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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.3

**10th Maths Exercise 7.3 Samacheer Kalvi Question 1.**

A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter is 14 cm and the height of the vessel is 13 cm. Find the capacity of the vessel.

Solution:

Diameter = 14 cm

Radius = 7 cm

Total height = 13 cm

Height of the cylindrical part = 13 – 7

= 6 cm

∴ Capacity of the vessel = Capacity of the cylinder + Capacity of the hemisphere.

∴ The total volume = 924 + 718.67

The capacity of the vessel = 1642.67 cm^{3}

**Exercise 7.3 Class 10 Samacheer Kalvi Question 2.**

Nathan, an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of the model that Nathan made.

Solution:

Volume of the model = Volume of the cylinder + Volume of 2 cones.

The volume of the model that Nathan made = 66 cm^{3}

**Ex 7.3 Class 10 Samacheer Question 3.**

From a solid cylinder whose height is 2.4 cm and the diameter 1.4 cm, a cone of the same height and same diameter is carved out. Find the volume of the remaining solid to the nearest cm^{3}.

Solution:

Volume of the cylinder = πr^{2}h cu. units

Volume of the cone = \(\frac{1}{3} \pi r^{2} h\) cu. units

d = 1.4 cm, r = 0.7 cm = \(\frac{7}{10}\)

h = 2.4 cm = \(\frac{24}{10}\)

Volume of the cylinder:

Volume of cone carved out

∴ Volume of the remaining solid = Volume of the cylinder – Volume of the cone

= 3.696 – 1.232

= 2.464

= 2.46 cm^{3}

**10th Maths Exercise 7.3 Question 4.**

A solid consisting of a right circular cone of height 12 cm and radius 6 cm standing on a hemisphere of radius 6 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water displaced out of the cylinder, if the radius of the cylinder is 6 cm and height is 18 cm.

Solution:

Volume of water displaced out = Volume of the solid immersed in.

Volume of the solid = Volume of the cone + Volume of the hemisphere

∴ The volume of water displaced out = Volume of the solid

= (1) + (2)

= 905.14 cm^{3}

**9th Maths Exercise 7.3 Samacheer Kalvi Question 5.**

A capsule is in the shape of a cylinder with two hemisphere stuck to each of its ends. If the length of the entire capsule is 12 mm and the diameter of the capsule is 3 mm, how much medicine it can hold?

Solution:

Volume of medicine the capsule can hold = Volume of the cylinder + 2 volume of hemisphere

Volume of the cylinder part

∴ The total volume = 56.571 mm^{3}

∴ The volume of the medicine the capsule can hold = 56.57 mm^{3}

**10th Maths Mensuration Exercise 7.3 Question 6.**

As shown in figure a cubical block of side 7 cm is surmounted by a hemisphere. Find the surface area of the solid.

Solution:

Clearly, greatest diameter of the hemisphere is equal to the length of an edge of the cube is 7 cm.

Radius of the hemisphere = \(\frac{7}{2}\) cm

Now, total surface area of the solid = Surface area of the cube + Curved surface area of the hemisphere – Area of the base of the hemisphere.

**10th Maths 7.3 Question 7.**

A right circular cylinder just enclose a sphere of radius r units.

Calculate

(i) the surface area of the sphere

(ii) the curved surface area of the cylinder

(iii) the ratio of the areas obtained in (i) and (ii).

Solution:

(i) Surface area of sphere = 4πr^{2} sq. units

**Samacheer Kalvi 10th Maths Mensuration Question 8.**

A shuttle cock used for playing badminton has the shape of a frustum of a cone is mounted on a hemisphere. The diameters of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm. Find its external surface area.

Solution:

External surface area of the cock = Surface area of frustum + CSA of hemisphere