Tag: Exercise 11.2 Chapter 11 Volume and Surface Area Class 9th Maths NCERT Solutions

Q1. Find the volume of the right circular cone with

(i) radius $6$ cm, height $7$ cm
(ii) radius $3.5$ cm, height $12$ cm

Volume formula: ${\displaystyle V=\frac{1}{3}\pi r^{2}h}$

(i) $V={\displaystyle \frac{1}{3}}\pi (6^2)(7)={\displaystyle \frac{1}{3}}\pi \cdot 36\cdot 7=84\pi$
Using $\pi ={\displaystyle \frac{22}{7}}$: $84\pi =84\times \frac{22}{7}=264{\text{cm}}^3\mathrm{}$

(ii) $r^2={3.5}^2=12.25$
$V={\displaystyle \frac{1}{3}}\pi (12.25)(12)=49\pi$
With $\pi ={\displaystyle \frac{22}{7}}$: $49\pi =49\times \frac{22}{7}=154{\text{cm}}^3\mathrm{}$

Answers Q1: (i) $264{\text{cm}}^3$(ii) $154{\text{cm}}^3$

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Q2. Capacity in litres of a conical vessel with

(i) radius $7$ cm, slant height $25$ cm
(ii) height $12$ cm, slant height $13$ cm

First find height $h$ where needed using $l^2=r^2+h^2$

Convert cm³ → litres: $1\text{ L}=1000{\text{cm}}^3$

(i) $l^2-r^2=625-49=576\Rightarrow h=24$cm.
$V={\displaystyle \frac{1}{3}}\pi (7^2)(24)=392\pi$. With $\pi ={\displaystyle \frac{22}{7}}$: $V=1232{\text{cm}}^3=1.232\text{ L}\mathrm{}$

(ii) $r=\sqrt{l^2-h^2}=\sqrt{169-144}=5$ cm.
$V={\displaystyle \frac{1}{3}}\pi (5^2)(12)=100\pi$. With $\pi ={\displaystyle \frac{22}{7}}$: $V={\displaystyle \frac{2200}{7}}{\text{cm}}^3\approx 314.29{\text{cm}}^3$ $=0.3143\text{ L}\mathrm{}$

Answers Q2: (i) $1.232\text{ L}$. (ii) ${\displaystyle \frac{2200}{7}}{\text{cm}}^3\approx 0.3143\text{ L}\mathrm{.}$

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Q3. Height of cone $=15$ cm. If volume $=1570{\text{cm}}^3$, find radius. (Use $\pi =3.14$.)

Volume formula: $V={\displaystyle \frac{1}{3}}\pi r^{2}h$. Solve for $r^2={\displaystyle \frac{3V}{\pi h}}$

$$r^2=\frac{3\times 1570}{3.14\times 15}=\frac{4710}{47.1}=100\Rightarrow r=10\text{ cm}\mathrm{}$$

Answer Q3: $r=10\text{ cm}\mathrm{.}$

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Q4. If volume of a cone of height $9$ cm is $48\pi {\text{cm}}^3$, find the diameter of base.

$V={\displaystyle \frac{1}{3}}\pi r^{2}h={\displaystyle \frac{1}{3}}\pi r^2\cdot 9=3\pi r^2$. So $3\pi r^2=48\pi \Rightarrow r^2=16\Rightarrow r=4$
Diameter $=8$ cm.

Answer Q4: Diameter $=8\text{ cm}\mathrm{}$

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Q5. A conical pit of top diameter $3.5$ m is $12$ m deep. Capacity in kilolitres?

Radius $r=1.75$, height $h=12$
$V={\displaystyle \frac{1}{3}}\pi r^{2}h={\displaystyle \frac{1}{3}}\pi ({1.75}^2)(12)=\pi \cdot 12.25{\text{m}}^3\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$

Using $\pi ={\displaystyle \frac{22}{7}}$: $V=12.25\times \frac{22}{7}=38.5{\text{m}}^3\mathrm{}$

$1{\text{m}}^3=1$ kilolitre, so capacity $=38.5\text{ kL}\mathrm{}$

Answer Q5: $38.5\text{ kilolitres}\mathrm{}$

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Q6. Volume of cone $=9856{\text{cm}}^3$. Diameter of base $=28$ cm. Find

(i) height, (ii) slant height, (iii) curved surface area.

Base radius $r=14$ cm. Use $V={\displaystyle \frac{1}{3}}\pi r^{2}h\Rightarrow h={\displaystyle \frac{3V}{\pi r^2}}$

With $\pi ={\displaystyle \frac{22}{7}}$: $\pi r^2={\displaystyle \frac{22}{7}}\times 196=616$
$h={\displaystyle \frac{3\times 9856}{616}}={\displaystyle \frac{29568}{616}}=48\text{ cm}\mathrm{}$

Slant height $l=\sqrt{r^2+h^2}=\sqrt{{14}^2+{48}^2}=\sqrt{196+2304}=\sqrt{2500}=50\text{ cm}\mathrm{}$

Curved surface area $=\pi rl={\displaystyle \frac{22}{7}}\times 14\times 50=2200{\text{cm}}^2\mathrm{}$

Answers Q6: (i) $48\text{ cm}$. (ii) $50\text{ cm}$. (iii) $2200{\text{cm}}^2\mathrm{}$

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Q7. Right triangle with sides $5,12,13$ revolved about side $12$ cm. Volume of solid?

Revolving about side $12$ (a leg) generates a cone of radius $5$ and height $12$.

$V={\displaystyle \frac{1}{3}}\pi (5^2)(12)=100\pi ={\displaystyle \frac{2200}{7}}{\text{cm}}^3\approx 314.29{\text{cm}}^3\mathrm{}$

Answer Q7: $100\pi {\text{cm}}^3(\approx 314.29{\text{cm}}^3)$

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Q8. Same triangle revolved about side $5$ cm. Find volume and ratio of volumes (Q7 : Q8).

Revolving about side $5$ gives cone with radius $12$, height $5$:

$V={\displaystyle \frac{1}{3}}\pi ({12}^2)(5)=240\pi ={\displaystyle \frac{5280}{7}}{\text{cm}}^3\approx 754.29{\text{cm}}^3\mathrm{}$

Ratio $\text{(Q7:Q8)}=100\pi :240\pi =100:240=5:12$

Answers Q8: Volume $=240\pi {\text{cm}}^3(\approx 754.29{\text{cm}}^3)$. Ratio $=5:\mathrm{12<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$

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Q9. Heap of wheat in form of cone: diameter $=10.5$ m, height $=3$ m. Find (i) volume, (ii) area of canvas required to cover it (assume canvas covers curved surface).

Radius $r=5.25$ m. Volume:

$$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi ({5.25}^2)(3)=\pi ({5.25}^2)=\pi \times \mathrm{27.5625.}$$

With $\pi ={\displaystyle \frac{22}{7}}$: $V=27.5625\times \frac{22}{7}=86.625{\text{m}}^3\mathrm{}$

Slant height $l=\sqrt{r^2+h^2}=\sqrt{27.5625+9}=\sqrt{36.5625}=\frac{\sqrt{585}}{4}\approx 6.0467\text{ m}\mathrm{}$

Curved surface area $=\pi rl={\displaystyle \frac{22}{7}}\times 5.25\times 6.0466925\approx 99.77{\text{m}}^2\mathrm{}$

Answers Q9: Volume $=86.625{\text{m}}^3\mathrm{}$ Canvas (curved surface) Area $\approx 99.77{\text{m}}^2\mathrm{.}$