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

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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 = \(\frac{1}{3}\)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.
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 = \(\frac{1}{3}\)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)
Answer:
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.
Answer:
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 ?
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 = \(\frac{1}{3}\)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.
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 = \(\frac{1}{3}\)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.
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 = \(\frac{1}{3}\)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.
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 = \(\frac{1}{3}\)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.
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 = \(\frac{1}{3}\)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