Additional Questions - Chapter 7 - Mensuration - 10th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Questions
Question 1.
If the radii of the circular ends of a conical bucket which is $45 \mathrm{~cm}$ high are $28 \mathrm{~cm}$ and $7 \mathrm{~cm}$, find the capacity of the bucket. (Use $\pi=\frac{22}{7}$ )
Solution:
Clearly bucket forms frustum of a cone such that the radii of its circular ends are $r_1=28 \mathrm{~cm}, \mathrm{r}_2=7$ $\mathrm{cm}, \mathrm{h}=45 \mathrm{~cm}$.
Capacity of the bucket $=$ volume of the frustum
$
\begin{aligned}
& \Rightarrow \frac{1}{3} \times \pi h\left[r_1^2+r_2^2+r_1 r_2\right] \\
& \left.\Rightarrow \frac{1}{3} \times \frac{22}{7} \times 45\left[28^2+7^2+28 \times 7\right)\right] \\
& \Rightarrow 22 \times 15 \times(28 \times 4+7+28) \\
& \Rightarrow 330 \times 147 \mathrm{~cm}^3 \Rightarrow 48510 \mathrm{~cm}^3
\end{aligned}
$
Question 2.
Find the depth of a cylindrical tank of radius $28 \mathrm{~m}$, if its capacity is equal to that of a rectangular tank of size $28 \mathrm{~m} \times 16 \mathrm{~m} \times 11 \mathrm{~m}$.
Solution:

Volume of the cylindrical tank $=$ Volume of the rectangle tank
$
\begin{aligned}
& \pi r^2 h=28 \times 16 \times 11 \mathrm{~m}^3 \\
& \frac{22}{7} \times 28 \times 28 \times h=28 \times 16 \times 11 \\
& h=\frac{16 \times 11}{88}=2 \mathrm{~m}
\end{aligned}
$
Question 3.
What is the ratio of the volume of a cylinder, a cone, and a sphere. If each has the same diameter and same height?
Solution:
Volume of a cylinder $=\pi r^2 h$
Volume of a cone $=\frac{1}{3} \pi r^2 h$
Volume of a sphere $=\frac{4}{3} \pi r^3$
Their ratio $\mathrm{V}_1: \mathrm{V}_2: \mathrm{V}_3$
$
\begin{aligned}
& \pi r^2 h: \frac{1}{3} \pi r^2 h: \frac{4}{3} \pi r^2 \\
& h: \frac{h}{3}: \frac{4 r}{3} \\
& 3 h: h: 4 r \\
& 3 h: h: 2(2 r) \\
& \therefore \mathrm{V}_1: \mathrm{V}_2: \mathrm{V}_3=3: 1: 2 \quad \text { (where } 2 r=h \text { ) }
\end{aligned}
$
Question 4.
Find the number of coins, $1.5 \mathrm{~cm}$ is diameter and $0.2 \mathrm{~cm}$ thick, to be melted to form a right circular cylinder of height $10 \mathrm{~cm}$ and diameter $4.5 \mathrm{~cm}$.

Solution:
No. of coins required
$
\begin{aligned}
& =\frac{\text { Volume of the cylinder }}{\text { Volume of } 1 \text { coin }} \\
& =\frac{\pi r_1^2 h_1}{\pi r_2^2 h_2}=\frac{\pi \times \frac{45}{20} \times \frac{45}{20} \times 10}{\pi \times \frac{15}{20} \times \frac{15}{20} \times \frac{2}{10}} \\
& = \\
& =\frac{45 \times 45 \times 10}{20 \times 20} \times \frac{20}{15} \times \frac{20}{15} \times \frac{10}{2} \\
& =450
\end{aligned}
$
Question 5.
A spherical ball of iron has been melted and made into small balls. If the raidus of each smaller ball is one-fourth of the radius of the original one, how many such balls can be made?
Solution:
$
\begin{aligned}
\text { No. of balls required } & =\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi \times\left(\frac{r}{4}\right)^3} \\
& =r^3 \times \frac{4}{r} \times \frac{4}{r} \times \frac{4}{r}=64
\end{aligned}
$
Question 6.
A wooden article was made by scooping out a hemisphere from each end of a cylinder as shown in figure. If the height of the cylinder is $10 \mathrm{~cm}$ and its base is of radius $3.5 \mathrm{~cm}$ find the total surface area of the article.

Solution:
Radius of the cylinder be $\mathrm{r}$ Height of the cylinder be $\mathrm{h}$ Total surface area of the article $=\mathrm{CSA}$ of cylinder $+$ CSA of 2 hemispheres $=2 \pi r h+2 \pi r^2=2 \pi r(h+2 r)$
$
\begin{aligned}
& =2 \times \frac{22}{7} \times 3.5 \times(10+2 \times 3.5) \\
& =22 \times 17=374 \mathrm{~cm}^2
\end{aligned}
$