Exercise 7.2 - Chapter 7 Mensuration 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\mathbf{E x} 7.2$
Question 1.
A $14 \mathrm{~m}$ deep well with inner diameter $10 \mathrm{~m}$ is dug and the earth taken out is evenly spread all around the well to form an embankment of width $5 \mathrm{~m}$. Find the height of the embankment.
Solution:
Inner diameter $=10 \mathrm{~m}$
Inner radius $=5 \mathrm{~m}$
Inner height $=14 \mathrm{~m}$

Volume of the cylinder $=\pi r^2 \mathrm{~h}$ cubic units
$
\begin{aligned}
& =\frac{22}{7} \times 5 \times 5 \times 14 \\
& =1100 \mathrm{~m}^3
\end{aligned}
$
Volume of the hollow $=n\left(R^2-r^2\right) h$ cubic units
$
\begin{aligned}
& \mathrm{R}=10 \mathrm{~m} \\
& \mathrm{r}=5 \mathrm{~m} \\
& \Rightarrow \frac{22}{7} \times\left(10^2-5^2\right) h=1100 \mathrm{~m}^3
\end{aligned}
$

$(\because$ the earth taken out $=$ the earth spread all around)
$
\begin{aligned}
\Rightarrow \frac{22}{7} \times(100-25) h & =1100 \mathrm{~m}^3 \\
h & =1100 \times \frac{7}{22} \times \frac{1}{75} \\
& =4.66=4.67 \mathrm{~m}
\end{aligned}
$
The height of the $\cong 4.7 \mathrm{~m}$ embankment.
Question 2.
A cylindrical glass with diameter $20 \mathrm{~cm}$ has water to a height of $9 \mathrm{~cm}$. A small cylindrical metal of radius $5 \mathrm{~cm}$ and height $4 \mathrm{~cm}$ is immersed it completely. Calculate the raise of the water in the glass?
Solution:
The volume of the water raised $=$ Volume of the cylindrical metal.

$
\begin{aligned}
\pi r_2^2 h_2 & =\pi r_1^1 h_1 \\
\pi \times 10^2 \times h_2 & =\pi \times 5^2 \times 4 \\
h_2 & =\frac{100}{100}=1 \mathrm{~cm} .
\end{aligned}
$
$\therefore$ The height of the raised water in the glass $=1 \mathrm{~cm}$.
Question 3.

If the circumference of a conical wooden piece is $484 \mathrm{~cm}$ then find its volume when its height is $105 \mathrm{~cm}$.
Solution:
Circumference of the base of the cone $=484 \mathrm{~cm}$
height $=105 \mathrm{~cm}$

$\therefore 2 \pi r=484$
$
\begin{aligned}
& r=484 \times \frac{1}{2} \times \frac{7}{22} \\
&=77 \mathrm{~cm} \\
& \therefore \quad \text { Its volume }=\frac{1}{3} \pi r^2 h \text { cubic units } \\
&=\frac{1}{\not 3} \times \frac{22}{7} \times 77 \times 77 \times 105 \\
&=652190 \mathrm{~cm}^3 .
\end{aligned}
$
Question 4.
A conical container is fully filled with petrol. The radius is $10 \mathrm{~m}$ and the height is $15 \mathrm{~m}$. If the container can release the petrol through its bottom at the rate of $25 \mathrm{cu}$. meter per minute, in how many minutes the container will be emptied. Round off your answer to the nearest minute.
Solution:
Volume of the cone $=\frac{1}{3} \pi r^2 h$ cu. units.

Volume of the given conical container $=\frac{1}{3} \times \pi \times 10 \times 10 \times 15$
$
=500 \pi \mathrm{m}^3
$
To empty $25 \mathrm{~m}^3$, the time taken $=1 \mathrm{mt}$.
$
\text { To empty } \begin{aligned}
& 500 \pi \mathrm{m}^3 \text { the time taken }=500 \times \frac{22}{7} \times 1 \\
&=62.857 \mathrm{mts} . \frac{25}{7} \\
& \cong 63 \text { minutes.(approx.) }
\end{aligned}
$

Question 5.
A right angled triangle whose sides are $6 \mathrm{~cm}, 8 \mathrm{~cm}$ and $10 \mathrm{~cm}$ is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two solids so formed.

Solution:

When the triangle $\mathrm{ABC}$ is rotated about $\mathrm{AB}$, the

When the $\triangle \mathrm{ABC}$ i rotated about $\mathrm{BC}$,
$
\begin{aligned}
& r=8 \mathrm{~cm}, h=6 \mathrm{~cm} \\
& \mathrm{~V}_2=\frac{1}{3} \times \frac{22}{7} \times 6 \times 8 \times 8=\frac{2816}{7}=402.29 \\
& \therefore \text { Difference in volume }=\mathrm{V}_2-\mathrm{V}_1 \\
&=402.29-301.71 \\
&=100.58 \mathrm{~cm}^3
\end{aligned}
$
Question 6.
The volumes of two cones of same base radius are $3600 \mathrm{~cm}^3$ and $5040 \mathrm{~cm}^3$. Find the ratio of heights.

Solution:
$
\begin{aligned}
\mathrm{V}_1 & =3600 \mathrm{~cm}^2, r_1=r_2 \\
\mathrm{~V}_2 & =5040 \mathrm{~cm}^3 \\
\frac{1}{3} \pi r_1^2 h_1 & =\frac{3600}{5040} \\
\frac{1}{3} \pi r_2^2 h_2 & \\
\frac{h_1}{h_2}=\frac{90}{126} & =\frac{45}{63}=\frac{15}{21}=\frac{5}{7} \\
\therefore h_1: h_2 & =5: 7
\end{aligned}
$
Question 7.
If the ratio of radii of two spheres is $4: 7$, find the ratio of their volumes.
Solution:
$
\begin{aligned}
& \frac{r_1}{r_2}=\frac{4}{7} \\
& \therefore \text { Ratio of the volume of two spheres }=\frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} \\
& =\frac{\frac{4}{3} \times \pi \times 4^3}{\frac{4}{3} \times \pi \times 4^3}=\frac{64}{343} \\
& \therefore \quad \mathrm{V}_1: \mathrm{V}_2=64: 343 \\
&
\end{aligned}
$
Question 8.
A solid sphere and a solid hemisphere have equal total surface area. Prove that the ratio of their volume is $3 \sqrt{3}: 4$

Solution:
Sphere : Hemisphere
$
\begin{aligned}
& 4 \pi r_1^2: 3 \pi r_2^2 \\
& 4 r_1^2: 3 r_2^2 \\
& \therefore \frac{r_1^2}{r_2}=\frac{3}{4} \Rightarrow \frac{r_1}{r_2}=\frac{\sqrt{3}}{2} \\
& \therefore \text { Their volumes }=\frac{4}{3} \pi r_1^3: \frac{4}{3} \pi r_2^3 \\
& \Rightarrow \frac{4}{3} \pi(\sqrt{3})^3: \frac{4}{3} \pi \times 2^3
\end{aligned}
$
The ratio of their volume $=3 \sqrt{3}: 8$
Hence proved.
Question 9.
The outer and the inner surface areas of a spherical copper shell are $576 \pi \mathrm{cm}^2$ and $324 \pi \mathrm{cm}^2$ respectively. Find the volume of the material required to make the shell.
Solution:

$
\begin{aligned}
& 4 \pi \mathrm{R}^2=576 \pi \mathrm{cm}^2 \\
& 4 \pi r^2=324 \pi \mathrm{cm}^2 \\
& 4 \pi \mathrm{R}^2=576 \pi \\
& \mathrm{R}^2=144 \mathrm{~cm}^2 \Rightarrow \mathrm{R}=12 \mathrm{~cm} \\
& 4 \pi r^2=324 \pi \\
& r^2=81 \mathrm{~cm}^2 \Rightarrow r= 9 \mathrm{~cm} \\
& \therefore \text { Volume of the hollow sphere } \\
&=\frac{4}{3} \pi\left(\mathrm{R}^3-r^3\right) \mathrm{cu} . \text { units } \\
&=\frac{4}{3} \times \frac{22}{7} \times\left(12^3-9^3\right) \\
&=\frac{4}{3} \times \frac{22}{7} \times(1728-729) \\
&=\frac{4}{3} \times \frac{22}{7} \times 999 \\
&=\frac{29304}{7} \\
&=4186.285 \\
&=4186.29 \mathrm{cu} . \mathrm{cm} .
\end{aligned}
$
$\therefore$ The volume of the material needed $=4186.29 \mathrm{~cm}^3$.
Question 10.
A container open at the top is in the form of a frustum of a cone of height $16 \mathrm{~cm}$ with radii of its lower and upper ends are $8 \mathrm{~cm}$ and $20 \mathrm{~cm}$ respectively. Find the cost of milk which can completely fill a container at the rate of $\square 40$ per litre.
Solution:
Volume of the frustum

$\begin{aligned}
& =\frac{1}{3} \pi\left(\mathrm{R}^2+\mathrm{R} r+r^2\right) h \text { cubic units } \\
& =\frac{1}{3} \times \frac{22}{7}\left(20^2+20 \times 8+8^2\right) \times 16 \\
& =\frac{1}{3} \times \frac{22}{7}(400+160+64) \times 16 \\
& =\frac{1}{3} \times \frac{22}{7} \times 624 \times 16 \\
& =\frac{73216}{7} \\
& =10459.428 \mathrm{~cm}^3 \\
& 1000 \mathrm{~cm}^3=1 \text { litre } \\
& \therefore 10459.428 \mathrm{~cm}^3=10.459 \text { litres. } \\
& \text { The cost of milk@₹40 per litre } \\
& =₹ 418.36 \text {. } \\
&
\end{aligned}$