# JAC Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.7

Jharkhand Board JAC Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.7 Textbook Exercise Questions and Answers.

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

Page-233

Question 1.
Find the volume of the right circular cone with:
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm.
(i) Radius (r) = 6 cm
Height (h) = 7 cm
∴ Volume e the cone = $$\frac{1}{3}$$πr2h
= $$\frac{1}{3} \times \frac{22}{7} \times 6^2 \times 7$$
= 264 cm3.

(ii) Radius (r) = 3.5 cm
Height (h) = 12 cm
∴ Volume of the cone = $$\frac{1}{3}$$πr2h
= $$\frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 12$$ cm3
= 154 cm3

Question 2.
Find the capacity in litres of a conical vessel with :
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm.
(i) Radius (r) = 7 cm
Slant height (l) = 25 cm
Let h be the height of the conical vessel.
h = $$\sqrt{l^{2} – r^{2}}$$
⇒ h = $$\sqrt{25^{2} – 7^{2}}$$
⇒ h = $$\sqrt{576}$$
⇒ h = 24 cm

Volume of the cone = $$\frac{1}{3}$$πr2h
= $$\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24$$ cm3
= 1232 cm3

Capacity of the vessel = = $$\frac{1232}{1000}$$ litres
= 1.232 litres.

(ii) Height (l) = 12 cm
Slant height (h) = 13 cm
Let r be the radius of the conical vessel.
r = $$\sqrt{l^{2} – h^{2}}$$
⇒ r2 = $$\sqrt{13^{2} – 12^{2}}$$
⇒ r = $$\sqrt{25}$$
⇒ r = 5 cm

Volume of the cone = $$\frac{1}{3}$$πr2h
= $$\frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12$$ cm3
= $$\frac{2200}{7}$$ cm3

Capacity
= $$\frac{2200}{7 \times 1000} \text { litres }$$
= $$\frac{11}{35} \text { litres }$$

Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use π = 3.14)
Height of the cone (h) = 15 cm
Volume of the cone = 1570 cm3
Let the radius of base of a cone be r сm.

∴ Volume of the cone = 1570 cm3
⇒ $$\frac{1}{3}$$πr2h = 1570
⇒ $$\frac{1}{3}$$ × 3.14 × r2 × 15 = 1570
⇒ 15.7 × r2 = 1570
⇒ r2 = $$\frac{1570}{3.14 × 5}$$ = 100
⇒ r2 = 100
⇒ r = 10 cm

Question 4.
If the volume of a right circular cone of height 9 cm is 48π cm3, find the diameter of its base.
Height (h) = 9 cm
Volume = 48π cm3
Let the radius of the base of the cone be r сm.

∴ Volume = 48π cm3
⇒ $$\frac{1}{3}$$πr2h = 48π
⇒ $$\frac{1}{3}$$ × r2 × 9 = 48
⇒ 3r2 = 48
⇒ r2 = $$\frac{48}{3}$$ = 16
⇒ r = 4 cm
∴ Diameter of the base of the cone = 2 × r = 2 × 4 cm
= 8 cm

Question 5.
A conical pit of top diameter 3.5 m is 12m deep. What is its capacity in kilolitres ?
Diameter of the top of the conical pit (d) = 3.5 m
Radius (r) = $$\frac{3.5}{2}$$ = 1.75 m
Depth of the pit (h) = 12 m

Volume = $$\frac{1}{3}$$πr2h
= $$\frac{1}{3} \times \frac{22}{7} \times1.75 \times1.75 \times 12$$ = 38.5 m3
∵ 1 m3 = 1 kilolitres
Capacity of the pit = 38.5 kilolitres.

Question 6.
The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find :
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone.
(i) Diameter of the base of the cone (d) = 28 cm
Radius (r) = $$\frac{28}{2}$$ cm = 14 cm
Let the height of the cone be h cm

The volume of the cone = $$\frac{1}{3}$$πr2h
= 9856 cm3
⇒ $$\frac{1}{3}$$πr2h = 9856
⇒ $$\frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times h$$ = 9856
⇒ h = $$\frac{9856 \times 3}{\frac{22}{7} \times 14 \times 14}$$
⇒ h = 48 cm

(ii) Radius (r) = 14 cm
Height (h) = 48 cm
Let l be the slant height of the cone.
l2 = h2 + r2
⇒ l2 = 482 + 142
⇒ l2 = 2304 + 196
⇒ l2 = 2500
⇒ l = $$\sqrt{2500}$$
⇒ l = 50 cm.

(iii) Radius (r) = 14 cm
Slant Height (l) = 50 cm

Curved surface area = πrl
= $$\frac{22}{7}$$ × 14 × 50 cm2
= 2200 cm2.

Question 7.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
On revolving the ΔABC alone the side 12 cm, a cone of the radius (r) 5 cm, height (h) 12 cm and slant height (l) 13 cm will be formed
Volume of the solid so obtained = $$\frac{1}{3}$$πr2h
= $$\frac{1}{3}$$ × π × 5 × 5 cm3 = 100 π cm3

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.
On revolving the ΔABC alone the side 5 cm, a cone of the radius (r) 12 cm, height (h) 5 cm and slant height (l) 13 cm will be formed
Volume of the solid so obtained = $$\frac{1}{3}$$πr2h
= $$\frac{1}{3}$$ × π × 12 × 12 × 5 = 240π cm3

Ratio of the Volume = $$\frac{100π}{240π}$$ = $$\frac{5}{12}$$
= 5 : 12

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 covered by canvas to protect it from rain. Find the area of the canvas required.
Diameter of the base of the conical heap (d) = 10.5 m
Radius (r) = $$\frac{10.5}{2}$$ = 5.25 m
Height (h) of the cone = 3 m

Volume of the heap = $$\frac{1}{3}$$πr2h
= $$\frac{1}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 3$$ m3
= 86.625 m3

Also,
l2 = h2 + r2
⇒ l2 = 32 + (5.25)2
⇒ l2 = 9 + 27.5625
⇒ l2 = 36.5625
⇒ l = $$\sqrt{36.5625}$$ = 6.05 cm.

Area of canvas = Curved surface area = πrl
= $$\frac{22}{7}$$ × 5.25 × 6.05 m2
= 99.825 m2