## MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.7

Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.

Find the volume of the right circular cone with

- radius 6 cm, height 7 cm
- radius 3.5 cm, height 12 cm.

Solution:

1. Here, radius of the cone r = 6 cm

height (h) = 7 cm

Volume = \(\frac{1}{3}\) x πr^{2}h

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x 6 x 6 x 7 cm^{3}

= 22 x 2 x 6 cm^{3}

= 264 cm^{3}

2. Here, radius of the cone (r) ;

= 3.5 cm = \(\frac{35}{10}\) cm

Height (h) = 12 m

Volume of the cone

Question 2.

Find the capacity in litres of a conical vessel with

- radius 7 cm, slant height 25 cm
- height 12 cm, slant height 13 cm

Solution:

1. Here, r = 7 and l = 25 cm

Thus, the required capacity of the conical vessel is 1.232 l.

2. Here, height (h) – 12 cm and l = 13 cm

Thus, the required capacity of the conical vessel is \(\frac{11}{35}\) l.

Question 3.

The height of a cone is 15 cm. If its volume is 1570 cm^{3}, find the radius of the base. (Use 71 = 3.14)

Solution:

Here, height of the cone (h) = 15 cm

Volume of the cone (v) = 1570 cm^{3}

Let the radius of the base be ‘r’ cm.

Question 4.

If the volume of a right circular cone of height 9 cm is 48 JI cm3, find the diameter of its base.

Solution:

Volume of cone = \(\frac{1}{3}\) πr^{2}h

\(\frac{1}{3}\) x πr^{2} x 9 = 48π

r^{2} = \(\frac{48π}{9π}\) x 3

r^{2} = 16

r = 4 cm

Diameter = 2 x 4 = 8 cm

Question 5.

A conical pit of top diameter 3,5 m is 12 m deep. What is its capacity in kilolitres?

Solution:

d = 3.5 m

r = 1.75 m

h = 12m

Volume of the pit = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 1. 75 x 1.75 x 12

= 38.5 m^{3} = 38.5 kl (1 kl= 1 m^{3})

Question 6.

The volume of a right circular cone is 9856 cm^{3}. If the diameter of the base 28 cm, find.

(i) height of the cone.

(ii) slant height of the cone.

(iii) curved surface area of the cone.

Solution:

V = 9856 cm^{2}

d =28 cm

r = 14 cm

Question 7.

A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the sides 12 cm. Find the volume of the solid 30 obtained.

Solution:

h = 12cm

r = 5cm

Volume of solid, V_{1} = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3}\) x π x 5 x 5 x 12

= 100π cm^{3}

Question 8.

If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Solution:

h = 5 cm

r = 12 cm

Volume of solid, V_{2} = \(\frac{1}{3}\) x π x 12 x 12 x 5

= 240π

Question 9.

A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be by covered canvas to protect it from rain. Find the area of the canvas required.

Solution:

d = 10.5 m ⇒ r = 5.25 m

h = 3m

Volume of heap of wheat = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 5.25 x 5.25 x 3

= 86.625 m^{3}

l^{2} = 3^{2} + (5.25)^{2}

l = \(\sqrt{9+27.56}\) = \(\sqrt{36.56}\) = 6.04 cm

Area of canvas required = CSA of cone –

= \(\frac{22}{7}\) x 5.25 x 6.04 = 99.66 m^{2}