Exercise 7.4 - Chapter 7 Mensuration 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $7.4$
Question 1.
An aluminium sphere of radius $12 \mathrm{~cm}$ is melted to make a cylinder of radius $8 \mathrm{~cm}$. Find the height of the cylinder.
Solution:
Sphere $-$ Radius $r_1=12 \mathrm{~cm}$
Cylinder $-$ Radius $r_2=8 \mathrm{~cm}$
$\mathrm{h}_2=?$
Volume of cylinder $=$ Volume of sphere melted
$
\pi r_2^2 h_2=\frac{4}{3} \pi r_1^3
$

Question 2 .
Water is flowing at the rate of $15 \mathrm{~km}$ per hour through a pipe of diameter $14 \mathrm{~cm}$ into a rectangular tank which is $50 \mathrm{~m}$ long and $44 \mathrm{~m}$ wide. Find the time in which the level of water in the tanks will rise by $21 \mathrm{~cm}$.
Solution:
In cylinder,
$
\begin{aligned}
& \mathrm{r}=7 \mathrm{~cm}=0.7 \mathrm{~m} \\
& 1=15 \mathrm{~km}=15000 \mathrm{~m}
\end{aligned}
$
In tank
$
1=50 \mathrm{~m}
$

$
\begin{aligned}
& \mathrm{b}=44 \mathrm{~m} \\
& \mathrm{~h}=0.21 \mathrm{~m} \\
& \text { Volume of water in tank }=1 \mathrm{lb} \\
& =50 \times 44 \times 0.21 \\
& =462 \mathrm{~m}^3 \\
& \text { Height of cylinderical pipe }=\frac{\text { Volume }}{\pi r^2} \\
& =\frac{462}{(0.07)^2\left(\frac{22}{7}\right)} \\
& =\frac{462}{0.0154} \\
& =30000 \mathrm{~m} \\
& \text { Time }=\frac{30000}{15000}=2 \text { hours. } \\
&
\end{aligned}
$
Question 3.
A conical flask is full of water. The flask has base radius $\mathrm{r}$ units and height $\mathrm{h}$ units, the water poured into a cylindrical flask of base radius xr units. Find the height of water in the cylindrical flask.
Solution:
The volume of water poured $=$ The volume of the water in the conical flask.

$
\begin{aligned}
\pi r_2^2 h_2 & =\frac{1}{3} \pi r_1^2 h_1 \\
\frac{22}{7} \times(x r)(x r) h_2 & =\frac{1}{3} \times \frac{22}{7} \times r \times h \\
x^2 h_2 & =\frac{h}{3} \\
\therefore h_2 & =\frac{h}{3 x^2} \text { units }
\end{aligned}
$
$\therefore$ The height of the water in the cylindrical flask $=\frac{h}{3 x^2}$ units.
Question 4.
A solid right circular cone of diameter $14 \mathrm{~cm}$ and height $8 \mathrm{~cm}$ is melted to form a hollow sphere. If the external diameter of the sphere is $10 \mathrm{~cm}$, find the internal diameter.
Solution:
Volume of the hollow sphere made $=$ Volume of the cone melted.

$
\begin{aligned}
\frac{4}{3} \pi\left(\mathrm{R}^3-r^3\right) & =\frac{1}{3} \pi r^2 h \\
\frac{4}{3} \pi\left[5^3-\mathrm{r}^3\right] & =\frac{1}{3} \pi \times 7 \times 7 \times 8 \\
\frac{4}{3} \times \frac{22}{7}\left(125-r^3\right) & =\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 8 \\
125-r^3 & =392^{98} \times \frac{1}{4} \\
-r^3 & =98-125 \\
r^3 & =27 \Rightarrow r=3 \\
\text { The internal diameter } & =2 r=2 \times 3 \\
& =6 \mathrm{~cm} .
\end{aligned}
$
Question 5.
Seenu's house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (underground tank) which is in the shape of a cuboid. The sump has dimensions $2 \mathrm{~m} \times$ $1.5 \mathrm{~m} \times 1 \mathrm{~m}$. The overhead tank has its radius of $60 \mathrm{~cm}$ and height $105 \mathrm{~cm}$. Find the volume of the water left in the sump after the overhead tank has been completely filled with water from the sump which has been full, initially.
Solution:
Volume of water in the sump $=\mathrm{lbh}$
$
=2 \times 1.5 \times 1=200 \times 150 \times 100=3000000 \mathrm{~cm}^3
$

Volume of water in the overhead tank $=\pi r^2 h$
$
\begin{aligned}
&=\frac{22}{7} \times 60 \times 60 \times 105 \\
&= 1188000 \mathrm{~cm}^3 \\
& \therefore \text { The volume of water left in the sump }=3000000-1188000 \\
&= 1812000 \mathrm{~cm}^3
\end{aligned}
$
Question 6.
The internal and external diameter of a hollow hemispherical shell are $6 \mathrm{~cm}$ and $10 \mathrm{~cm}$ respectively. If it is melted and recast into a solid cylinder of diameter $14 \mathrm{~cm}$, then find the height of the cylinder.
Solution:

$
\begin{aligned}
& \mathrm{D}=10 \mathrm{~cm} \\
& \mathrm{R}=5 \mathrm{~cm} \\
& \mathrm{~d}=6 \mathrm{~cm} \\
& \mathrm{r}=3 \mathrm{~cm}
\end{aligned}
$

Volume of the cylinder made $=$ Volume of hemisphere melted.

$\pi r^2 h=\frac{2}{3} \pi\left(\mathrm{R}^3-r^3\right)$ cubic units
$
\begin{aligned}
\frac{2 z}{7} \times 7 \times 7 \times h & =\frac{2}{3} \times \frac{2 z}{7}\left(5^3-3^3\right) \\
h & =\frac{2}{3}(125-27) \times \frac{1}{7} \times \frac{1}{7} \\
& =\frac{2}{3} \times 98 \times \frac{1}{7} \times \frac{1}{7}
\end{aligned}
$
$\therefore$ Height of the cylinder made $=1.33 \mathrm{~cm}$.
Question 7.
A solid sphere of radius $6 \mathrm{~cm}$ is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is $5 \mathrm{~cm}$ and its height is $32 \mathrm{~cm}$, then find the thickness of the cylinder.
Solution:
Volume of the hollow cylinder made $=$ Volume of the sphere melted.
$
\begin{aligned}
\frac{22}{7}\left(5^2-r^2\right) 32 & =\frac{4}{3} \times \frac{22}{7} \times 6 \times 6 \times 6 \\
25-r^2 & =\frac{4}{3} \times 6 \times 6 \times 6 \times \frac{1}{32} \\
25-r^2 & =9 \\
-r^2 & =9-25=-16 \\
r^2 & =16 \Rightarrow r=4 \mathrm{~cm} .
\end{aligned}
$
$\therefore$ The thickness $=$ External radius $-$ Internal radius
$
=5-4=1
$
Question 8.
A hemispherical bowl is filled to the brim with juice. The juice is poured into a cylindrical vessel whose radius is $50 \%$ more than its height. If the diameter is same for both the bowl and the cylinder then find the percentage of juice that can be transferred from the bowl into the cylindrical vessel.
Solution:
Diameter of the bowl $=$ Diameter of the cylinder

Volume of the hemispherical bowl $=\frac{2}{3} \pi r_1^3$
Volume of the cylindrical vessel
$
\begin{aligned}
& =\pi r_2^2 h \\
& =\frac{22}{7} \times\left(\frac{3}{2} h\right)^2 \times h=\frac{22}{7} \times \frac{9}{4} h^3 \\
& =\frac{22}{7} \times \frac{9}{4}\left(\frac{2}{3} r\right)^3 \\
& =\frac{22}{7} \times \frac{9}{4} \times \frac{g_1}{2 r_2} r^3=\frac{2}{3} \pi r_1^3
\end{aligned}
$
$\therefore$ Both volumes are equal.
$\therefore 100 \%$ of juice that can be transferred from the bowl into the cylindrical vessel.