Exercise 7.5 - Chapter 7 Mensuration 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $7.5$
Multiple choice questions.
Question 1.
The curved surface area of a right circular cone of height $15 \mathrm{~cm}$ and base diameter $16 \mathrm{~cm}$ is
(1) $60 \pi \mathrm{cm}^2$
(2) $68 \pi \mathrm{cm}^2$
(3) $120 \pi \mathrm{cm}^2$
(4) $136 \pi \mathrm{cm}^2$
Answer:
(4) $136 \pi \mathrm{cm}^2$
Hint:
$
\begin{aligned}
l & =\sqrt{h^2+r^2}=\sqrt{15^2+8^2} \\
& =\sqrt{225+64}=\sqrt{289}=17 \mathrm{~cm} \\
\pi r l & =\frac{22}{7} \times 8 \times 17=136 \pi \mathrm{cm}^2
\end{aligned}
$
Question 2.
If two solid hemispheres of same base radius $r$ units are joined together along with their bases, then the curved surface area of this new solid is
(1) $4 \pi r^2$ sq. units
(2) $67 \pi r^2$ sq. units
(3) $3 \pi r^2$ sq. units
(4) $8 \pi r^2$ sq. units
Solution:
(1) $47 \pi r^2$ sq. units]

Question 3.
The height of a right circular cone whose radius is $5 \mathrm{~cm}$ and slant height is $13 \mathrm{~cm}$ will be
(1) $12 \mathrm{~cm}$
(2) $10 \mathrm{~cm}$
(3) $13 \mathrm{~cm}$
(4) $5 \mathrm{~cm}$
Answer:
(1) $12 \mathrm{~cm}$
Hint:

Question 4.
If the radius of the base of a right circular cylinder is halved keeping the same height, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is
(1) $1: 2$
(2) $1: 4$
(3) $1: 6$
(4) $1: 8$
Solution:
(2) $1: 4$

Hint:

$
\begin{aligned}
\mathrm{V}_2 & =\pi\left(\frac{r}{2}\right)^2 h=\frac{\pi r^2 h}{4} \\
\mathrm{~V}_1 & =\pi r^2 h=\pi r^2 h \\
\frac{\mathrm{V}_2}{\mathrm{~V}_1} & =\frac{\pi r^2 h}{\pi r^2 h}=\frac{\pi r^2 h}{4} \times \frac{1}{\pi r^2 h} \\
& =\frac{1}{4}=1: 4
\end{aligned}
$

Question 5.
The total surface area of a cylinder whose radius is $\frac{1}{3}$ of its height is
(1) $\frac{9 \pi h^2}{8}$ sq.units
(2) $24 \pi h^2$ sq.units
(3) $\frac{8 \pi h^2}{9}$ sq.units
(4) $\frac{56 \pi h^2}{9}$ sq.units
Solution:
(3) $\frac{8 \pi h^2}{9}$ sq. units
$
\begin{aligned}
\text { TSA of a cylinder } & =2 \pi r(h+r), r=\frac{1}{3} h \\
& =2 \pi \frac{1}{3} h\left(h+\frac{1}{3} h\right) \\
& =\frac{2}{3} \pi h^2+\frac{2}{3} \pi \frac{h^2}{3} \\
& =\frac{2}{3} \pi h^2+\frac{2}{3} \pi \frac{h^2}{3} \\
& =\frac{6 \pi h^2+2 \pi h^2}{9}=\frac{8 \pi h^2}{9}
\end{aligned}
$
Question 6.
In a hollow cylinder, the sum of the external and internal radii is $14 \mathrm{~cm}$ and the width is $4 \mathrm{~cm}$. If its height is $20 \mathrm{~cm}$, the volume of the material in it is
(1) $560 \pi \mathrm{cm}^3$
(2) $1120 \pi \mathrm{cm}^3$
(3) $56 \pi \mathrm{cm}^3$
(4) $360 \pi \mathrm{cm}^3$
Answer:
(2) $1120 \pi \mathrm{cm}^3$
Hint:
$\mathrm{R}+\mathrm{r}=14 \mathrm{~cm}$
$\mathrm{w}=4 \mathrm{~cm}$
$\mathrm{h}=90 \mathrm{~cm}$

$
\begin{aligned}
\text { Volume } & =\pi\left(\mathrm{R}^2-r^2\right) h \\
& =\pi(\mathrm{R}+r)(\mathrm{R}-r) h \\
& =\pi(14) 4 \times 20 \\
& =1120 \pi \mathrm{cm}^3
\end{aligned}
$
Question 7.
If the radius of the base of a cone is tripled and the height is doubled then the volume is
(1) made 6 times
(2) made 18 times
(3) made 12 times
(4) unchanged
Solution:
(2) made 18 times
Hint:
$
\begin{aligned}
r & =3 r \\
h & =2 h \\
v & =\frac{1}{3} \pi(3 r)^2 \times(2 h) \\
& =\frac{1}{3} \times \pi 9 r^2 \times 2 h \\
& =6 \pi r^2 h=18 \times \frac{1}{3} \pi r^2 h \\
& =\text { made } 18 \text { times. }
\end{aligned}
$
Question 8.

The total surface area of a hemisphere is how many times the square of its radius

(1) $\pi$
(2) $4 \pi$
(3) $3 \pi$
(4) $2 \pi$
Answer:
(3) $3 \pi$
Hint:

$
\text { TSA }=3 \pi r^2
$
Question 9.
A solid sphere of radius $\mathrm{x} \mathrm{cm}$ is melted and cast into a shape of a solid cone of same radius. The height of the cone is
(1) $3 \mathrm{xcm}$
(2) $\mathrm{x} \mathrm{cm}$
(3) $4 \mathrm{x} \mathrm{cm}$
(4) $2 \mathrm{xcm}$
Solution:
(3) $4 \mathrm{xcm}$
Hint:

$
\begin{aligned}
\frac{1}{3} \pi r^2 h & =\frac{4}{3} \pi r^3 \\
\frac{1}{\not} \pi \chi^2 h & =\frac{4}{\not 3} \pi \chi^x x \\
h & =4 x \mathrm{~cm}
\end{aligned}
$
Question 10.
A frustum of a right circular cone is of height $16 \mathrm{~cm}$ with radii of its ends as $8 \mathrm{~cm}$ and $20 \mathrm{~cm}$.Then, the volume of the frustum is
(1) $3328 \pi \mathrm{cm} 3$
(2) $3228 \pi \mathrm{cm} 3$
(3) $3240 \pi \mathrm{cm} 3$
(4) $3340 \pi \mathrm{cm} 3$
Answer:
(1) $3328 \pi \mathrm{cm} 3 \mathrm {Hint}:$

$
\begin{aligned}
\mathrm{V} & =\frac{\pi h}{3}\left[\mathrm{R}^2+r^2+\mathrm{Rr}\right] \\
& =\pi \times \frac{16}{3}\left[20^2+8^2+20 \times 8\right] \\
& =\pi \times \frac{16}{3}[400+64+160] \\
& =\pi \times \frac{16}{3} \times 624=3328 \pi \mathrm{cm}^3
\end{aligned}
$
Question 11.
A shuttlecock used for playing badminton has the shape of the combination of
(1) a cylinder and a sphere
(2) a hemisphere and a cone
(3) a sphere and a cone
(4) frustum of a cone and a hemisphere
Solution:
(4) frustum of a cone and a hemisphere
Question 12.
A spherical ball of radius $r_1$ units is melted to make 8 new identical balls each of radius $r_2$ units.
Then $r_1: r_2$ is
(1) $2: 1$
(2) $1: 2$
(3) $4: 1$
(4) $1: 4$
Solution:

Question 13.
The volume (in $\mathrm{cm}^3$ ) of the greatest sphere that can be cut off from a cylindrical log of wood of base radius $1 \mathrm{~cm}$ and height $5 \mathrm{~cm}$ is
Solution:
(1) $\frac{4}{3} \pi$
$
\frac{4}{3} \pi \times 1=\frac{4}{3} \pi
$
Question 14.
The height and radius of the cone of which the frustum is a part are $h_1$ units and $r_1$ units respectively. Height of the frustum is $h_2$ units and the radius of the smaller base is $r_2$ units. If $h_2: h_1$ $=1: 2$ then $r_2: r_1$ is
(1) $1: 3$
(2) $1: 2$
(3) $2: 1$
(4) $3: 1$
Solution:
(2) $1: 2$

$\begin{aligned}
& h_2: h_1=1: 2 \\
& r_2: r_1 \\
& \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{1}{\frac{1}{3} \pi r_1^2\left(h_1\right)} \\
& \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{r_2^2 h_1-r_2^2 h_2}{r_1^2 h_1} \\
& =\frac{r_2^2 2 h_2-r_2^2 h_2}{r_1^2 2 h_2}=\frac{r_2^2\left(2 h_2-h_2\right)}{r_1^2 2 h_2} \\
& =\frac{r_2^2 \cdot h_2}{r_1^2 2 h_2}=\frac{r_2^2}{r_1^2}=\frac{1}{2}
\end{aligned}$

Question 15.
The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is
(1) $1: 2: \overline{3}$
(2) $2: 1: 3$
(3) $1: 3: 2$
(4) $3: 1: 2$
Answer:
(4) $3: 1: 2$
Hint:

$h: \frac{h}{3}: \frac{4}{3} r$