,
Mass of a person = 60 kg
Velocity of the car = 50 ms^-1
Radius of curvature = 10 m
Centrifugal force F = \(\frac{m \mathrm{V}^{2}}{r}\) = \(\frac{60 \times(50)^{2}}{10}\) = 15,000 N

Question 15.
A long stick rests on the surface. A person standing 10 m away from the stick. With what minimum speed an object of mass 0.5 kg should he thrown so that it hits the stick. (Assume the coefficient of kinetic friction is 0.7).
Answer:
Given,
Distance (s) = 10 m
Mass of the object (m) = 0.5 kg
Coefficient of kinetic friction (µ) = 0.7
Work done in moving a body in horizontal surface ω = µ_R x s = µmg x s
This work done is equal to initial kinetic energy of the object
\(\frac{1}{2} m v^{2}\) = µ mg s
\(\left|v^{2}\right|\) = 2 µgs = 2 x 0.7 x 9.8 x 10
v² = 14 x 9.8 = 137. 2
v = 11. 71 ms^-1

Samacheer Kalvi 11th Physics Laws of Motion Additional Questions Solved

Samacheer Kalvi 11th Physics Laws of Motion Multiple Choice Questions

Question 1.
The concept “force causes motion” was given by –
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(b) Aristotle

Question 2.
Who decoupled the motion and force?
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(a) Galileo

Question 3.
The inability of objects to move on its own or change its state of motion is called as –
(a) force
(b) momentum
(c) inertia
(d) impulse
Answer:
(c) inertia

Question 4.
Inertia means –
(a) inability
(b) resistance to change its state
(c) movement
(d) inertial frame
Answer:
(b) resistance to change its state

Question 5.
When a bus starts to move from rest, the passengers experience a sudden backward push is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(c) Inertia of rest

Question 6.
If the brake is applied in the moving bus suddenly, passengers move forward is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(a) Inertia of motion

Question 7.
In whirling motion, if the string is cut suddenly, the stone moves tangential to circle is an –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(b) Inertia of direction

Question 8.
Newtons laws are applicable in –
(a) Inertial frame
(b) non inertial frame
(c) in any frame
(d) none
Answer:
(a) Inertial frame

Question 9.
The accelerated train is an example for –
(a) inertial frame
(b) non-inertial frame
(c) both (a) and (b)
(d) none of the above
Answer:
(b) non-inertial frame

Question 10.
Rate of change of momentum of an object is equal to –
(a) acceleration
(b) work done
(c) force
(d) impulse
Answer:
(c) force

Question 11.
The product of mass and velocity is –
(a) force
(b) impulse
(c) momentum
(d) acceleration
Answer:
(c) momentum

Question 12.
Unit of momentum –
(a) kg ms^-2
(b) kg ms^-1
(c) MLT^-2
(d) MLT^-1
Answer:
(b) kg ms^-1

Question 13.
According to Newton’s third law –
(a) F₁₂ = F₂₁
(*) F₁₂ = -F₂₁
(c) F₁₂ + F₂₁ = 0
(d) F₁₂ x F₂₁ = 0
Answer:
(a) F₁₂ = F₂₁

Question 14.
According to Newton’s third law –
(a) \(\overrightarrow{\mathrm{F}_{12}}=\overrightarrow{\mathrm{F}_{21}}\)
(b) \(\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}\)
(c) \(\mathrm{F}_{12}+\mathrm{F}_{21}\) = 0
(d) \(\mathrm{F}_{12}x\mathrm{F}_{21}\) = 0
Answer:
(b) \(\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}\)

Question 15.
The law which is valid in both inertial and non-inertial frame is –
(a) Newton’s first law
(b) Newton’s second law
(c) Newton’s third law
(d) none
Answer:
(c) Newton’s third law

Question 16.
When a force is applied on a body, it can change –
(a) velocity
(b) momentum
(c) direction of motion
(d) all the above
Answer:
(d) all the above

Question 17.
The rate of change of velocity is 1 ms^-2 when a force is applied on the body of mass 75 gm the force is –
(a) 75 N
(b) 0.75 N
(c) 0.075 N
(d) 0.75 x 10^-3 N
Answer:
(c) Force is given by
F = m a
= 75 gm x 1 cm s^-2 = 75 x 10^-3 x 1 = 75 x 10^-3 = 0.075 N

Question 18.
The action and reaction forces acting on –
(a) same body
(b) different bodies
(c) either same or different bodies
(d) none of the above
Answer:
(b) different bodies

Question 19.
Newton’s first law of motion gives the concept of –
(a) velocity
(b) energy
(c) momentum
(d) Inertia
Answer:
(d) Inertia

Question 20.
Inertia of a body has direct dependence on –
(a) velocity
(b) area
(c) mass
(d) volume
Answer:
(c) mass

Question 21.
If a car and a scooter have the same momentum, then which one is having greater speed?
(a) scooter
(b) car
(c) both have same velocity
(d) data insufficient
Answer:
(a) scooter

Question 22.
Newton’s second law gives –
(a) \(\overrightarrow{\mathrm{F}} \propto \frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(b) \(\overrightarrow{\mathrm{F}}=\frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(c) \(\overrightarrow{\mathrm{F}}=m \vec{a}\)
(d) all the above
Answer:
(d) all the above

Question 23.
1 dyne is –
(a) 10⁵N
(b) 10^-5N
(c) 1N
(d) 10^-3N
Answer:
(b) 10^-5N

Question 24.
If same force is acting on two masses m₁ and m₂, and the accelerations of two bodies are a₁ and a₂ respectively, then –
(a) \(\frac{a_{2}}{a_{1}}=\frac{m_{2}}{m_{1}}\)
(b) \(\frac{a_{1}}{a_{2}}=\frac{m_{1}}{m_{2}}\)
(c) \(\frac{a_{1}}{a_{2}}=\frac{m_{2}}{m_{1}}\)
(d) m₁ a₁ + m₂a₂ = 0
Answer:
(c) \(\frac{a_{1}}{a_{2}}=\frac{m_{2}}{m_{1}}\)

Question 25.
If a force \(\overline{\mathrm{F}}\) = 3\(\hat{i}\) – 4\(\hat{j}\) N produces an acceleration of 10 ms^-2 on a body, then the mass of a body is –
(a) 10 kg
(b) 9 kg
(c) 0.9 kg
(d) 0.5 kg
Answer:
\(\overline{\mathrm{F}}\) = 3\(\hat{i}\) – 4\(\hat{j}\)
Magnitude:
|\(\overline{\mathrm{F}}\)| = \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5N
F = ma
⇒ m = \(\frac{|\mathrm{F}|}{a}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\) = 0.5 kg

Question 26.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop?
(a) 0.75 s
(b) 1.33 s
(c) 6 s
(d) 35 s
Answer:
Acceleration a = \(\frac{-F}{m}\) = \(\frac{50}{20}\) = – 2.5 ms^-2
u = l5 ms^-1
v = 0
t = ?
v = u + at
0 = 15 – 2.5t
t = \(\frac{15}{2.5}\) = 6s

Question 27.
Rain drops come down with –
(a) zero acceleration and non zero velocity
(b) zero velocity with non zero acceleration
(c) zero acceleration and non zero net force
(d) none
Answer:
(a) zero acceleration and non zero velocity

Question 28.
If force is the cause then the effect is –
(a) mass
(b) potential energy
(c) acceleration
(d) Inertia
Answer:
(c) acceleration

Question 29.
In free body diagram, the object is represented by a –
(a) line
(b) arrow
(c) circle
(d) point
Answer:
(d) point

Question 30.
When an object of mass m slides on a friction less surface inclined at an angle 0, then normal force exerted by the surface is –
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg tan θ
Answer:
(b) mg cos θ

Question 31.
The acceleration of the sliding object in an inclined plane –
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg sin θ
Answer:
(c) g sin θ

Question 32.
The speed of an object sliding in an inclined plane at the bottom is –
(a) mg cos θ
(b) \(\sqrt{2 s g sin θ}\)
(c) \(\sqrt{2 s g cos θ}\)
(d) \(\sqrt{2 s g tan θ}\)
Answer:
(b) \(\sqrt{2 s g sin θ}\)

Question 33.
The acceleration of two bodies of mass m₁ and m₂ in contact on a horizontal surface is –
(a) \(a=\frac{\mathbf{F}}{m_{1}}\)
(b) \(a=\frac{F}{m_{2}}\)
(c)  \(a=\frac{\mathrm{F}}{m_{1}+m_{2}}\)
(d) \(a=\frac{\mathrm{F}}{m_{1} m_{2}}\)
Answer:
(c)  \(a=\frac{\mathrm{F}}{m_{1}+m_{2}}\)

Question 34.
Two blocks of masses m₁ and m₂ (m₁ > m₂) in contact with each other on frictionless, horizontal surface. If a horizontal force F is given on m₁, set into motion with acceleration a, then reaction force on mass m₁ by m₂, is –
(a) \(\frac{\mathrm{F} m_{1}}{m_{1}+m_{2}}\)
(b) \(\frac{m_{1} m_{2}}{\mathrm{F} m_{1}}\)
(c) \(\frac{m_{1} m_{2}}{\mathrm{F} m_{2}}\)
(d) \(\frac{\mathrm{F} m_{2}}{m_{1}+m_{2}}\)
Answer:
(d) \(\frac{\mathrm{F} m_{2}}{m_{1}+m_{2}}\)

Question 35.
If two masses m₁ and m₂ (m₁ > m₂) tied to string moving over a frictionless pulley, then acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 36.
if two masses m₁ and m₂ (m₁ > m₂)tied to string moving over a frictionless pulley, then acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 37.
Three massses is in contact as shown. If force F is applied to mass m₁, the acceleration of three masses is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Answer:
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Question 38.
Three masses in contact is as shown above. If force F is applied to mass m₁ then the contact force acting on mass m₂ is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)

Question 39.
Three masses is contact as shown. It force F is applied to mass m₁, then the contact force acting on mass m₃ is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Question 40.
Two masses connected with a string. When a force F is applied on mass m₂. The acceleration produced is –

(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}}\)
(b) \(\frac{\mathbf{F}}{m_{1}-m_{2}}\)
(c) \(\frac{m_{1}+m_{2}}{\mathrm{F}}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}}\)

Question 41.
Two masses connected with a string. When a force F is applied on mass m₂. The force acting on m₁ is –
(a) \(\frac{m_{1} \mathrm{F}}{m_{1}+m_{2}}\)
(b) \(\frac{m_{2} \mathrm{F}}{m_{1}+m_{2}}\)
(c) \(\frac{m_{1}+m_{2}}{m_{1}} \mathbf{F}\)
(d) \(\frac{m_{1}+m_{2}}{m_{2}} \mathbf{F}\)
Answer:
(b) \(\frac{m_{2} \mathrm{F}}{m_{1}+m_{2}}\)

Question 42.
If a block of mass m lying on a frictionless inclined plane of length L height h and angle of inclination θ, then the velocity at its bottom is –
(a) g sin θ
(b) g cos θ
(c) \(\sqrt{2 g h}\)
(d) \(\sqrt{2 a sin θ}\)
Answer:
(c) \(\sqrt{2 g h}\)

Question 43.
If a block of mass m lying on a frictionless inclined plane of length L, height h and angle of inclination θ, then the time take taken to reach the bottom is –
(a) g sing θ
(b) sin θ \(\sqrt{\frac{2 h}{g}}\)
(c) sin θ \(\sqrt{\frac{g}{h}}\)
(d) \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)
Answer:
(d) \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)

Question 44.
A rocket works on the principle of conservation of –
(a) energy
(b) mass
(c) angular momentum
(d) linear momentum
Answer:
(b) mass

Question 45.
A bomb at rest explodes. The total momentum of all its fragments is –
(a) zero
(b) infinity
(c) always 1
(d) always greater then 1
Answer:
(a) zero

Question 46.
A block of mass m₁ is pulled along a horizontal friction-less surface by a rope of mass m₂ If a force F is given at its free end. The net force acting on the block is –
(a) \(\frac{m_{1} \mathrm{F}}{m_{1}-m_{2}}\)
(b) F
(c) \(\frac{m_{2} \mathrm{F}}{\left(m_{1}+m_{2}\right)}\)
(d) \(\frac{m_{1} \mathrm{F}}{\left(m_{1}+m_{2}\right)}\)
Answer:
(b) F

Question 47.
A block of mass m is pulled along a horizontal surface by a rope. The tension in the rope will be same at all the points –
(a) if the rope is accelerated
(b) if the rope is mass less
(c) always
(d) none of the above
Answer:
(b) if the rope is mass less

Question 48.
The lines of forces act at a common point is called as –
(a) concurrent forces
(b) co-planar forces
(c) equilibrium
(d) resultant
Answer:
(a) concurrent forces

Question 49.
If the lines of forces act in the same plane, they can be –
(a) concurrent forces
(b) coplanar forces
(c) either concurrent force or coplanar forces
(d) Lami’s force
Answer:
(d) concurrent forces

Question 50.
Lami’s theorem is applicable only when the system of forces are is –
(a) same plane
(b) different plane
(c) equilibrium
(d) none of the above
Answer:
(c) equilibrium

Question 51.
Due to the action of internal forces of the system, the total linear momentum of the system is –
(a) a variable
(b) a constant
(c) always zero
(d) always infinity
Answer:
(c) always zero

Question 52.
The velocity with which a gun suddenly moves backward after firing is –
(a) linear velocity
(b) positive velocity
(c) recoil velocity
(d) v₁ + v₂
Answer:
(c) recoil velocity

Question 53.
If a very large force acts on an object for a very short duration, then the force is called as –
(a) Newtonian force
(b) impulsive force
(c) concurrent force
(d) coplanar force
Answer:
(A) impulsive force

Question 54.
The unit of impulse is –
(a) Nm
(b) Ns
(c) Nm²
(d) Ns^-2
Answer:
(b) Ns

Question 55.
The force which always opposes the relative motion between an object and the surface where it is placed is –
(a) concurrent force
(b) frictional force
(c) impulsive force
(d) coplanar force
Answer:
(b) frictional force

Question 56.
The force which opposes the initiation of motion of an object on the surface is –
(a) static friction
(b) kinetic friction
(c) friction
(d) zero
Answer:
(d) static friction

Question 57.
When the object is at rest, the resultant of gravitational force and upward normal force is –
(a) Static force
(b) zero
(c) one
(d) infinity
Answer:
(b) zero

Question 58.
The magnitude of static frictional force d lies between –
(a) 0 ≤ f ≤ µ_sN
(b) 0 ≥f ≥ µ_sN
(c) 0 and 1
(d) 0 and minimal static frictional force.
Answer:
(a) 0 ≤ f ≤ µ_sN

Question 59.
The unit of co-efficient of static friction is –
(a) N
(b) N m
(c) N s
(d) no unit
Answer:
(d) no unit

Question 60.
If the object is at rest and no external force is applied on the object, the static friction acting on the object is –
(a) µ_sN
(b) zero
(c) one
(d) infinity
Answer:
(d) no unit

Question 61.
When object begins to slide, the static friction acting on the object attains –
(a) zero
(b) minimum
(c) maximum
(d) infinity
Answer:
(c) maximum

Question 62.
The static friction does not depend upon –
(a) the area of contact
(b) normal force
(c) the magnitude of applied force
(d) none of the above
Answer:
(a) the area of contact

Question 63.
Which of the following pairs of materials has minimum amount of coefficient of static friction is –
(a) Glass and glass
(b) wood and wood
(c) ice and ice
(d) steel and steel
Answer:
(c) ice and ice

Question 64.
Kinetic friction is also called as –
(a) sliding friction
(b) dynamic friction
(c) both (a) and (b)
(d) static friction
Answer:
(c) both (a) and (b)

Question 65.
The unit of coefficient of kinetic friction is/has –
(a) Nm
(b) Ns
(c) Nm²
(d) no unit
Answer:
(d) no unit

Question 66.
The nature of materials in mutual contact decides –
(a) µs
(b) µk
(c) µs or µk
(d) none
Answer:
(c) µs or µk

Question 67.
Coefficient of kinetic friction is less than –
(a) O
(b) one
(c) µs
(d) µsN
Answer:
(c) µs

Question 68.
The static friction –
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(a) increases linearly

Question 69.
The kinetic friction –
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(b) is constant

Question 70.
Kinetic friction is independent of –
(a) nature of materials
(b) temperature of the surface
(c) applied force
(d) none of the above
Answer:
(c) applied force

Question 71.
The angle between the normal force and the resultant force of normal force and maximum frictional force is –
(a) angle of friction
(b) angle of repose
(c) angle of inclination
(d) none of the above
Answer:
(a) angle of friction

Question 72.
The angle friction θ is given by –
(a) tan µ_s
(b) tan^-1 µ_s
(c) \(\frac{f S^{\mathrm{max}}}{N}\)
(d) sin^-1 µ_s
Answer:
(b) tan^-1 µ_s

Question 73.
The angle of inclined plane with the horizontal such that an object placed on it begins to slide is –
(a) angle of friction
(b) angle of repose
(c) angle of response
(d) angle of retardation
Answer:
(b) angle of repose

Question 74.
Comparatively, which of the following has lesser value than others?
(a) static friction
(b) kinetic friction
(c) Rolling friction
(d) skiping friction
Answer:
(c) Rolling friction

Question 75.
The origin of friction is –
(a) electrostatic interaction
(b) electromagnetic interaction magnetic
(c) photon interaction
(d) interaction
Answer:
(b) electromagnetic interaction

Question 76.
Friction can be reduced by –
(a) polishing
(b) lubricating
(c) using ball bearings
(d) all the above
Answer:
(c) using ball bearings

Question 77.
For a particle revolving in a circular path, the acceleration of the particle is –
(a) along the tangent
(b) along the radius
(c) along the circumference of the circle
(d) zero
Answer:
(b) along the radius

Question 78.
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 of the above
Answer:
(b) zero

Question 79.
A bullet hits and gets embedded in a solid block resting on a horizontal frictionless table. Which of the following is conserved?
(a) Momentum and kinetic energy
(b) kinetic energy alone
(c) Momentum alone
(d) potential energy alone
Answer:
(c) Momentum alone

Question 80.
The origin of the centripetal force can be –
(a) gravitational force
(b) frictional force
(c) coulomb force
(d) all the above
Answer:
(d) all the above

Question 81.
Centripetal acceleration is –
(a) \(\frac{m v^{2}}{r}\)
(b) \(\frac{v^{2}}{r}\)
(c) r v²
(d) rω
Answer:
(b) \(\frac{v^{2}}{r}\)

Question 82.
Centripetal acceleration is –
(a) \(\frac{m v^{2}}{r}\)
(b) r ω²
(c) rv²
(d) rω
Answer:
(c) rω²

Question 83.
The centripetal force is –
(a) \(\frac{m v^{2}}{r}\)
(b) rω²
(c) both (a) and (b)
(d) none
Answer:
(c) both (a) and (b)

Question 84.
When a car is moving on a circular track the centripetal force is due to –
(a) gravitational force
(b) frictional force
(c) magnetic force
(d) elastic force
Answer:
(b) frictional force

Question 85.
If the road is horizontal then the normal force and gravitational force are –
(a) equal and along the same direction
(b) equal and opposite
(c) unequal and along the same direction
(d) unequal and opposite
Answer:
(b) equal and opposite

Question 86.
The velocity of a car for safe turn on leveled circular road –
(a) \(v \leq \sqrt{\mu_{s} r g}\)
(b) \(v \geq \sqrt{\mu_{s} r g}\)
(c) \(v=\sqrt{\mu_{s} rg}\)
(d) \(v \leq \mu_{s} rg\)
Answer:
(a) \(v \leq \sqrt{\mu_{s} r g}\)

Question 87.
In a leveled circular road, skidding mainly depends on –
(a) µ_s
(b) µ_k
(c) acceleration
(d) none
Answer:
(a) µ_s

Question 88.
The speed of a car to move on the banked road so that it will have safe turn is –
(a) µ_srg
(b) \(\sqrt{r g \tan \theta}\)
(c) rg tan θ
(d) r²g tan θ
Answer:
(b) \(\sqrt{r g \tan \theta}\)

Question 89.
Centrifugal force is a –
(a) pseudo force
(b) real force
(c) forced acting towards center
(d) none of the above
Answer:
(a) pseudo force

Question 90.
Origin of centrifugal force is due to –
(a) interaction between two
(b) inertia
(c) electromagnetic interaction
(d) inertial frame
Answer:
(b) inertia

Question 91.
Centripetal force acts in –
(a) inertial frame
(b) non inertial frame
(c) both (a) and (h)
(d) linear motion
Answer:
(c) both (a) and (b)

Question 92.
Centrifugal force acts in –
(a) inertial frame
(b) non inertial frame
(c) both (a) and (b)
(d) linear motion
Answer:
(b) non inertial frame

Question 93.
A cricket ball of mass loo g moving with a velocity of 20 ms-¹ is brought to rest by a player in 0.05s the impulse of the ball is –
(a) 5 Ns
(b) – 2 Ns
(c) – 2.5 Ns
(d) zero
Answer:
(b) – 2 Ns
mass = 0.1 kg
Initial velocity t = 20 ms^-1
Final velocity y = 0
Change in momentum in impulse = m(v – u) = 0.1(0 – 20) = – 2 Ns

Question 94.
If a stone tied at the one end of a string of length 0.5 m is whirled in a horizontal circle with a constant speed 6 ms^-1  then the acceleration of the shone is –
(a) 12 ms^-2
(b) 36 ms^-2
(c) 2π² ms^-2
(d) 72 ms^-2
Answer:
(d) Centripetal acceleration = \(\frac{v^{2}}{r}\) = \(\frac{6^{2}}{0.5}\) = \(\frac{36}{0.5}\) = 72 ms^-2

Question 95.
A block of mass 3 kg is at rest on a rough inclined plane with angle of inclination 30° with horizontal. If .is 0.7, then the frictional force is –
(a) 17.82 N
(b) 1.81 N
(c) 3.63 N
(d) 2.1 N
Answer:
(a) Frictional force = µmg cos θ = 0.7 x 3 x 9.8 cos 30° = 17.82 N

Question 96.
Two masses 2 kg and 4 kg are tied at the ends of a mass less string and which is passing over a friction-less pulley. The tension in the string is –
(a) 3.68 N
(b) 78.4 N
(c) 26 N
(d) 13.26 N
Answer:
(c) Tension in the string T = \(\frac{2 m_{1} m_{2}}{\left(m_{1}+m_{2}\right)}\)g
T = \(\frac{2 x 2 x 4}{2 + 4}\) x 9.8 = \(\frac{16}{6}\) x 9.8 = 26.13 N

Question 97.
A bomb of 10 kg at rest explodes into two pieces of mass 4 kg and 6 kg. if the velocity of 4 kg mass is 6 ms-¹ then the velocity of 6 kg is –
(a) – 4 ms^-1
(b) – 6 ms^-1
(c) – 24 ms^-1
(d) – 2.2 ms^-1
Answer:
(a) According to law of conservation of momentum
m₁v₁ + m₂v₂ = 0
v₂ = –\(\frac{m_{1} v_{1}}{m_{2}}\) = \(\frac{4 x 6}{6}\) = -4ms^-1

Question 98.
A body is subjected under three concurrent forces and it is in equilibrium. The resultant of any two forces is –
(a) coplanar with the third force
(b) is equal and opposite to third force
(c) both (a) and (b)
(d) none of the above
Answer:
(c) both (a) and (b)

Question 99.
An impulse is applied to a moving object with the force at an angle of 20° with respect to velocity vector. The angle between the impulse vector and the change in momentum vector is –
(a) 0°
(b) 30°
(c) 60°
(d) 120°
Answer:
(a) Impulse and change in momentum are in same direction. So the angle is zero.

Question 100.
A bullet of mass m and velocity v₁ is fired into a large block of wood of mass M. The final velocity of the system is-
(a) \(\frac{v_{1}}{m+\mathrm{M}}\)
(b) \(\frac{m v_{1}}{m+\mathrm{M}}\)
(c) \(\frac{m+m}{m} v_{1}\)
(d) \(\frac{m+m}{m-M} v_{1}\)
Answer:
(b) \(\frac{m v_{1}}{m+\mathrm{M}}\)

Question 101.
A block of mass 2 kg is placed on the floor. The co – efficient of static friction is 0.4. The force of friction between the block and floor is –
(a) 2.8 N
(b) 7.8 N
(c) 2 N
(d) zero
Answer:
(b) The force required to move = = µR = µmg = 0.4 x 2 x 9.8 = 7.84 N

Question 102.
A truck weighing 1000 kg is moving with velocity of 50 km/h on smooth horizontal roads. A mass of 250 kg is dropped into it. The velocity with which it moves now is –
(a) 12.5 km/h
(b) 20 km/h
(c) 40 km/h
(d) 50 km/h
Answer:
(c) According to law of conservation of linear momentum
m₂ v₂ = (m₁ + m₂)v₂
v₂ = \(\frac{m_{1} v_{1}}{m_{1}+m_{2}}\) = \(\frac{1000 \times 50}{1250}\) = 40 km/h

Question 103.
A body of mass loo g is sliding from an inclined plane of inclination 30°. if u = 1.7, then the frictional force experienced is –
(a) \(\frac{3.4}{\sqrt{3}}\)N
(b) 1.47 N
(c) \(\frac{\sqrt{3}}{3.4}\)N
(d) 1.38 N
Answer:
(b) Frictional force F = µ mg cos θ = 1.7 x 0.1 x 10 cos 30°= \(\frac{1.7}{2}\) x \(\sqrt{3}\) = 1.47 N

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (1 Mark)

Question 1.
A passenger sitting in a car at rest, pushes the car from within. The car doesn’t move, why?
Answer:
For motion, there should be external force.

Question 2.
Give the magnitude and directions of the net force acting on a rain drop falling with a constant speed.
Answer:
as \(\overline{\mathrm{a}}\) = 0 so \(\overline{\mathrm{F}}\) = 0.

Question 3.
Why the passengers in a moving car are thrown outwards when it suddenly takes a turn?
Answer:
Due to inertia of direction.

Question 4.
You accelerate your car forward. What is the direction of the frictional force on a package resting on the floor of the car?
Answer:
The package in the accelerated car (a non inertial frame) experiences a Pseudo force in a direction opposite to that of the motion of the car. The frictional force on the package which acts opposite to this pseudo force is thus in the same direction (forward) as that of the car.

Question 5.
What is the purpose of using shockers in a car?
Answer:
To decrease the impact of force by increasing the time for which force acts.

Question 6.
Why are types made of rubber not of steel?
Answer:
Since coefficient of friction between rubber and road is less than the coefficient of friction between steel and road.

Question 7.
Wheels are made circular. Why?
Answer:
Rolling friction is less than sliding friction.

Question 8.
If a ball is thrown up in a moving train, it comes back to the thrower’s hands. Why?
Answer:
Both during its upward and downward motion, the ball continues to move inertia of motion with the same horizontal velocity as the train. In this period, the ball covers the same horizontal distance as the train and so it comes back to the thrower’s hand.

Question 9.
Calculate the force acting on a body which changes the momentum of the body at the rate of 1 kg-m/s² .
Answer:
As F = rate change of momentum
F = 1 kg-m/s² = 1N

Question 10.
On a rainy day skidding takes place along a curved path. Why?
Answer:
As the friction between the types and road reduces on a rainy day.

Question 11.
Why does a gun recoils when a bullet is being fired?
Answer:
To conserve momentum.

Question 12.
Why is it difficult to catch a cricket ball than a tennis ball even when both are moving with the same velocity?
Answer:
Being heavier, cricket ball has higher rate of change of momentum during motion so more force sumed.

Question 13.
The distance travelled by a moving body is directly proportional to time. Is any external force acting on it?
Answer:
As s ∝ t, so acceleration a = 0, therefore, no external force is acting on the body.

Question 14.
Calculate the impulse necessary to stop a 1500 kg car moving at a speed of 25 ms^-1.
Answer:
Use formula I = change in momentum = m(v – u) (Impulse – 37500 Ns)

Question 15.
Lubricants are used between the two parts of a machine. Why?
Answer:
To reduce friction and so to reduce wear and tear.

Question 16.
What provides the centripetal force to a car taking a turn on a level road?
Answer:
Force of friction between the type and road provides centripetal force.

Question 17.
A body is acted upon by a number of external forces. Can it remain at rest?
Answer:
Yes, if the external forces acting on the body can be represented in magnitude and direction by the sides of a closed polygon taken in the same order.

Question 18.
Bodies of larger mass need greater initial effort to put them in motion. Why?
Answer:
As F = ma so for given a, more force will be required to put a large mass in motion.

Question 19.
An athlete runs a certain distance before taking a long jump Why?
Answer:
So that inertia of motion may help him in his muscular efforts to take a longer jump.

Question 20.
Action and reaction forces do not balance each other. Why?
Answer:
As they acts on different bodies.

Question 21.
The wheels of vehicles are provided with mudguards. Why?
Answer:
When the wheel rotates at a high speed, the mud sticking to the wheel flies off tangentially, this is due to inertia of direction. If order that the flying mud does not spoil the clothes of passer by the wheels are provided with mudguards.

Question 22.
China wares are wrapped in straw paper before packing. Why?
Answer:
The straw paper between the China ware increases the Time of experiencing the jerk during transportation. Hence impact of force reduces on China wares.

Question 23.
Why is it difficult to walk on a sand?
Answer:
Less reaction force.

Question 24.
The outer edge of a curved road is generally raised over the inner edge Why?
Answer:
In addition to the frictional force, a component of reaction force also provides centripetal force.

Question 25.
Explain why the water doesn’t fall even at the top of the circle when the bucket full of water is upside down rotating in a vertical circle?
Answer:
Weight of the water and bucket is used up in providing the necessary centripetal force at the top of the circle.

Question 26.
Why does a speedy motor cyclist bends towards the center of a circular path while taking a turn on it?
Answer:
So that in addition of the frictional force, the horizontal component of the normal reaction also provides the necessary centripetal forces.

Question 27.
An impulse is applied to a moving object with a force at an angle of 20° wr.t. velocity vector, what is the angle between the impulse vector and change in momentum vector ?
Answer:
Impulse and change in momentum are along the same direction. Therefore angle between these two vectors is zero.

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (2 Marks)

Question 28.
A man getting out of a moving bus runs in the same direction for a certain distance. Comment.
Answer:
Due to inertia of motion.

Question 29.
If the net force acting upon the particle is zero, show that its linear momentum remains constant.
Answer:
As F x \(\frac {dp}{dt}\)
when F = 0, \(\frac {dp}{dt}\) = 0 so P = constant

Question 30.
A force of 36 dynes is inclined to the horizontal at an angle of 60°. Find the acceleration in a mass of 18 g that moves in a horizontal direction.
Answer:
F = 36 dyne at an angle of 60°
F_x = F cos 60° = 18 dyne
F_x = ma_x
So a_x = \(\frac{F_{x}}{m}\) = 1 cm /s²

Question 31.
The motion of a particle of mass m is described by h = ut + \(\frac {1}{2}\) gt². Find the force acting on particle.
Answer:
a = ut + \(\frac {1}{2}\) gt²
find a by differentiating h twice w.r.t.
a = g
As F = ma so F = mg (answer)

Question 32.
A particle of mass 0.3 kg is subjected to a force of F = -kx with k= 15 Nmr^-1. What will be its initial acceleration if it is released from a point 20 cm away from the origin?
Answer:
As F = ma so F = -kx = ma
a = \(\frac {-kx}{m}\)
for x = 20 cm, ⇒ a = -10 m/s².

Question 33.
A 50 g bullet is fired from a 10 kg gun with a speed of 500 ms-¹. What is the speed of the recoil of the gun?
Answer:
Initial momentum = 0
Using conservation of linear momentum mv + MV = 0
V = \(\frac {-mv}{M}\) ⇒ V = 2.5 m/s

Question 34.
Smooth block is released at rest on a 45° incline and then slides a distance d. If the time taken of slide on rough incline is n times as large as that to slide than on a smooth incline. Show that coefficient of friction, µ = \(\left(1-\frac{1}{n^{2}}\right)\)
Answer:
When there is no friction, the block slides down the inclined plane with acceleration. a = g sin θ
when there is friction, the downward acceleration of the block is a’ = g (sin θ – µ cos θ)
As the block Slides a distance d in each case so
d = \(\frac {1}{2}\) at² = \(\frac {1}{2}\) a’t’²
\(\frac{a}{a^{\prime}}=\frac{t^{\prime 2}}{t^{2}}=\frac{(n t)^{2}}{t^{2}}\) = n²
or \(\frac {g sin θ}{g(sin θ – µ cos θ)}\) = n²
Solving, we get (Using θ = 45°)
µ = 1 – \(\frac{1}{n^{2}}\)

Question 35.
A spring balance is attached to the ceiling of a lift. When the lift is at rest spring balance reads 49 N of a body hang on it. If the lift moves:

1. Downward
2. upward, with an acceleration of 5 ms²
3. with a constant velocity.

What will be the reading of the balance in each case?
Answer:
1.  R = m(g – a) = 49 N
so = m = \(\frac {49}{9.8}\) = 5 kg
R = 5 (9.8 – 5)
R = 24 N

2. R = m(g + a)
R = 5 (9.8 + 5)
R = 74 N

3.  as a = 0 so R = mg = 49 N

Question 36.
A bob of mass 0.1 kg hung from the ceiling of room by a string 2 m long is oscillating. At its mean position the speed of a bob is 1 ms^-1. What is the trajectory of the ‘oscillating bob if the string is cut when the bob is –

1. At the mean position
2. At its extreme position.

Answer:

1. Parabolic
2. vertically downwards

Question 37.
A block placed on a rough horizontal surface is pulled by a horizontal force F. Let f be the force applied by the rough surface on the block. Plot a graph of f versus F.
Answer:

Unto point A, f = F (50 Long as block is stationary) beyond A, when F increases, block starts moving f remains constant.

Question 38.
A mass of 2 kg is suspended with thread AB. Thread CD of the same type is attached to the other end of 2 kg mass.

• Lower end of the lower thread is pulled gradually, hander and hander is the downward direction so as to apply force on AB Which of the thread will break & why?
• If the lower thread is pulled with a jerk, what happens?

Answer:

• Thread AB breaks down
• CD will break.

Question 39.
A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is p and the acceleration due to gravity is g, calculate the minimum force required to be applied by the finger to held the block against the wall?
Answer:

For the block not to fall f = Mg
But f = µR = µF so
µF = Mg
F = \(\frac {Mg}{µ}\)

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (3 Marks) & Numericals

Question 40.
A block of mass 500 g is at rest on a horizontal table. What steady force is required to give the block a velocity of 200 cm s^-2 in 4 s?
Answer:
Use F – ma
a = \(\frac {v – u}{ t }\) = \(\frac {200- 0}{ 4 }\) = 50 cm/s²
F = 500 x 50 = 25,000 dyne.

Question 41.
A force of 98 N is just required to move a mass of 45 kg on a rough horizontal surface. Find the coefficient of friction and angle of friction?
Answer:
F = 48 N,R = 45 x 9.8 = 441 N
µ = \(\frac {F’}{ R}\) = 0.22
Angle of friction θ = tan^-1 0.22 = 12°24′

Question 42.
Calculate the force required to move a train of 2000 quintal up on an incline plane of 1 in 50 with an acceleration of 2 ms^-2. The force of friction per quintal is 0.5 N.
Answer:
Force of friction = 0.5 N per quintal
f = 0.5 x 2000 = 1000 N
m = 2000 quintals = 2000 x 100 kg
sin θ = \(\frac {1}{50}\), a – 2 m/s²
In moving up an inclined plane, force required against gravity
mg sin θ = 39200 N
And force required to produce acceleration = ma
= 2000 x 100 x 2 = 40,0000 N
Total force required = 1000 + 39,200 + 40,0000 = 440200 N.

Question 43.
A force of 100 N gives a mass m₁, an acceleration of 10 ms^-2 and of 20 ms^-2 to a mass m₂.
What acceleration must be given to it if both the masses are tied together?
Answer:
Suppose, a = acceleration produced if m₁ and m₂ are tied together,
F = 100 N
Let a₁ and a₂ be the acceleration produced in m₁ and m₂ respectively.
∴ a₁ and a₂ = 20ms^-2 (given)
Again m₁ = \(\frac{\mathrm{F}}{a_{1}}\) and m₂ = \(\frac{\mathrm{F}}{a_{2}}\)
⇒ m₁ = \(\frac {100}{10}\) = 10kg
and m₂ = \(\frac {100}{20}\) = 5kg
∴ m₁ + m₂ = 10 + 5 = 15
so, a = \(\frac{\mathbf{F}}{m_{1}+m_{2}}\) = \(\frac {100}{15}\) = \(\frac {20}{3}\) = 6.67 ms²

Question 44.
The pulley arrangement of figure are identical. The mass of the rope is negligible. In (a) mass m is lifted up by attaching a mass (2m) to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2 mg. In which case, the acceleration of m is more?

Answer:
Case (a):
a = \(\frac {2m – m}{2m + m}\) g = a = \(\frac {g}{3}\)
Case (b):
FBD of mass m
ma’ = T – mg
ma’ = 2 mg – mg
⇒ ma’ = mg
a’ = g
So in case (b) acceleration of m is more.

Question 45.
Figure shows the position-time graph of a particle of mass 4 kg. What is the

(a) Force on the particle for t < 0, t > 4s, 0 < t < 4s?
(b) Impulse at t = 0 and t = 4s?
(Consider one dimensional motion only)
Answer:
(a) For t < 0. No force as Particles is at rest. For t > 4s, No force again particle comes at rest.
For 0 < t < 4s, as slope of OA is constant so velocity constant i.e., a = 0, so force must be zero.

(b) Impulse at t = 0
Impulse = change in momentum
I = m(v – w) = 4(0 – 0.75) = 3 kg ms^-1
Impulse at t = 4s
1 = m(v – u) = 4 (0 – 0.75) = -3 kg ms^-1

Question 46.
What is the acceleration of the block and trolley system as the figure, if the coefficient of kinetic friction between the trolley and the surface is 0.04? Also Calculate friction in the string: Take g = 10 m/s², mass of the string is negligible.

Answer:
Free body diagram of the block
30 – T = 3a
Free body diagram of the trolley

T – f_k = 20 a ………….(2)
where f_k = µ_k= 0.04 x 20 x 10 = 8 N
Solving (i) & (ii), a = 0.96 m/s² and T = 27.2 N

Question 47.
Three blocks of masses ml = 10 kg, m² = 20 kg are connected by strings on smooth horizontal surface and pulled by a force of 60 N. Find the acceleration of the system and frictions in the string.

Solution:
All the blocks more with common acceleration a under the force F = 60 N.
F = (m₁ + m₂ + m₃)a
a = \(\frac{\mathrm{F}}{\left(m_{1}+m_{2}+m_{3}\right)}\) = 1 m/s²
to determine, T₁ →Free body diagram of m₁.

T₁ = m₁a = 10 x 1 = 10 N
to determine, T₂ →Free body diagram of m₃

F – T₂ = m₃a
Solving, we get T₂ = 30 N

Question 48.
The rear side of a truck is open and a box of 40 kg mass is placed 5m away from the open end. The coefficient of friction between the box and the surface below it is 0.15 on a straight road, the truck starts from rest and accelerates with 2 m/s². At what distance from the starting point does the box fall off the truck ? (ignore the size of the box)
Answer:
Force on the box due to accelerated motion of the truck
F = ma = 40 x 2 = 80 N (in forward direction)
Reaction on the box, F’ = F = 80 N (in backward direction)
Force of limiting friction, f = µR = 0 .15 x 40 x 10 = 60 N
Net force on the box in backward direction is P = F’ f = 80 – 60 = 20 N
Backward acceleration in the box = a= \(\frac {p}{m}\) = \(\frac {20}{40}\) = 0.5 ms^-2
t = time taken by the box to travel s = 5 m and falls off the truck, then from
s = ut + \(\frac {1}{2}\) at²
5 = 0 x t + \(\frac {1}{2}\) x 0.5 x t²
t = 4.47
If the truck travels a distance x during this time
then x = 0 x 4.34 +\(\frac {1}{2}\) x 2 x (4.471)²
x = 19.98 m

Question 49.
A block slides down as incline of 30° with the horizontal. Starting from rest, it covers 8 m in the first 2 seconds. Find the coefficient of static friction.
Use s = ut + \(\frac {1}{2}\) at²
a = \(\frac{2 s}{t^{2}}\) at² as u = 0
µ = \(\frac{g sin θ – a}{g Cos θ }\)
Putting the value and solving, µ = 0.11

Question 50.
A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m/s² . The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the:
(a) Force on the floor of the helicopter by the crew and passenger.
(b) Action of the rotor of the helicopter on the surrounding air
(c) Force on the helicopter due to the surrounding air (g = 10 m/s² )
Answer:
(a) Force on the floor of the helicopter by the crew and passengers
= apparent weight of crew and passengers
= 500(10+ 15)
=12500 N

(b) Action of rotor of helicopter on surrounding air is Obviously vertically downwards, because helicopter rises on account of reaction of this force. Thus force of action
= (2000 + 500) (10 + 15)
= 2500 x 25
= 62,500 N

(c) Force on the helicopter due to surrounding air is obviously a reaction. As action and reaction are equal and opposite, therefore
Force of reaction F’ = 62,500 vertically upwards.

Question 51.
A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is (µ). Let the mass of the box be m.

1.  At what angle of inclination θ of the plane to the horizontal will the box just start to slide down the plane ?
2. What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to a > θ.
3. What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed ?
4. What is the force needed to be applied upwards along the plane to pk kg f make the box move up the plane with acceleration a ?

Answer:
1. When the box just starts sliding
µ = tanθ
or 0 = tan^-1 µ

2. Force acting on the box down the plane
= mg (sin a – µ cos a)

3. Force needed mg (sin a + µ cos a)

4. Force needed = mg (sin a + µ cos a) + ma.

Question 52.
Two masses of 5 kg and 3 kg are suspended with help of mass less in extensible string as shown. Calculate T₁ and T₂ when system is going upwards with acceleration m/s². (Use g 9.8 m/s²)
Answer:
According Newton’s second law of motion
(1) T₁ – (m₁ + m₂)g = (m₁ + m₂)a
T₁ = (m₁ + m₂)(a + g) = (5 + 3) (2 + 9.8)
T₁ = 94.4 N

(2) T₂ – m₂g = m₂a
T₂ = m₂ (a + g)
T₂ = 3(2 + 9.8)
T₂ = 35.4 N

Question 53.
There are few forces acting at a Point P produced by strings as shown, which is at rest. Find the forces F₁ & F₁

Answer:
Using Resolution of forces IN and 2N and then applying laws of vector addition. Calculate for F₁ & F₁.
F₁ = \(\frac{1}{\sqrt{2}}\) N, F₂ = \(\frac{3}{\sqrt{2}}\)N

Question 54.
A hunter has a machine gun that can fire 50g bullets with a velocity of 150 ms A 60 kg tiger springs at him with a velocity of 10 ms^-1. How many bullets must the hunter fire into the target so as to stop him in his track?
Answer:
Given m = mass of bullet = 50 gm = 0.50 kg
M = mass of tiger = 60 kg
v = Velocity of bullet – 150 m/s
V = Velocity of tiger = – 10 m/s
(v It is coming from opposite direction n = no. of bullets fired per second at the tiger so as to stop it.)
P_i = 0, before firing ……..(i)
P_f = n (mv) + MV …………(ii)
∴ From the law of conservation of momentum,
P_i = P_f
⇒ 0 = n (mv) + MV
n = \(\frac{MV}{mv}\) = \(\frac{-60 \times(-10)}{0.05 \times 150}\) = 80

Question 55.
Two blocks of mass 2 kg and 5 kg are connected by an ideal string passing over a pulley. The block of mass 2 kg is free to slide on a surface inclined at an angle of 30° with the horizontal whereas 5 kg block hangs freely. Find the acceleration of the system and the tension in the string.

Let a be the acceleration of the system and T be the Tension in the string. Equations of motions for 5 kg and 2 kg blocks are
5g – T = 5a
T – 2g sin θ – f = 2a
where f = force of limiting friction
= µR = µ mg cos θ = 0.3 x 2 g x cos 30°
Solving (1) & (2)
a = 4.87 m/s²

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Tamilnadu Samacheer Kalvi 11th English Solutions Poem Chapter 5 Everest is not the Only Peak

Check out the topics covered in Poem Chapter 5 Everest is not the Only Peak Questions and Answers before you start your preparation. Improve your grammar skills with the help of Samacheer Kalvi 11th English Book Solutions Questions and Answers pdf links. The solutions for Tamilnadu State Board 11th English Textbook are prepared by the English experts. So, if you follow Tamilnadu State Board Solutions 11th English Textbook Solutions you can cover all the topics in Poem Chapter 5 Everest is not the Only Peak Questons and Answers. This helps to improve your communication skills.

Warm up

Identify the following personalities and their fields of achievement.

Answer:

Everest Is Not The Only Peak Question 1.
Mention a remarkable achievement of any of these personalities
Answer:
Kailash Satyarthi has rescued thousands of child labourers and gave them new life.

Everest Is Not The Only Peak Poem Summary Question 2.
What quality do you admire the most in each of these achievers?
Answer:
They were persistent and never accepted failure.

Everest Is Not The Only Peak Summary Question 3.
What are the qualities that you may share with them?
Answer:
Each one has concern for the country.

Everest Is Not The Only Peak Poem Line By Line Explanation Question 4.
Name a few more popular personalities who have made our nation proud.
Answer:
(a) Rabindranath Tagore
(b) Mother Teresa
(c) Sarojini Naidu
(d) Sania Mirza
(e) Mary Kom
(f) Virat Kohli
(g) Saina Nehwal
(h) Abhinav Bindra
(i) MS Dhoni
(j) AR Rahman

Samacheer Kalvi 11th English Everest is not the Only Peak Textual Questions

A. Based on your understanding of the poem, answer the following questions in a sentence or two each.

We are proud and feel so tall,
Our virtues though be few and small
Our nature it is that whatever we try
We do with devotion deep and true.

Defeat we repel, courage our fort;
Cringing from others we haven’t done,
To seek a gain we adore none:
We are proud and feel so tall.

We deem it our duty and mission in life,
To bless and praise the deserving ones;
Never shall we fail in what we commit,
Shall nourish the ones that nourish the world.

We are proud of the position we
Hold; humble as we are,
Our pride springs from the way we live.
Ours is a path of dignity and honour,
A life that knows no kneeling and bending.
We are proud and feel so tall.

Everest is not the only peak,
Every hillock has a summit to boast!
The height you reach is not that we care;
He, who does not stoop, is a king we adore.
We bow before competence and merit;
The ones that are true and stand on their own
Are really the ladder for the rise of Man.
Honour is a property, common to all:
In dignity and pride no one need to be poor.
We are proud and feel so tall.

Everest Is Not The Only Peak Poem Figures Of Speech Question 1.
Which line is repeated in the poem? What is the effect created by this repetition?
Answer:
The line, “We are proud and feel so tall” is repeated often in the poem. This establishes beyond a doubt the poet’s pride in dignity of iabour and pride of hardwork experienced by ordinary folks in life.

Everest Is Not The Only Peak Poem Summary In English Question 2.
Who are the ‘deserving ones’?
Answer:
Those who have merit and competence are the deserving ones.

Summary Of The Poem Everest Is Not The Only Peak Question 3.
Which quality does the speaker wish to nourish? What is his mission?
Answer:
The speaker wishes to nourish love to the mankind. The poet loves to nourish the ones who nourish the world. His mission is to bless and praise the deserving ones.

Everest Is Not The Only Peak Poem Question 4.
Which path should we follow in life?
Answer:
We should follow the path of dignity and honour. We should never stoop before others for any ‘gains’.

Everest Is Not The Only Peak Poem Warm Up Answers Question 5.
What does ‘Everest’ in the title stand for?
Answer:
‘Everest’ means the greatest achievement in life or the highest point one can reach in life.

Everest Is Not The Only Peak Poem Theme Question 6.
What does ‘hillock’ refer to in the line ‘Every hillock has a summit to boast!’?
Answer:
Hillock is a small hill. Everyone need not become Tenzing. Each one can achieve some ordinary pursuit and be proud of his achievements.

Everest Is Not The Only Peak Poem Explanation Question 7.
Why does the speaker say ‘Everest is not the only peak’?
Answer:
Everyone is not made to be a mountain climber or a captain. Each one has an important role in this life however small it may appear to be. The poet respects every small achievement in every walk of life. So, he says, “Everest is not the only peak”.

Everest Is Not The Only Peak Poem Paragraph Question 8.
What does the ladder symbolize?
Ladder symbolizes help given to enable others to climb up to a higher position in life.

B. Read the given lines and answer the questions that follow.

1. Our nature it is that whatever we try
We do with devotion deep and true.

Everest Is Not The Only Peak Meaning In Tamil Question (a)
Who does ‘we’ refer to?
Answer:
‘We’refers to ordinary people.

Everest Is Not The Only Peak Meaning Question (b)
How should we carry out our duties?
Answer:
We must do our duty with sincerity and deep devotion.

2. Defeat we repel, courage our fort;

Everest Is Not The Only Peak Meaning In English Question (a)
How do we react to defeat?
Answer:
We defy defeat.

(b) Which is considered as our stronghold?
Answer:
Courage is our stronghold.

3. We are proud of the position we hold;
humble as we are,

Everest Is Not The Only Peak Poem Analysis Question (a)
What is the speaker proud of?
Answer:
The speaker is proud of the position people hold on to.

Everest Is Not The Only Peak Theme Question (b)
How is the speaker both humble and proud?
Answer:
The ordinary position they hold keeps them humble. But the path of self-dignity and honour they tread, makes them feel proud.

Question (c)
Pick out the alliteration in these lines.
Answer:
proud, position, hold, humble are the words which alliterate.

4. He, who does not stoop, is a king we adore.
We bow before competence and merit;

Question (a)
Who is adored as a king?
Answer:
An upright or straight forward person is adored as a king.

Question (b)
What is the figure of speech used in the first line?
Answer:
Metaphor

5. Honour is a property, common to all:
In dignity and pride no one need to be poor.

Question (a)
Who are considered rich?
Answer:
Those who possess dignity and pride are considered rich.

Question (b)
What is their asset?
Answer:
Honour is their asset.

C. Answer the following questions in a paragraph of 100-150 words each.

Question 1.
In what way is every hillock similar to Everest?
Answer:
The poet does not compare rare feats of athletes, mountaineers or horsemen. He does not attach great value to positions or possessions. He scoffs at those who pull strings to achieve their ends. The means must justify the ends. One should not stoop to underhand dealings to achieve their desired goals in life. Those who reach great heights in life like Everest due to their hardwork, perseverance and competence are adorable. At the same time those who trek any small hillocks can’t be underestimated. The efforts made in reaching even the smallest positions in life, if done with sincerity of purpose and deep devotion, is worthy of hearty appreciation.

One who holds a humble position, but upright and serves as a ladder for fellow humans to reach great heights deserve our respect. The poet admits that he is proud of people’s humble positions because their pride springs not from positions or possessions but the way they live. Their life knows no bending. The poet just doesn’t bother the height of the peak one reaches. It could even be a hillock. What matters is how one reaches that spot. If merit and competence have paved the way for their success and positions, however humble they are, the poet admires them.

“Take on risks and ride the journey called life with no regrets. ”

Question 2.
The poem does not focus on the destination but the journey towards it. Discuss.
Answer:
The poet discusses the merits of efforts, duty and devotion and values of honesty, uprightness and service-mindedness. He does not have any special appreciation to those who reach great peaks like Himalayas. He appreciates the process, the journey and not the destination. When the whole world has a perspective of seeking glory using any foul method or underhand dealing, the poet differs from it. For him the means is more important than the end. However modest may be one’s position is, it is adorable if attained by competence and merit. Pride is not in heights one reaches but in a life that knows no bending or kneeling. The poet respects one who does not stoop as a king. Thus the poet pays importance to the journey of life not the destination.

“The journey of life is not meant to be feared and planned;It is meant to be travelled and enjoyed. ”

Creative Activity

D. Write eight words you associate with success.

• Use the words to write eight lines that mean success to you or how success makes you feel.
• Arrange your lines into a poem.
• Share your poem with the class and post a copy on the notice board.
  1. strive
  2. flaunt
  3. determination
  4. destination
  5. wise
  6. want
  7. success
  8. kind

Succes
Poem:

Real success is when you strive
for what you want
when you have that guts to flaunt
Not thinking about the world and wise
when you can alone suffice
With upright will and determination
where finally you reach your destination.
Real success is hard to find. But its one of a kind.
(OR)

1. brave
2. hunt
3. learn
4. mistakes
5. encounter
6. failure
7. never
8. If

Poem

You’ll never be brave.
If you don’t get hurt
you’ll never learn.

If you don’t make mistakes
you’ll never be successful.
If you don’t enounter failure.

Speaking Activity

E. Discuss the following topics in groups of five and choose a representative to sum up the views and share them with the class.

Question (a)
To succeed in life, one must have a single-minded devotion to duty.
Answer:
Singleminded determination is necessary to achieve success in life. Legend says Dronacharya was training Pandavas to shoot arrows in the jungle. Once guru saw a bird at the top branch of a tall tree. He assigned the task of shooting the bird on its right eye.
Bhim, Nakulan, Sakadevan and Dharman were denied the chance to aim at it because they all told Dronacharya that they saw the whole jungle, tree, sky and the bird respectively. But it was Arjun who said that he saw only the right eye of the bird and nothing else. Indo-Pak war was in progress. Four bombers were prowling into Indian airspace.

Indian fighter bomber pilot realized all the bombs were exhausted. He had singleminded determination to prevent the four bombers from bombing India harm and force them to surrender. He tried a trick. He called them and spoke to them. He informed them that he was thousand feet above them with his finger on a missile. If they just surrender, their lives would be spared. He radioed to his chief that they should welcome four prisoners of war along with their fighter bombers.

Napoleon Boneparte was once watching a battle from a hillock. One of his soldiers came with an urgent message. Napoleon looking at the badly wounded soldier thought that the war was lost. He called one of his aides and gave instructions for pullout. But the soldier said, “Sire, we‘ve won”. Then he gave a slip from another pocket. Being a practical leader, he had alternate plans always at hand. Such leaders never accept failure as permanent. People who pursue their goals in life with single minded determination always win.

Question (b)
‘Success is not final, failure is not fatal.’ It is the courage and perseverance that Counts.
Answer:
India’s cricket team was beaten in Test series in Africa. Infact, it was routed. The team’s morale was a little down. The team captain Virat Kohli told his team to focus on what they were good at (i.e) sterling performance in One Day Internationals. He believed in the youngest bowlers and told them to play the game, the way they loved to play. He still believed in them. They would have to prove who are the masters of the game. Gathering their broken hopes and courage, the Indian team players, snatched the ODI series from the overconfident South Africans. They went on to win the T20 series too against the hosts. Thus they proved that success is not final and failure never fatal. One may bounce back from failure if one persists long enough.

A king had lost a battle. All his soldiers had been scattered across the country. Heart broken king Bruce was hiding in a cave. He saw a spider failing a number of times to spin a web. But it made it after about 20 attempts. This bolstered the confidence of the king. He refused to be controlled by failure. He defied defeat. He gathered his soldiers again and won the battle. These incidents throw much light on the truth that perseverance and courage counts for success in fife.

Question (c)
Successful people neither brood over the past nor worry about the future.
Answer:
Mahatma Gandhi and his followers were arrested and jailed many times. Gandhi’s followers were brutally lathi charged. Gandhi had decided to silence the guns of the British with Ahimsha passive non-violent resistance. The brutal suppression of the struggle for freedom did not dishearten Gandhi. He did not brood about the strength of British army and weaknesses of unarmed peasants who believed in his leadership. He was a Karmayogi. What ever the duty to be done it must be done with steadfast devotion and sincerity. Other leaders got worried. Some angry young men resorted to violence.

They burnt down a police station at Chauri chaura too. But Gandhi declared a fast unto death. He plunged into action. If he had worried about the unpleasant developments, he wouldn’t have launched Quit India Movement or Salt Satyagraha effectively.

Thomas Alva Edison was not able to find the element that would glow if electricity was passed. He had failed 1000 times to invent the bulb. But he said, “The light bulb was an invention with 1000 steps”. Each step taught him what did not work. He lost his hearing capacity. He had many failures. His teachers believed him to be mad and unteachable.

His entire schooling was only a few years. His mom taught him and made him believe in himself. This man who had been ill-treated in school and faced many challenges had no time to brood. He went on to make 1093 inventions and got them patented. Those who are busy building facilities for transforming the world have neither the time nor the inclination to brood about failures or about possibilities of success in future.

Everest is not the Only Peak About the poet:

Prof. V.C. Kulandtiisamy ( 14th July 1929 10th Dec 2016) is known as Kulotliimgan. He i is a prolific Tamil writer with six volumes of poems and seven of prose essays. He has 1 been conferred Thiruvalluvar Award by Government of Tamil Nadu (1999) and Sahitva j Academy’s Award in 1988 for his outstanding masterpiece “Vaazhum Valluvam”. Ills i poems, as evident in the poem “Everest is not the only Peak” deal with the gamut of human [ progress and all pervasive human effort to better the world.

Everest is not the Only Peak Summary

This poem discusses in depth the virtues that make people greater than those who scale Himalayas. We have a few virtues which make us feel proud and tall. By nature we do every assigned duty with deep and true devotion.

We repel defeats because courage is our fort. We have never cringed from others. We don’t lick others’ shoes to achieve personal gains. We are proud and tall.

We deem it a duty and a mission to appreciate those who deserve appreciation. We shall carry on doing unfailingly what we are committed to do. We shall nourish those who nourish the world (i.e) those who better the world with their services, inventions and discoveries.

Despite living a simple and humble life, we have pride over our path of dignity and honour. We feel proud and tall for we never bend before the mighty.

We value individual differences. Every small effort to succeed in life matters. Peaks alone don’t matter. Even a hillock is a symbol of human achievement. We respect those who don’t stoop to conquer. We are proud to be rich in honour, dignity and pride. Those who stand on their own merit and serve as ladder for progress of fellowmen are worthy of our appreciation. We are proud and feel so tall.

Not the worldly riches, power and positions impress us. We derive our happiness in serving those who serve the nation. We respect all who are upright in their dealings. So, as honest citizens, we are proud.

Everest is not the Only Peak Glossary

Textual:
adore – worship someone
competence – the ability to do something efficiently
cringing – behaving in an excessively humble or servile way
devotion – loyal commitment towards a particular activity
merit -the quality of being particularly good or worthy
nourish – to help the growth and development of someone
repel – hate or detest
stoop – yield or submit, to descend from dignity
summit – the highest point of a hill or mountain peak
virtues – good qualities

Additional:
dignity – deserves respect
fort – huge building for protection
hillock – small hill
humble -modest
mission – vocation
praise – appreciate
property – possession
proud – feeling important
springs – arises

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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 11th Computer Science Solutions Chapter 2 Number Systems

Samacheer Kalvi 11th Computer Science Number Systems Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

11th Computer Science Chapter 2 Book Back Answers Question 1.
Which refers to the number of bits processed by a computer’s CPU?
(a) Byte
(b) Nibble
(c) Word length
(d) Bit
Answer:
(c) Word length

11th Computer Science Chapter 2 Workshop Answers Question 3.
Expansion for ASCII ………………..
(а) American School Code for Information Interchange
(b) American Standard Code for Information Interchange
(c) All Standard Code for Information Interchange
(d) American Society Code for Information Interchange
Answer:
(b) American Standard Code for Information Interchange

11th Computer Science 2nd Lesson Book Back Answers Question 4.
2^50 is referred as
(a) Kilo
(b) Tera
(c) Peta
(d) Zetta
Answer:
(c) Peta

Class 11 Computer Science Chapter 2 Solutions Question 5.
How many characters can be handled in Binary Coded Decimal System?
(a) 64
(b) 255
(c) 256
(d) 128
Answer:
(a) 64

11th Computer Science 2nd Lesson Question 6.
For 1101₂ what is the Hexadecimal equivalent?
(a) F
(b) E
(c) D
(d) B
Answer:
(c) D

Computer Science Chapter 2 Question 7.
What is the 1’s complement of 00100110?
(a) 00100110
(b) 11011001
(c) 11010001
(d) 00101001
Answer:
(b) 11011001

Number System Class 11 Pdf Question 8.
Which amongst this is not an Octal number?
(a) 645
(b) 234
(c) 876
(d) 123
Answer:
(c) 876

II. Short Answers

Number System Class 11 Computer Science Pdf Question 1.
What is data?
Answer:
The term data comes from the word datum which means a raw fact. The data is a fact about people, places or some objects.
Example: Rajesh, 16, XI

Samacheer Kalvi Guru 11th Computer Science Question 2.
Write the 1’s complement procedure.
Answer:
Step 1 : convert given decimal number into Binary
Step 2 : if the binary bit contains 8 bits if less add 0 at the left most bit, to make it as 8 bits.
Step 3 : Invert all the bits, (i.e.) change 1 as 0 and 0 as 1.

Class 11 Computer Science Chapter 2 Question Answer Question 3.
Convert (46)₁₀ into Binary number
Answer:
(46)₁₀ into binary = 101110₂

Question 4.
We cannot find 1’s complement for (28)₁₀. State reason.
Answer:
Since it is a positive number. 1 ’s complement will come only for negative number.

Question 5.
List the encoding systems for characters in memory.
Answer:

1. BCD – Binary Coded Decimal
2. EBCDIC – Extended Binary Coded Decimal Interchange Code
3. ASCII – American Standard Code for Information Interchange
4. Unicode
5. ISCII – Indian standard code for Information interchange

III. Explain in Brief

Question 1.
What is radix of a number system? Give example.
Answer:
The radix refers to the base of a number system: the total number of possible digits. The decimal number system that we all use is base ten, as it has ten distinct digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9).
Example : Binary system – Radix 2

Question 2.
Write note on binary number system.
Answer:
In Binary Number System, there are only two digits namely 0 and 1. The numbers in the binary system are represented to the base 2. In the Binary Number, the left most bit is the Most Significant Bit (MSB) and the right most bit is the Least Significant Bit (LSB). MSB has largest positional weight and LSB has smallest positional weight.
Example : 110010₂

Question 3.
Convert (150)₁₀ into Binary, then convert that Binary number to Octal.
Step 1:
Change it to Binary.

(150)₁₀ = 10010110

Step – 2:
Change it to Octal.

Answer:
Binary 10010110₂
Octal 226_g

Question 4.
Write short note on ISCII.
Answer:
ISCII – Indian Standard Code for Information Interchange (ISCII) is the system of handling the character of Indian local languages. This is a 8 – bit coding system. Therefore it can handle 256 (2⁸) characters. It is recognized by Bureau of Indian Standards (BIS). It is integrated with Unicode.

The supported scripts are:
Assamese, Bengali (Bangla), Devanagari, Gujarati, Gurmukhi, Kannada, Malayalam, Oriya, Tamil, and Telugu. ISCII does not encode the writing systems of India based on Arabic, but its writing system switching codes none the less provide for Kashmiri, Sindhi, Urdu, Persian, Pashto and Arabic. The Arabic – based writing systems were subsequently encoded in the PASCII encoding.

ISCII is an 8 – bit encoding. The lower 128 code points are plain ASCII, the upper 128 code points are ISCII – specific. In addition to the code points representing characters, ISCII makes use of a code point with mnemonic ATR that indicates that the following byte contains one of two kinds of information. One set of values changes the writing system until the next writing system indicator or end – of – line.

Question 5.
Add (a) – 22₁₀ + 15₁₀
(b) 20₁₀ + 25<sub>10

Answer:
(a) – 2210 = 101102
Binary Equivalent = 101102

Answer:
111110012</sub>

(b) 20₁₀ + 25₁₀
Answer:

Answer:
00101101₂

IV. Detail Answers

Question 1.
(a) Write the procedure to convert fractional Decimal to Binary.
(b) Convert (98.46)₁₀ to Binary
Answer:
(а) Procedure to convert fractional Decimal to Binary.
Step 1 : Multiply the decimal fraction by 2 and note the integer part. The integer part is either Oor 1.
Step 2 : Discard the integer part of the previous product. Multiply the fractional part of the previous product by 2. Repeat step 1 until the same fraction repeats or terminates (0).
Step 3 : The resulting integer part forms a sequence of 0’s and 1 ’s that becomes the binary equivalent of decimal fraction.
Step 4 : The final answer is to be written from first integer part obtained till the last integer part obtained.

(b) Convert (98.46)₁₀ to Binary

Answer:
1100010.01110 … ₂

Question 2.
Find 1’s Complement and 2’s Complement for the following Decimal number (a) – 98 (b) – 135
Answer:

98₁₀ = 01100010
8 bit format = 01100010
1’s complement = 10011101
Add 1 bit = + 1
= 10011110

135₁₀ = 10000111
1’st complement = 01111000
Add 1 bit = + 1
= 01111001

Question 3.
(a) Add 1101010₂ + 101101₂
(b) Subtract 1101011₂ – 111010₂
Answer:

PART – 1
I. Choose the correct answer

Question 1.
Which is a basic electronic circuit which operates on one or more signals?
(a) Boolean algebra
(b) Gate
(c) Fundamental gates
(d) Derived gates
Answer:
(b) Gate

Question 2.
Which gate is called as the logical inverter?
(a) AND
(b) OR
(c) NOT
(d) XNOR
Answer:
(c) NOT

Question 3.
A + A = ?
(a) A
(b) O
(c) 1
(d) A
Answer:
(a) A

Question 4.
NOR is a combination of?
(a) NOT(OR)
(b) NOT(AND)
(c) NOT(NOT)
(d) NOT(NOR)
Answer:
(a) NOT(OR)

Question 5.
NAND is called as ……………… Gate.
(a) Fundamental Gate
(b) Derived Gate
(c) Logical Gate
(d) Electronic gate
Answer:
(b) Derived Gate

PART – 2
II. Short Answers

Question 1.
What is Boolean Algebra?
Answer:
Boolean algebra is a mathematical discipline that is used for designing digital circuits in a digital computer. It describes the relation between inputs and outputs of a digital circuit.

Question 2.
Write a short note on NAND Gate.
Answer:
The NAND gate is the combination of NOT and AND gates. The NAND gate is generated by inverting the output of a AND gate. The algebraic expression of the NAND is Y = \(\overline{\mathrm{A} . \mathrm{B}}\)

Question 3.
Draw the truth table for XOR gate.
Answer:
The truth table for XOR gate is

Question 4.
Write the associative laws?
Answer:
A + (B + C) = (A + B) + C
A.(B.C) = (A.B).C

Question 5.
What are derived gates?
Answer:
Derived Gates are the gates which are derived from the fundamental gates.
Example : NAND, NOR, XOR, XNOR

PART – 3
III. Explain in Brief

Question 1.
Write the truth table of fundamental gates.
Answer:
(a) AND gate – Truth Table

(b) OR gate – Truth Table

(c) NOT Gate – Truth Table

Question 2.
Write a short note on XNOR gate.
The XNOR (exclusive – NOR) gate is a combination of XOR gate followed by an inverter. Its output is “true” if the inputs are same and false if the inputs are different.
The output of XNOR is C = AB + \(\overline{\mathrm{A}}\) \(\overline{\mathrm{B}}\)
= A \(\odot\) B
Where \(\odot\) indicates included dot.
Truth Table:

Logic symbol:

Logic circuit:

Question 3.
Reason out why the NAND an NOR are called universal gates?
Answer:
NAND and NOR gates are called as Universal gates because the fundamental logic gates can be realized through them.

Question 4.
Give the truth table of XOR gate.
Answer:
XOR – Truth Table

Question 5.
Write the De Morgan’s law.
Answer:
De Morgan’s \(\overline{\mathrm{A}+\mathrm{B}}\) = \(\overline{\mathrm{A}}\) . \(\overline{\mathrm{B}}\)
(\(\overline{\mathrm{A}+\mathrm{B}}\)) = \(\overline{\mathrm{A}}\) + \(\overline{\mathrm{B}}\)

PART – 4
IV. Explain in Detail

Question 1.
Explain the fundamental gates with expression and truth table.
Answer:
A gate is a basic electronic circuit which operates on one or more signals to produce an output signal.
The three fundamental gates are AND, OR and NOT gates.

AND Gate:
The AND gate can have two or more input signals and produce an output signal. The output is “true” only when both inputs are “true”, otherwise, the output is “false”. In other words the output will be 1 if and only if both inputs are 1; otherwise the output is 0. The output of the AND gate is represented by a variable say C, where A and B are two and if input boolean variables. In boolean algebra, a variable can take either of the values ‘0’ or ‘1’. The logical symbol of the AND gate is

The truth table for AND gate is

OR Gate:
The OR gate gets its name from its behaviour like the logical inclusive “OR”. The output is “true” if either or both of the inputs are “true”. If both inputs are “false” then the output is “false”. In other words the output will be 1 if and only if one or both inputs are 1; otherwise, the output is 0. The logical symbol of the OR gate is

The truth table for OR gate is

NOT Gate:
The NOT gate, called a logical inverter, has only one input. It reverses the logical state. In other words the output C is always the complement of the input. The logical symbol of the NOT gate is

The truth table for NOT gate is

Question 2.
How AND and OR can be realized using NAND and NOR gate.
Bubbled AND Gate
(i) Realized of and using only AND gate:
The Boolean function for AND is C = AB. The same can be realized using only NAND gates.

(ii) Realization of or using only NAND’s:
The Boolean function of OR is C = A + B. The same can be realized using only NAND gates.

(iii) Realization of AND using NOR:
By using only the NOR gates, we can get the output equivalent to the output of AND gate.
C = A.B

(iv) Realization of OR using NOR’s:
By using only NOR gates we are getting the output equivalent to OR gate.

Question 3.
Explain the Derived gates with expression and truth table.
Answer:
The logic gates like NAND, NOR, XOR and XNOR are derived gates which are derived from the fundamental gates AND, OR NOT.
(i) NAND gate:
The NAND is the combination of NOT and AND. The NAND is generated by inverting the output of an AND operation.
Output: y = \(\overline{\mathrm{A}\mathrm{B}}\)
Logic circuit

Logic symbol:

Truth table:

Inputs Outputs
(ii) NOR gate:
The NOR is the combination of NOT and OR. The NOR is generated by inverting the output of an OR operation.
Logic function: y = \(\overline{\mathrm{A}\mathrm{B}}\)
Logic circuit:

logic symbol:

Truth table:

(iii) XOR gate:
The XOR (exclusive – OR) gate acts in the same way as the logical either /or.
Logic symbol:

Logic circuit:

Truth table:

(iv) XNOR gate:
XNOR gate (exclusive – NOR) gate is a combination of XOR gate followed by an inverter.
Logic function: y = \(\overline{\mathrm{A}\mathrm{B}}\)

logic symbol:

Truth table:

Boolean function:

Samacheer kalvi 11th Computer Science Number Systems Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
The simplest method to represent negative binary number is called ………………..
(a) signed magnitude
(b) sign bit or parity bit
(c) binary
(d) decimal
Answer:
(a) signed magnitude

Question 2.
The term data comes from the word ………………..?
(a) number
(b) datum
(c) nibble
(d) bit
Answer:
(b) datum

Question 3.
Expansion for BCD ………………..
(a) Binary coded decimal
(b) binary complement decimal
(c) binary computer decimal
(d) binary convert decimal
Answer:
(a) Binary coded decimal

Question 4.
……………….. scheme is denoted by hexadecimal numbers.
(a) binary
(b) Unicode
(c) word length
(d) data
Answer:
(b) Unicode

Question 5.
The ……………….. operator is defined in boolean algebra by the use of the dot (.) operator.
(a) AND
(b) OR
(c) NOT
(d) NAND
Answer:
(a) AND

Question 6.
A ……………….. number is represented using base 16.
(a) Hexadecimal
(b) octal
(c) binary
(d) decimal
Answer:
(a) Hexadecimal

Question 7.
The convert (65)₁₀ into its equivalent octal number ………………..
(a) (101)₈
(b) (101)₁₀
(c) (101)₁₂
(d) (101)₄
Answer:
(a) (101)₈

Question 8.
Octal number system uses digits ………………..
(a) 7
(b) 5
(c) 8
(d) 10
Answer:
(c) 8

Question 9.
……………….. is the general idea behind positional numbering system.
(a) Radix
(b) Computer memory
(c) Binary number
(d) Decimal number
Answer:
(a) Radix

Question 10.
The NAND gate operates an AND gate followed by a ……………….. gate.
(a) AND
(b) OR
(c) NOT
(d) XOR
Answer:
(c) NOT

Question 11.
Bit means ………………..
(a) nibble
(b) byte
(c) word length
(d) binary digit
Answer:
(d) binary digit

Question 12.
Expand BIT.
(a) Basic Input Term
(b) Binary Input Term
(c) Binary Digit
(d) Binary Inverse Digit
Answer:
(c) Binary Digit

Question 13.
The computer can understand ……………….. languages.
(a) computer
(b) machine
(c) post
(d) pre
Answer:
(b) machine

Question 14.
Identify the wrong pair.
1. 1 YB = 2^80
2. 1 BM = 2^70
3. 1 MB = 2^20
4. 1 TM = 2^40
(a) 2
(b) 3
(c) 4
(d) 1
Answer:
(a) 2

Question 15.
How many bytes does 1 zetta byte contains?
(a) 2⁹⁰
(b) 2⁸⁰
(c) 2⁷⁰
(d) 2⁶⁰
Answer:
(c) 2⁷⁰

Question 16.
The collection of 4 bits is ………………..
(a) bit
(b) byte
(c) nibble
(d) KB
Answer:
(c) nibble

Question 17.
Match the following.

(a) 4 12 3
(b) 12 3 4
(c) 4 3 2 1
(d) 4 2 1 3
Answer:
(a) 4 12 3

Question 18.
1 kilo byte represents ……………….. bytes.
(a) 512
(b) 256
(c) 1024
(d) 64
Answer:
(c) 1024

Question 19.
How many mega bytes does 1 GB contains?
(a) 2²⁰
(b) 2¹⁰
(c) 2³⁰
(d) 2⁴⁰
Answer:
(b) 2¹⁰

Question 20.
What is the decimal value of 1111₂?
(a) 10
(b) 11
(c) 14
(d) 15
Answer:
(d) 15

Question 21.
What is the 1’ s complement of 11001.
(a) 11100110
(b) 01010101
(c) 11110000
(d) 100100111
Answer:
(a) 11100110

Question 22.
The decimal value of Binary number 10 is ………………..
(a) 101010
(b) 2
(c) 100
(d) A
Answer:
(b) 2

Question 23.
The hexadecimal equivalent of 15 is ………………..
(a) A
(b) B
(c) E
(d) F
Answer:
(d) F

Question 24.
Which of the following are data?
(a) Alphabet
(b) Special character
(c) Number
(d) All of these
Answer:
(d) All of these

Question 25.
The radix of hexadecimal number is ………………..
(a) 2
(b) 8
(c) 16
(d) 10
Answer:
(c) 16

Question 26.
Pick the odd one.
(a) BCD
(b) ENIAC
(c) ASCII
(d) EBCDIC
Answer:
(b) ENIAC

Question 27.
The most commonly used number system is ………………..
(a) binary
(b) decimal
(c) octal
(d) hexadecimal
Answer:
(b) decimal

Question 28.
Unicode can handles how many characters?
(a) 64
(b) 128
(c) 256
(d) 65536
Answer:
(d) 65536

Question 29.
What does MSB means?
(a) Major sign bit
(b) Most sign bit
(c) Minor sign bit
(d) Most significant bit
Answer:
(d) Most significant bit

Question 30.
Which one is the right most bit?
(a) MSB
(b) LSB
(c) USB
(d) USRB
Answer:
(b) LSB

Question 31.
The binary equivalent of hexadecimal number B is ………………..
(a) 1011
(b) 1100
(c) 1001
(d) 1010
Answer:
(a) 1011

Question 32.
The left most bit of a positive binary number in signed notation is ………………..
(a) 0
(b) 1
(c) 2
(d) A
Answer:
(a) 0

Question 33.
What is the range of ASCII values for lower case alphabets?
(a) 65 to 90
(b) 65 to 122
(c) 97 to 122
(d) 98 to 122
Answer:
(c) 97 to 122

Question 34.
The radix for octal number system is ………………..
(a) 2
(b) 8
(c) 1
(d) 16
Answer:
(b) 8

Question 35.
What is the ASCII value for blank space?
(a) 8
(b) 2
(c) 18
(d) 32
Answer:
(d) 32

Question 36.
Which one of the following company have formulated EBCDIC?
(a) Microsoft
(b) 1 BM
(c) Sun
(d) Apple
Answer:
(b) 1 BM

Question 37.
Which one of the following bit has smallest positional weight?
(a) MSB
(b) LSB
(c) UPS
(d) USB
Answer:
(b) LSB

Question 38.
The base value of hexadecimal number is ………………..
(a) 2
(b) 8
(c) 16
(d) 18
Answer:
(c) 16

Question 39.
Name the person who proposed the basic principles of Boolean Algebra?
(a) Wiliam Boole
(b) George Boole
(c) James Boole
(d) Boolean George
Answer:
(b) George Boole

Question 40.
How many truth values are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 41.
What is the other name for logical statement?
(a) Truth values
(b) Truth functions
(c) Truth table
(d) Truth variables
Answer:
(b) Truth functions

Question 42.
The variables which can store the truth values are called as ………………..
(a) logical variable
(b) binary valued variable
(c) boolean variables
(d) all of these
Answer:
(d) all of these

Question 43.
The NOT operator is represented by the symbol.
(a) over bar
(b) single apostrophe
(c) a and b
(d) plus
Answer:
(c) a and b

Question 44.
Which is not a logical operator?
(a) dot
(b) plus
(c) over bar
(d) command
Answer:
(d) command

Question 45.
The output for the AND operator is ………………..
(a) A + B
(b) –
(c) A.B
(d) AB + C
Answer:
(c) A.B

Question 46.
Which symbol is used to in OR operator?
(a) –
(b) •
(c) *
(d) +
Answer:
(d) +

Question 47.
Which gate takes only one input?
(a) OR
(b) AND
(c) NOT
(d) XOR
Answer:
(c) NOT

Question 48.
Which among the following can be replaced by a bubbled AND gate?
(a) AND
(b) NAND
(c) OR
(d) not
Answer:
(b) NAND

Question 49.
Which is not a derived date?
(a) AND
(b) NAND
(c) NOR
(d) XOR
Answer:
(a) AND

Question 50.
Find the universal gates from the following.
(a) XOR
(b) XNOR
(c) a and b
(d) NOR
Answer:
(d) NOR

Question 51.
The statement “C equal the complement of A or B” means
(a) C = A + B
(b) C = \(\overline{\mathrm{A}}+\overline{\mathrm{B}}\)
(c) C = \(\overline{\mathrm{A}}\) + \(\overline{\mathrm{B}}\)
(d) C = \(\overline{\mathrm{A}\mathrm{B}}\)
Answer:
(a) C = A + B

Question 52.
Which symbol is used in XOR gate?
(a) \(\odot\)
(b) \(\otimes\)
(c) \(\oplus\)
(d) –
Answer:
(c) \(\oplus\)

Question 53.
Included dot means ………………..
(a) \(\odot\)
(b) [•]


Answer:
(a) \(\odot\)

Question 54.
What is the output of XOR gate?
(a) C = A% B
(b) C = A \(\otimes\) A
(c) C = A \(\odot\) B
(d) C = A \(\oplus\) B
Answer:
(d) C = A \(\oplus\) B

Question 55.
Identify the statement which is wrong.
(a) A . 1 = A
(b) A . A = A
(c) A + O = A
(d) A . 1 = 0
Answer:
(b) A . A = A

Question 56.
Find A + \(\overline{\mathrm{A}}\) .B = ………………..
(a) A + B
(b) A.B
(c) \(\overline{\mathrm{A}}\).B
(d) A.\(\overline{\mathrm{B}}\)
Answer:
(d) A.\(\overline{\mathrm{B}}\)

Question 57.
Identify the statements which are true.
(i) A + 0 = A
(ii) A.A = A
(iii) A + \(\overline{\mathrm{A}}\) = 1
(iv) A. O = O

(a) (iii) (iv) are true
(b) (i) (ii) are true
(c) (i) (ii) (iii) are true.
(d) all are true
Answer:
(d) all are true

Question 58.
Find the wrong pair from the following:
(a) Null element : A + 1 = 1
(b) Involution : \(\overset { = }{ A }\) = A
(c) Demorgan’s : \(\overline{\mathrm{A+B}}\) =\(\overline{\mathrm{A}}\) . \(\overline{\mathrm{A}}\)
(d) Commutative : A + B = B . A
Answer:
(d) Commutative : A + B = B . A

Question 59.
Match the following

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

Question 60.
With 2 inputs in the truth table, how many set of values will be obtained.
(a) 4
(b) 8
(c) 2
(d) 1
Answer:
(a) 4

PART – 2
II. Short Answers

Question 1.
What is nibble?
Answer:
Nibble is a collection of 4 bits. A nibble is a half a byte.

Question 2.
Expand ASCII.
Answer:
American Standard Code for Information Interchange

Question 3.
What is radix?
Answer:
The base value of a number is also known as the radix.

Question 4.
What is MSB and LSB?
Answer:
MSB means Most Significant Bit; LSB-Least Significant Bit

Question 5.
Expand: BCD, EBCDIC, ASCII
Answer:
BCD – Binary Coded Decimal; EBCDIC – Extended Binary Coded Decimal Interchange Code; ASCII – American Standard Code for Information Interchange.

Question 6.
Define word length?
Answer:
Word length refers to the number of bits processed by a computers CPU.

Question 7.
What are the methods of converting a number from decimal to binary.
Answer:

1. Repeated division by two.
2. Sum of powers of 2.

Question 8.
How will you convert a number from octal to Binary.
Answer:
For each octal digit in the given number write its 3 digits binary equivalent using positional notation.

Question 9.
What are the various ways for Binary representation of signed numbers?
Answer:

1. Signed magnitude representation
2. 1’s complement
3. 2’s complement

Question 10.
Write a short note on BCD.
Answer:

1. BCD stands for Binary Coded Decimal system.
2. BCD is 26 bit encoding system.
3. IT can handle 64 characters.

PART – 3
III. Explain in Brief

Question 1.
What is binary number system?
Answer:
There are only two digits in the Binary system, namely, 0 and 1. The numbers in the binary system are represented to the base 2 and the positional multipliers are the powers of 2. The left most bit in the binary number is called as the Most Significant Bit (MSB) and it has the largest positional weight. The right most bit is the Least Significant Bit (LSB) and has the smallest positional weight.

Example:
The binary sequence (11101)₂ has the decimal equivalent:
(1101)₂ = 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2⁰
= 8 + 4 + 0 + 1
= (13)₁₀

Question 2.
What is octal number system?
Answer:
Octal number system uses digits 0, 1, 2, 3, 4, 5, 6 and 7 (8 digits): Each octal digit has its own positional value or weight as a power of 8.
Example:
The Octal sequence (547)₈ has the decimal equivalent:

Question 3.
What is decimal number system?
Answer:
It consists of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9(10 digits). It is the oldest and most popular number system used in our day to day life. In the positional number system, each decimal digit is weighted relative to its position in the number. This means that each digit in the number is multiplied by 10 raised to a power corresponding to that digit’s position.
Example:
(123)₁₀ = (1 x 10²) + (2 x 10¹) + (3 x 10⁰)
= 100 + 20 + 3
= (123)₁₀

Question 4.
Write the distributive law.
Answer:
A . (B + C) = A . B + A . C
A + (B . C) = (A + B) . (A + C)

Question 5.
What is truth table?
Answer:
A truth table represents all the possible values of logical variable or statements along with ail the possible results of given combination of truth values.

PART – 4
IV. Explain in Detail

Question 1.
Explain the different types of number systems?
Answer:
Different Types of Number Systems

A numbering system is a way of representing numbers. The most commonly used numbering system in real life is Decimal number system. Other number systems are Binary, Octal, Hexadecimal number system. Each number system is uniquely identified by its base value or radix. Radix or base is the count of number of digits in each number system. Radix or base is the general idea behind positional numbering system.

Decimal Number System:
It consists of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9(10 digits). It is the oldest and most popular number system used in our day to day life. In the positional number system, each decimal digit is weighted relative to its position in the number. This means that each digit in the number is multiplied by 10 raised to a power corresponding to that digit’s position.
Example:
(123)₁₀ = 1 x 10² + 2 x 10¹ + 3 x 10⁰
= 100 + 20 + 3
= (123)₁₀

Binary Number System:
There are only two digits in the Binary system, namely, 0 and 1. The numbers in the binary system are represented to the base 2 and the positional multipliers are the powers of 2. The left most bit in the binary number is called as the Most Significant Bit (MSB) and it has the largest positional weight. The right most bit is the Least Significant Bit (LSB) and has the smallest positional weight.

Example:
The binary sequence (1101)₂ has the decimal equivalent:
(1101)₂ = 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2⁰
= 8 + 4 + 0 + 1
= (13)₁₀

Octal Number System:
Octal number system uses digits 0,1,2,3,4,5,6 and 7 (8 digits). Each octal digit has its own positional value or weight as a power of 8.
Example:
The Octal sequence (547)₈ has the decimal equivalent:
(547)₈ = 5 x 8² + 4 x 8² + 7 x 8⁰
= 5 x 64 + 4 x 8 + 7 x 1
= 320 + 32 + 7
= (359)₁₀

Hexadecimal Number System:
A hexadecimal number is represented using base 16. Hexadecimal or Hex numbers are used as a shorthand form of binary sequence. This system is used to represent data in a more compact manner. Since 16 symbols are used, 0 to F, the notation is called hexadecimal. The first 10 symbols are the same as in the decimal system, 0 to 9 and the remaining 6 symbols are taken from the first 6 letters of the alphabet sequence, A to F, where A represents 10, B is 11, C is 12, D is 13, E is 14 and F is 15.

Question 2.
Explain the octal to decimal conversion and hexadecimal to decimal conversion.
Answer:
Octal to Decimal Conversion:
To convert Octal to Decimal, we can use positional notation method.

1. Write down the Octal digits and list the powers of 8 from right to left(Positional Notation).
2. For each positional notation of the digit write the equivalent weight.
3. Multiply each digit with its corresponding weight.
4. Add all the values.

Example:
Convert (1265)₈ to equivalent Decimal number

(1265)₈ = 512 x 1 + 64 x 2 + 8 x 6 + 1 x 5
= 512 + 128 + 48 + 5
(1265)₈ = (693)₁₀

Hexadecimal to Decimal Conversion:
To convert Hexadecimal to Decimal we can use positional notation method.

1. Write down the Hexadecimal digits and list the powers of 16 from right to left (Positional Notation).
2. For each positional notation written for the digit, now write the equivalent weight.
3. Multiply each digit with its corresponding weight.
4. Add all the values to get one final value.

Example:
Convert (25F)₁₆ into its equivalent Decimal number.

Question 3.
Explain the binary addition and binary subtraction
Answer:
Binary Addition:
The following table is useful when adding two binary numbers.

Example Add: 1011₂ + 1001₂

Example : Perform Binary addition for the
following : 23₁₀ + 12₁₀
Step 1 : Convert 23 and 12 into binary form

Step 2 : Binary addition of 23 and 12:

Binary Subtraction:
The table for Binary Subtraction is as follows:

When subtracting 1 from 0, borrow 1 from the next Most Significant Bit, when borrowing from the next Most Significant Bit, if it is 1, replace it with 0. If the next Most Significant Bit is 0, you must borrow from a more significant bit that contains 1 and replace it with 0 and Os upto that point become Is.
Example : Subtract 1001010₂ – 10100₂

Example Perform binary addition for the following: (-21)₁₀ + (5)₁₀
Step 1 : Change – 21 and 5 into binary form

Step 2:

Step 3.
Binary Addition of – 21 and 5:

Question 4.
Explain the theorems of boolean algebra.
Answer:
Theorems of Boolean Algebra:
Identity:
A + 0 = A
A. 1 = A

Complement:
A + \(\overline{\mathrm{A}}\) = 1
A. \(\overline{\mathrm{A}}\) = 0

Commutative:
A + B = B + A
A. B = B .A

Associative:
A + (B + C) = (A + B) + C
A. (B . C) = (A. B). C

Distributive
A. (B + C) = A B + A. C
A + (B . C) = (A + B). (A + C)

Null Element:
A + 1 = 1
A. 0 = 0

Involution
(\(\overset { = }{ A }\)) = A

Indempotence:
A +A = A
A.A = A

Absorption:
A + (A . B) = A
A . (A + B) = A

3rd Distributive:
A + \(\overline{\mathrm{A}}\). B = A + B

De Morgan’s:
\(\overline{\mathrm{A+B}}\) = \(\overline{\mathrm{A}}\).\(\overline{\mathrm{B}}\)
(\(\overline{\mathrm{A.B}}\)) = \(\overline{\mathrm{A}}\) + \(\overline{\mathrm{B}}\)

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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: