Exercise 7.3 - Chapter 7 Mensuration 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

E x 7.3

Question 1.
A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter is 14 $\mathrm{cm}$ and the height of the vessel is $13 \mathrm{~cm}$. Find the capacity of the vessel.
Solution:
Diameter $=14 \mathrm{~cm}$
Radius $=7 \mathrm{~cm}$
Total height $=13 \mathrm{~cm}$
Height of the cylindrical part $=13-7$
$=6 \mathrm{~cm}$
$\therefore$ Capacity of the vessel $=$ Capacity of the cylinder $+$ Capacity of the hemisphere.

Volume of the cylinder $=\pi r^2 h$
$
=\frac{22}{7} \times 7 \times 7 \times 6
$
Volume of the hemisphere
$
\begin{aligned}
& =\frac{2}{3} \pi r^3=\frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 \\
& =\frac{2156}{3}=718.67
\end{aligned}
$
$\therefore$ The total volume $=924+718.67$
The capacity of the vessel $=1642.67 \mathrm{~cm}^3$
Question 2.
Nathan, an engineering student was asked to make a model shaped like a cylinder with two cones attached at its two ends. The diameter of the model is $3 \mathrm{~cm}$ and its length is $12 \mathrm{~cm}$. If each cone has a height of $2 \mathrm{~cm}$, find the volume of the model that Nathan made.
Solution:
Volume of the model $=$ Volume of the cylinder $+$ Volume of 2 cones.

$
=\pi r^2 h+2 \frac{1}{3} \pi r^2 h
$
Volume of the cylinder part
$
\begin{aligned}
& =\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 8 \\
& =\frac{396}{7}=56.57 \mathrm{~cm}^3
\end{aligned}
$
Volume of the conical parts
$
\begin{aligned}
& =\not 2 \times \frac{1}{\not 3} \times \frac{22}{7} \times \frac{\not 3}{\not 2} \times \frac{3}{\not 2} \times \not 2 \\
& =9.42 \mathrm{~cm}^3 \\
\therefore \quad \text { Total volume } & =56.57+9.42 \\
& =65.99 \mathrm{~cm}^3
\end{aligned}
$
The volume of the model that Nathan made $=66 \mathrm{~cm}^3$
Question 3.
From a solid cylinder whose height is $2.4 \mathrm{~cm}$ and the diameter $1.4 \mathrm{~cm}$, a cone of the same height and same diameter is carved out. Find the volume of the remaining solid to the nearest $\mathrm{cm}^3$.
Solution:
Volume of the cylinder $=\pi^2 \mathrm{~h}$ cu. units
Volume of the cone $=\frac{1}{3} \pi r^2 h$ cu. units

$
\begin{aligned}
& \mathrm{d}=1.4 \mathrm{~cm}, \mathrm{r}=0.7 \mathrm{~cm}=\frac{7}{10} \\
& \mathrm{~h}=2.4 \mathrm{~cm}=\frac{24}{10}
\end{aligned}
$
Volume of the cylinder:
$
\begin{aligned}
& =\frac{22}{7} \times \frac{7}{10} \times \frac{7}{10} \times \frac{24}{10} \\
& =\frac{3696}{1000}=3.696 \mathrm{~cm}^3 .
\end{aligned}
$
Volume of cone carved out
$
\begin{aligned}
& =\frac{1}{\not 3} \times \frac{22}{7} \times \frac{7}{10} \times \frac{7}{10} \times \frac{24^8}{10} \\
& =\frac{1232}{1000}=1.232 \mathrm{~cm}^3
\end{aligned}
$
$\therefore$ Volume of the remaining solid $=$ Volume of the cylinder $-$ Volume of the cone
$
=3.696-1.232
$
$
\begin{aligned}
& =2.464 \\
& =2.46 \mathrm{~cm}^3
\end{aligned}
$
Question 4.
A solid consisting of a right circular cone of height $12 \mathrm{~cm}$ and radius $6 \mathrm{~cm}$ standing on a hemisphere of radius $6 \mathrm{~cm}$ is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water displaced out of the cylinder, if the radius of the cylinder is $6 \mathrm{~cm}$ and height is $18 \mathrm{~cm}$.

Solution:
Volume of water displaced out $=$ Volume of the solid immersed in. Volume of the solid = Volume of the cone $+$ Volume of the hemisphere

$
\begin{aligned}
\therefore \text { Volume of the cone } & =\frac{1}{3} \pi r^2 h \\
& =\frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 12 \\
& =\frac{3168}{7}=452.57 \mathrm{~cm}^3 \ldots(1)
\end{aligned}
$
Volume of the hemisphere $=\frac{2}{3} \pi r^3$
$
\begin{aligned}
& =\frac{2}{3} \times \frac{22}{7} \times 6 \times 6 \times 6 \\
& =\frac{3168}{7}=452.57 \mathrm{~cm}^3
\end{aligned}
$
$\therefore$ The volume of water displaced out $=$ Volume of the solid

$
\begin{aligned}
& =(1)+(2) \\
& =905.14 \mathrm{~cm}^3
\end{aligned}
$
Question 5.
A capsule is in the shape of a cylinder with two hemisphere stuck to each of its ends. If the length of the entire capsule is $12 \mathrm{~mm}$ and the diameter of the capsule is $3 \mathrm{~mm}$, how much medicine it can hold?
Solution:
Volume of medicine the capsule can hold = Volume of the cylinder $+2$ volume of hemisphere Volume of the cylinder part
$
\begin{aligned}
=\pi r^2 h & =\frac{22^{11}}{7} \times \frac{3}{2} \times \frac{3}{2} \times 6^3 \\
& =\frac{297}{7}=42.428 \mathrm{~mm}^3
\end{aligned}
$
Volume of 2 hemispherical parts

$\therefore$ The total volume $=56.571 \mathrm{~mm}^3$
$\therefore$ The volume of the medicine the capsule can hold $=56.57 \mathrm{~mm}^3$

Question 6.
As shown in figure a cubical block of side $7 \mathrm{~cm}$ is surmounted by a hemisphere. Find the surface area of the solid.

Solution:
Clearly, greatest diameter of the hemisphere is equal to the length of an edge of the cube is $7 \mathrm{~cm}$. Radius of the hemisphere $=\frac{7}{2} \mathrm{~cm}$
Now, total surface area of the solid $=$ Surface area of the cube $+$ Curved surface area of the hemisphere - Area of the base of the hemisphere.
$
\begin{aligned}
& =\left(6 \times 7^2+2 \times \frac{22}{7} \times\left(\frac{7}{2}\right)^2-\frac{22}{7} \times\left(\frac{7}{2}\right)^2\right) \mathrm{cm}^2 \\
& =\left(294+77-\frac{77}{2}\right) \mathrm{cm}^2 \\
& =\left(294+\frac{77}{2}\right) \mathrm{cm}^2 \\
& =332.5 \mathrm{~cm}^2
\end{aligned}
$
Question 7.
A right circular cylinder just enclose a sphere of radius $\mathrm{r}$ units. Calculate
(i) the surface area of the sphere
(ii) the curved surface area of the cylinder
(iii) the ratio of the areas obtained in (i) and (ii).
Solution:

(i) Surface area of sphere $=4 \pi r^2$ sq. units
(ii) C.S.A of cylinder

(iii) Ratio $=\frac{4 \pi r^2}{4 \pi r^2}=1: 1$
Question 8.
A shuttle cock used for playing badminton has the shape of a frustum of a cone is mounted on a hemisphere. The diameters of the frustum are $5 \mathrm{~cm}$ and $2 \mathrm{~cm}$. The height of the entire shuttle cock is $7 \mathrm{~cm}$. Find its external surface area.
Solution:
External surface area of the cock $=$ Surface area of frustum $+$ CSA of hemisphere

$\begin{aligned}
& \text { CSA of frustum }=\pi(\mathrm{R}+r) l \text { sq. units. } \\
& \text { Here } \mathrm{R}=\frac{5}{2} \mathrm{~cm} \\
&
\end{aligned}$

$\begin{aligned}
r & =\frac{2}{2}=1 \mathrm{~cm} . \\
l & =\sqrt{(\mathrm{R}-r)^2+h^2} \\
& =\sqrt{(2.5-1)^2+6^2} \\
& =\sqrt{1.5^2+36} \\
& =\sqrt{2.25+36} \\
& =\sqrt{38.25} \\
& \cong 6.18
\end{aligned}$

$\therefore$ CSA of the frustum
$
\begin{aligned}
&=\frac{22}{7} \times 3.5 \times 6.1=\frac{469.7}{7} \\
&=67.1 \mathrm{~cm}^2 \\
& \text { CSA of hemisphere }=2 \pi r^2 \\
&=2 \times \frac{22}{7} \times 1 \times 1 \\
&=6.28 \mathrm{~cm}^2 \\
& \therefore \text { Total external surface area } \\
&=67.1+6.28 \\
&=73.38 \mathrm{~cm}^2 \\
&=73.39 \mathrm{~cm}^2 \text { (approx.) }
\end{aligned}
$