Exercise 11.3 (Revised) - Chapter 13 - Surface Areas & Volumes - Ncert Solutions class 9 - Maths

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Chapter 11: Surface Areas & Volumes - NCERT Solutions for Class 9 Maths

Assume $\pi=\frac{22}{7}$ unless stated otherwise.
Ex 11.3 Question 1.

Find the volume of the right circular cone with:
(i) Radius $\mathbf{6 c m}$, Height $7 \mathrm{~cm}$
(ii) Radius $3.5 \mathrm{~cm}$, Height $12 \mathrm{~cm}$

Answer.

(i) Given: $r=6 \mathrm{~cm}, h=7 \mathrm{~cm}$

Volume of cone $=\frac{1}{3} \pi r^2 h$
$
\begin{aligned}
& =\frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 7 \\
& =264 \mathrm{~cm}^3
\end{aligned}
$$
(ii) Given: $r=3.5 \mathrm{~cm}, h=12 \mathrm{~cm}$

Volume of cone $=\frac{1}{3} \pi r^2 h$
$
\begin{aligned}
& =\frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 12 \\
& =154 \mathrm{~cm}^3
\end{aligned}
$
Ex 11.3 Question 2.

Find the capacity of a conical vessel with:
(i) Radius $\mathbf{7 c m}$, Slant height $\mathbf{2 5 ~ c m}$
(ii) Height $12 \mathrm{~cm}$, Slant height $13 \mathrm{~cm}$

Answer.

(i) Given: $r=7 \mathrm{~cm}, l=25 \mathrm{~cm}$
$
h=\sqrt{l^2-r^2}=\sqrt{(25)^2-(7)^2}=\sqrt{625-49}=\sqrt{576}=24 \mathrm{~cm}
$

Capacity of conical vessel $=\frac{1}{3} \pi r^2 h$
$
\begin{aligned}
& =\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24=1232 \mathrm{~cm}^3 \\
& =1.232 \text { liters }\left[\because 1000 \mathrm{~cm}^3=1 \text { liter }\right]
\end{aligned}
$

(ii) Given: $h=12 \mathrm{~cm}, l=13 \mathrm{~cm}$
$
\begin{aligned}
& r=\sqrt{l^2-h^2}=\sqrt{(13)^2-(12)^2} \\
& =\sqrt{169-144} \\
& =\sqrt{25}=5 \mathrm{~cm}
\end{aligned}
$

Capacity of conical vessel $=\frac{1}{3} \pi r^2 h$
$
\begin{aligned}
& =\frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12=\frac{2200}{7} \mathrm{~cm}^3 \\
& =\frac{2200}{7} \times \frac{1}{1000} \text { liters } \\
& {\left[\because 1000 \mathrm{~cm}^3=1 \text { liter }\right]} \\
& =\frac{11}{35} \text { liter }
\end{aligned}
$

Ex 11.3 Question 3.

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

Answer.

Height of the cone ${ }^{(h)}=15 \mathrm{~cm}$

Volume of cone $=1570 \mathrm{~cm}^3$
$
\begin{aligned}
& \Rightarrow \frac{1}{3} \pi r^2 h=1570 \\
& \Rightarrow \frac{1}{3} \times \frac{22}{7} \times r^2 \times 15=1570 \\
& \Rightarrow 15.70 r^2=1570 \\
& \Rightarrow r^2=1570 \times \frac{100}{1570}=100 \\
& \Rightarrow r=10 \mathrm{~cm}
\end{aligned}
$

Hence required radius of the base is $10 \mathrm{~cm}$.

Ex 11.3 Question 4.

If the volume of a right circular cone of height $9 \mathrm{~cm}$ is $48 \pi \mathrm{cm}^3$, find the diameter of the base.

Answer.

Height of the cone ${ }^{(h)}=9 \mathrm{~cm}$

Volume of cone $=48 \pi \mathrm{cm}^3$
$
\begin{aligned}
& \Rightarrow \frac{1}{3} \pi r^2 h=48 \pi \\
& \Rightarrow \frac{1}{3} \pi r^2 \times 9=48 \pi \\
& \Rightarrow 3 r^2=48 \\
& \Rightarrow r^2=\frac{48}{3}=16 \\
& \Rightarrow r=4 \mathrm{~cm}
\end{aligned}
$
$\therefore$ Diameter of base $=2 r=2 \times 4=8 \mathrm{~cm}$

Ex 11.3 Question 5.

A conical pit of top diameter $3.5 \mathrm{~m}$ is $12 \mathrm{~m}$ deep. What is its capacity in kiloliters?

Answer.

Diameter of pit $=3.5 \mathrm{~m}$

$\therefore$ Radius of pit $=\frac{3.5}{2}=1.75 \mathrm{~m}$

Depth of pit ${ }^{(h)}=12 \mathrm{~m}$

Capacity of pit $=\frac{1}{3} \pi r^2 h$
$=\frac{1}{3} \times \frac{22}{7} \times 1.75 \times 1.75 \times 12$
$=\frac{1}{3} \times \frac{22}{7} \times \frac{175}{100} \times \frac{175}{100} \times 12$
$=\frac{1}{3} \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times 12$
$=22 \times \frac{7}{4}=\frac{77}{2} \mathrm{~m}^3=35.8 \mathrm{~m}^3$
$=38.5 \mathrm{kl}\left[\because 1 \mathrm{~m}^3=1 \mathrm{kl}\right]$

Ex 11.3 Question 6.

The volume of a right circular cone is $9856 \mathrm{~cm}^3$. If the diameter of the base if $\mathbf{2 8} \mathbf{~ c m}$, find:
(i) Height of the cone
(ii) Slant height of the cone
(iii) Curved surface area of the cone.

Answer.

(i) Diameter of cone $=28 \mathrm{~cm}$

$\therefore$ Radius of cone $=14 \mathrm{~cm}$

Volume of cone $=9856 \mathrm{~cm}^3$
$\Rightarrow \frac{1}{3} \pi r^2 h=9856$
$\Rightarrow \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times h=9856$
$
\Rightarrow h=\frac{9856 \times 3 \times 7}{22 \times 14 \times 14}=48 \mathrm{~cm}
$
(ii) Slant height of cone ${ }^{(l)}=\sqrt{r^2+h^2}$
$
\begin{aligned}
& =\sqrt{(14)^2+(48)^2} \\
& =\sqrt{196+2304} \\
& =\sqrt{2500}=50 \mathrm{~cm}
\end{aligned}
$
(iii) Curved surface area of cone $=\pi r l={ }^{\frac{22}{7} \times 14 \times 50}=2200 \mathrm{~cm}^2$

Ex 11.3 Question 7.

A right triangle $A B C$ with sides $5 \mathrm{~cm}, 12 \mathrm{~cm}$ and $13 \mathrm{~cm}$ is revolved about the side $12 \mathrm{~cm}$. Find the volume of the solid so obtained. (Use $\pi=3.14$ )

Answer.

When right angled triangle $\mathrm{ABC}$ is revolved about side $12 \mathrm{~cm}$, then the solid formed is a cone.

In that cone, Height ${ }^{(h)}=12 \mathrm{~cm}$
And radius $(r)=5 \mathrm{~cm}$

Therefore, Volume of cone $=\frac{1}{3} \pi r^2 h$
$
\begin{aligned}
& =\frac{1}{3} \pi \times 5 \times 5 \times 12 \\
& =100 \pi \mathrm{cm}^3
\end{aligned}
$

Ex 11.3 Question 8.

If the triangle $A B C$ in question 7 above is revolved about the side $5 \mathrm{~cm}$, then find the volume of the solid so obtained. Find, also, the ratio of the volume of the two solids obtained.

Answer.

When right angled triangle $\mathrm{ABC}$ is revolved about side $5 \mathrm{~cm}$, then the solid formed is a cone.

In that cone, Height ${ }^{(h)}=5 \mathrm{~cm}$

And radius $(r)=12 \mathrm{~cm}$

Therefore, Volume of cone $=\frac{1}{3} \pi r^2 h$
$
\begin{aligned}
& =\frac{1}{3} \pi \times 12 \times 12 \times 5 \\
& =240 \pi \mathrm{cm}^3
\end{aligned}
$

Volume of cone in Q. No. 7
Now, Volume of vone in Q. No. 8
$
=\frac{100 \pi}{240 \pi}=\frac{5}{12}
$
$\therefore$ Required ratio $=5: 12$

Ex 11.3 Question 9.

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

Answer.

Radius $(r)$ of heap
$
=\left(\frac{10.5}{2}\right) \mathrm{m}=5.25 \mathrm{~m}
$

Height $(h)$ of heap $=3 \mathrm{~m}$

Volume of heap
$
\begin{aligned}
& =\frac{1}{3} \pi r^2 h \\
& =\left(\frac{1}{3} \times \frac{22}{7} \times(5.25)^2 \times 3\right) \mathrm{m}^3 \\
& =86.625 \mathrm{~m}^3
\end{aligned}
$

Therefore, the volume of the heap of wheat is $86.625 \mathrm{~m}^3$.

Area of canvas required $=$ CSA of cone
$
\begin{aligned}
& =\pi r l=\pi r \sqrt{r^2+h^2} \\
& =\left[\frac{22}{7} \times 5.25 \times \sqrt{(5.25)^2+(3)^2}\right] \mathrm{m}^2 \\
& =\left(\frac{22}{7} \times 5.25 \times 6.05\right) \mathrm{m}^2 \\
& =99.825 \mathrm{~m}^2
\end{aligned}
$

$\text { Therefore, } 99.825 \mathrm{~m}^2 \text { canvas will be required to protect the heap from rain. }$