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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Unit Exercise 7
Question 1.
The barrel of a fountain-pen cylindrical in shape, is $7 \mathrm{~cm}$ long and $5 \mathrm{~mm}$ in diameter. A full barrel of ink in the pen will be used for writing 330 words on an average. How many words can be written using a bottle of ink containing one fifth of a litre?
Solution:
Volume of ink in the bottle $=\frac{1}{5}$ litre $=\frac{1000}{5} \mathrm{~cm}^3$ $=200 \mathrm{~cm}^3$
With $1.375 \mathrm{~cm}^3$, no. of words can be written $=330$ words.

With $200 \mathrm{~cm}^3$, No. of words can be written
$
\begin{aligned}
& =\frac{200 \times 330 \times 1000}{1.375 \times 1000}=\frac{200 \times 330 \times 1000}{1375} \\
& =\frac{66000000}{1375}=48000 \text { words }
\end{aligned}
$

Question 2.
A hemi-spherical tank of radius $1.75 \mathrm{~m}$ is full of water. It is connected with a pipe which empties the tank at the rate of 7 litre per second. How much time will it take to empty the tank completely?

Solution:
Suppose the pipe takes $\mathrm{x}$ seconds to empty the tank. Then, volume of the water that flows out of the tank in $x$ seconds $=$ Volume of the hemispherical tank. Volume of the water that flows out of the tank in $\mathrm{x}$ seconds. $=$ Volume of hemispherical shell of radius $175 \mathrm{~cm}$.

Question 3.
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius $r$ units.

Solution:
Radius of the base of cone $=$ Radius of the hemisphere $=r$
Height of the cone $=$ Radius of the hemisphere

Question 4.
An oil funnel of tin sheet consists of a cylindrical portion $10 \mathrm{~cm}$ long attached to a frustum of a cone. If the total height is $22 \mathrm{~cm}$, the diameter of the cylindrical portion be $8 \mathrm{~cm}$ and the diameter of the top of the funnel be $18 \mathrm{~cm}$, then find the area of the tin sheet required to make the funnel.
Solution:
Slant height of the frustum

$
\begin{aligned}
l & =\sqrt{(\mathrm{R}-r)^2+h_1^2} \\
& =\sqrt{(9-4)^2+12^2} \\
& =\sqrt{5^2+12^2} \\
& =\sqrt{25+144} \\
& =\sqrt{169}=13 \mathrm{~cm}
\end{aligned}
$
Outer surface area $=2 \pi r h_2+\pi(\mathrm{R}+r) l$
$\therefore$ Area of the sheet required to make the funnel $\cong 782.57 \mathrm{~cm}^2$
Question 5.
Find the number of coins, $1.5 \mathrm{~cm}$ in diameter and $2 \mathrm{~mm}$ 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 larger cylinder }}{\text { Volume of } 1 \text { coin }} \\
& =\frac{\pi \mathrm{R}^2 \mathrm{H}}{\pi r^2 h} \\
& \longrightarrow=\frac{\frac{22}{7} \times \frac{45}{20} \times \frac{45}{20} \times 10}{\frac{1.5 \mathrm{~cm}}{22} \times \frac{15}{20} \times \frac{15}{20} \times \frac{2}{10}} \\
& =\frac{22}{7} \times \frac{45}{20} \times \frac{45}{20} \times 10 \times \frac{7}{22} \times \frac{20}{15} \times \frac{20}{15} \times \frac{10}{2} \\
& =450 \text { coins } \\
&
\end{aligned}
$
Question 6.
A hollow metallic cylinder whose external radius is $4.3 \mathrm{~cm}$ and internal radius is $1.1 \mathrm{~cm}$ and whole length is $4 \mathrm{~cm}$ is melted and recast into a solid cylinder of $12 \mathrm{~cm}$ long. Find the diameter of solid cylinder.
Solution:

Volume of the solid cylinder $=$ Volume of the hollow cylinder melted.

$
\begin{aligned}
\pi r_2^2 h & =\pi\left(\mathrm{R}^2-r^2\right) h \\
\frac{22}{7} \times r_2^2 \times 12 & =\frac{22}{7}\left((4.3)^2-(1.1)^2\right) 4 \\
r_2^2 & =\frac{2 z}{7}(18.49-1.21) \times 4 \times \frac{7}{22} \times \frac{1}{12_3} \\
& =\frac{17.28}{3} \\
r_2^2 & =5.76 \\
r_2 & =2.4
\end{aligned}
$
$\therefore$ The diameter of the solid cylinder
$
\begin{aligned}
& =2 \times 2.4 \mathrm{~cm} \\
& =4.8 \mathrm{~cm}
\end{aligned}
$
Question 7.
The slant height of a frustum of a cone is $4 \mathrm{~m}$ and the perimeter of circular ends are $18 \mathrm{~m}$ and $16 \mathrm{~m}$. Find the cost of painting its curved surface area at $\square 100$ per sq. $m$.

Solution:
$
2 \pi \mathrm{R}=18
$

$
\begin{aligned}
\mathrm{R} & =18 \times \frac{1}{2} \times \frac{7}{22}=2.86 \mathrm{~m} \\
2 \pi r & =16 \mathrm{~m} \\
r & =16 \times \frac{1}{2} \times \frac{7}{22} \\
& =2.54 \mathrm{~m}
\end{aligned}
$
$
l=4 \mathrm{~m}
$
C.S.A of frustum $=\pi l(\mathrm{R}+r)$
$
\begin{aligned}
& =\frac{22}{7} \times 4(2.86+2.54) \\
& =\frac{22}{7} \times 4 \times 5.4 \\
& =67.88 \mathrm{~m}^2 \cong 68 \mathrm{~m}^2
\end{aligned}
$
Cost of painting@₹100per sq.m
$
\begin{aligned}
& =6.8 \times 100 \\
& =₹ 6800
\end{aligned}
$
Question 8.
A hemi-spherical hollow bowl has material of volume $\frac{436 \pi}{3}$ cubic $\mathrm{cm}$. Its external diameter is 14 $\mathrm{cm}$. Find its thickness.

Solution:
$
\begin{aligned}
\text { Volume } & =\frac{2}{3} \pi\left(\mathrm{R}^3-r^3\right) \\
\mathrm{D} & =14 \mathrm{~cm}, \mathrm{R}=7 \mathrm{~cm} \\
\frac{2}{\not 3} \times \frac{22}{7} \times\left(7^3-r^3\right) & =\frac{436 \pi}{\not z} \\
\left(343-r^3\right) & =\frac{436}{2} \\
-r^3 & =218-343 \\
-r^3 & =1-125 \\
r & =5 \mathrm{~cm}
\end{aligned}
$

Thickness of the bowl
$
\begin{aligned}
& =\mathrm{R}-r=7-5 \\
& =2 \mathrm{~cm}
\end{aligned}
$
Question 9.
The volume of a cone is $1005 \frac{5}{7} \mathrm{cu} . \mathrm{cm}$. The area of its base is $201 \frac{1}{7} \mathrm{sq} . \mathrm{cm}$. Find the slant height of the cone.

Solution:
$
\begin{aligned}
\text { Volume of a cone } & =1005 \frac{5}{7} \mathrm{cu} . \mathrm{cm} \\
\frac{1}{3} \pi r^2 h & =\frac{7040}{7}
\end{aligned}
$

$
\frac{1}{3} \times \frac{1408}{7} \times h
$
$
\begin{aligned}
& =\frac{7040}{7} \times 3=15 \mathrm{~cm} \\
h & =\frac{7040}{1408}=5 \mathrm{~cm}
\end{aligned}
$
$\therefore$ The height of the cone $=15 \mathrm{~cm}$

$\begin{aligned}
\text { Slant height } l & =\sqrt{h^2+r^2} \\
& =\sqrt{15^2+8^2} \\
& =\sqrt{225+64} \\
& =\sqrt{289} \\
l & =17 \mathrm{~cm}
\end{aligned}$

Question 10.
A metallic sheet in the form of a sector of $\mathrm{T}$ a circle of radius $21 \mathrm{~cm}$ has central angle of $216^{\circ}$. The sector is made into a cone by bringing the bounding radii together. Find the volume of the cone formed.
Solution:
Length of the arc $=\frac{216}{360} \times 2 \times \frac{22}{7} \times 21$
$
=79.2 \mathrm{~cm}
$

Base perimeter of the cone
$
\begin{aligned}
& =\text { length of the arc } \\
2 \pi r & =79.2 \\
r & =79.2 \times \frac{1}{2} \times \frac{7}{22}
\end{aligned}
$

$\begin{aligned}
2 \pi r & =79.2 \\
r & =79.2 \times \frac{1}{2} \times \frac{7}{22} \\
& =12.6 \mathrm{~cm} \\
l & =21 \mathrm{~cm} \\
h & =\sqrt{l^2-r^2} \\
h & =\sqrt{21^2-12.6^2} \\
& =\sqrt{441-158.76} \\
& =\sqrt{282.24} \\
h & =16.8 \mathrm{~cm} \\
\therefore \text { Volume of the cone } & =\frac{1}{3} \pi r^2 h \\
& =\frac{1}{3} \times \frac{22}{7} \times 12.6 \times 12.6 \times 16.8 \\
& =2794.18 \mathrm{~cm}^3
\end{aligned}$