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.

Answer:

(i) Radius (r) = 6 cm

Height (h) = 7 cm

∴ Volume e the cone = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3} \times \frac{22}{7} \times 6^2 \times 7\)

= 264 cm^{3}.

(ii) Radius (r) = 3.5 cm

Height (h) = 12 cm

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

= \(\frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 12\) cm^{3}

= 154 cm^{3}

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.

Answer:

(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 = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24\) cm^{3}

= 1232 cm^{3}

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 = \(\sqrt{l^{2} – h^{2}}\)

⇒ r^{2} = \(\sqrt{13^{2} – 12^{2}}\)

⇒ r = \(\sqrt{25}\)

⇒ r = 5 cm

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

= \(\frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12\) cm^{3}

= \(\frac{2200}{7}\) cm^{3}

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 cm^{3}, find the radius of the base. (Use π = 3.14)

Answer:

Height of the cone (h) = 15 cm

Volume of the cone = 1570 cm^{3}

Let the radius of base of a cone be r сm.

∴ Volume of the cone = 1570 cm^{3}

⇒ \(\frac{1}{3}\)πr^{2}h = 1570

⇒ \(\frac{1}{3}\) × 3.14 × r^{2} × 15 = 1570

⇒ 15.7 × r^{2} = 1570

⇒ r^{2} = \(\frac{1570}{3.14 × 5}\) = 100

⇒ r^{2} = 100

⇒ r = 10 cm

Question 4.

If the volume of a right circular cone of height 9 cm is 48π cm^{3}, find the diameter of its base.

Answer:

Height (h) = 9 cm

Volume = 48π cm^{3}

Let the radius of the base of the cone be r сm.

∴ Volume = 48π cm^{3}

⇒ \(\frac{1}{3}\)πr^{2}h = 48π

⇒ \(\frac{1}{3}\) × r^{2} × 9 = 48

⇒ 3r^{2} = 48

⇒ r^{2} = \(\frac{48}{3}\) = 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 ?

Answer:

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 = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3} \times \frac{22}{7} \times1.75 \times1.75 \times 12\) = 38.5 m^{3}

∵ 1 m^{3} = 1 kilolitres

Capacity of the pit = 38.5 kilolitres.

Question 6.

The volume of a right circular cone is 9856 cm^{3}. 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.

Answer:

(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 = \(\frac{1}{3}\)πr^{2}h

= 9856 cm^{3}

⇒ \(\frac{1}{3}\)πr^{2}h = 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.

l^{2} = h^{2} + r^{2}

⇒ l^{2} = 48^{2} + 14^{2}

⇒ l^{2} = 2304 + 196

⇒ l^{2} = 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 cm^{2}

= 2200 cm^{2}.

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.

Answer:

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 = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3}\) × π × 5 × 5 cm^{3} = 100 π 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.

Answer:

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 = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3}\) × π × 12 × 12 × 5 = 240π cm^{3}

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.

Answer:

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 = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 3\) m^{3}

= 86.625 m^{3}

Also,

l^{2} = h^{2} + r^{2}

⇒ l^{2} = 3^{2} + (5.25)^{2}

⇒ l^{2} = 9 + 27.5625

⇒ l^{2} = 36.5625

⇒ l = \(\sqrt{36.5625}\) = 6.05 cm.

Area of canvas = Curved surface area = πrl

= \(\frac{22}{7}\) × 5.25 × 6.05 m^{2}

= 99.825 m^{2}