These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1

Question 1.

2 cubes each of volume 64 cm^{3} are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Volume of one cube = 64 cm^{3}

Let edge of one cube = a

Volume of the cube = (edge)^{3}

a^{3} = 64 ⇒ a = 4 cm

Similarly, edge of the another cube = 4 cm.

Now, both cubes are joined together and a cuboid is formed as shown in the figure.

Now, length of the cuboid (l) = 8 cm

breadth of the cuboid (b) = 4 cm

height of the cuboid (h) = 4 cm

Surface area of the cuboid so formed = 2 (lb + bh + hl)

= 2(8 x 4 + 4 x 4 + 4 x 8)

= 2(32 + 16 + 32) = 160 cm²

Question 2.

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Solution:

Given: diameter of the hemisphere = 14 cm

Radius = \(\frac { 14 }{ 2 }\) = 7 cm

Curved surface area of the hemisphere = 2πr² = 2 x \(\frac { 22 }{ 7 }\) x 7 x 7 cm²

= 14 x 22 cm² = 308 cm²

Here, total height of the vessel = 13 cm

Height of the cylinder = Total height – Height of the hemisphere = 13 cm – 7 cm = 6 cm

and radius of the cylinder = radius of the hemisphere = 7 cm

Inner surface area of the cylinder = 2πrh = 2 x \(\frac { 22 }{ 7 }\) x 7 x 6

= 2 x 22 x 6 = 264 cm²

Inner surface area of the vessel = Inner surface area of the cylinder + curved surface area of the hemisphere

= 264 cm² + 308 cm² = 572 cm².

Question 3.

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution:

Radius of the hemispheree, r = 3.5 cm

Radius of the base of the cone, r = 5 cm

Height of the cone, h = (15.5 – 3.5) cm = 12 cm

∴ total surface area of the toy = (curved surface area of the hemisphere + (curved surface area of the cone)

Hence, the total surface area of the toy is 214.5 cm².

Question 4.

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution:

Greatest diameter of hemisphere is = 7 cm (side of cube)

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

Surface area of solid

Question 5.

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution:

Diameter of hemisphere = l

So radius of hemisphere = \(\frac { l }{ 2 }\)

Surface area of cube = 6 (edge)² – 6l².

Surface area of remaining solid = Total surface area of cube-back area of hemisphere + Curved surface area of hemisphere.

Question 6.

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:

Radius of the cylinder = radius of the

hemispherical ends = r = \(\frac { 5 }{ 2 }\) mm

Height of the cylinder = h = 14 – 2 x \(\frac { 5 }{ 2 }\) = 9 mm

∴ Total surface area = curved surface area of the cylinder + surface area of two hemispherical ends

= (2πrh + 2 x πr²) mm²

= 2πr (h + 2r) cm²

= 2 x \(\frac { 22 }{ 7 }\) x \(\frac { 5 }{ 2 }\)(9 + 2 x \(\frac { 5 }{ 2 }\))

= \(\frac { 110 }{ 7 }\)(9 + 5)

⇒ = \(\frac { 110 }{ 7 }\) x 14

∴ Surface area = 220mm²

Question 7.

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)

Solution:

Area of the canvas used = Curved surface area of the cylinder + (Curved surface area of the cone)

Cost of the canvas of the tent at the rate ₹ 500 per m².

= ₹ (44 x 500) = ₹ 22,000.

Question 8.

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

Solution:

Height of solid cylinder is 2.4 cm

Diameter of cylinder is 1.4 cm

∴ Radius = \(\frac { 1.4 }{ 2 }\) cm = 0.7 cm

Total surface area of remaining solid

= 2πrh + πrl² + nrl … (i)

where πrl curved surface area of cone

Question 9.

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Solution:

Radius of the cylinder = Radius of the hemispherical ends = r = 3.5 cm

Height of the cylinder = 10 cm.

Total surface area of the article

= Curved surface area of the cylinder + Surface area of the two hemispherical ends.

= (2πrh + 2 x 2πr²) cm².

= 2πr (h + 2r) cm².

= 2 x \(\frac { 22 }{ 7 }\) x 3.5 (10 + 7)

= 22 (14) = 374 cm².