Exercise-11.4
Maths - Exercise
Q1. Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m
Given the radius of the sphere, we use the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
For (i) when the radius \( r = 7 \) cm, substituting the value and using \(\pi \approx \frac{22}{7}\), we get:
\[ \begin{aligned} V &= \frac{4}{3} \times \frac{22}{7} \times 7^3 \\\\&= \frac{4}{3} \times \frac{22}{7} \times 343 \end{aligned}\]
Simplifying, the factors \(7\) in the denominator and numerator cancel out:
\[ \begin{aligned} V &= \frac{4}{3} \times 22 \times 49 \\\\&= \frac{4}{3} \times 1078 \\\\&= 1437.33 \, \text{cm}^3 \end{aligned}\]
So, the volume of the sphere with radius 7 cm is approximately \(1437.33 \, \text{cm}^3\).
For (ii) when the radius \( r = 0.63 \) m, substituting into the volume formula:
\[ \begin{aligned} V &= \frac{4}{3} \pi (0.63)^3 \\\\&= \frac{4}{3} \times \frac{22}{7} \times 0.63^3 \end{aligned}\]
Calculating \(0.63^3 = 0.250047\) approximately, we have:
\[\begin{aligned} V &= \frac{4}{3} \times \frac{22}{7} \times 0.250047 \\\\&\approx 1.047 \, \text{m}^3 \end{aligned}\]
Hence, the volume of the sphere with radius 0.63 m is approximately \(1.05 \, \text{m}^3\).
Q2. Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m
Solution:
The amount of water displaced by a solid spherical ball is equal to the volume of the ball itself, since the ball displaces water equivalent to its own volume when submerged.
For the first case, the diameter of the ball is 28 cm. To find the radius, we divide the diameter by 2, so:
\[ r = \frac{28}{2} = 14 \text{ cm} \]
The volume \(V\) of the sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting \(r = 14\) cm and \(\pi \approx \frac{22}{7}\), the volume is:
\[\begin{aligned} V &= \frac{4}{3} \times \frac{22}{7} \times 14^3 \\\\&= \frac{4}{3} \times \frac{22}{7} \times 2744 \end{aligned}\]
Simplifying by canceling 7 in the denominator with part of 2744:
\[\begin{aligned} V &= \frac{4}{3} \times 22 \times 392 \\\\&= \frac{4}{3} \times 8624 \\\\&= 11498.67 \text{ cm}^3 \end{aligned}\]
This means the ball displaces approximately 11498.67 cubic centimeters of water.
In the second case, the diameter is 0.21 m, so the radius is:
\[ r = \frac{0.21}{2} = 0.105 \text{ m} \]
Using the volume formula again and \(\pi \approx \frac{22}{7}\), we calculate:
\[\begin{aligned} V &= \frac{4}{3} \times \frac{22}{7} \times (0.105)^3 \\\\&= \frac{4}{3} \times \frac{22}{7} \times 0.001157625 \end{aligned}\]
Numerically evaluating yields approximately:
\[ V \approx 0.004851 \text{ m}^3 \]
Therefore, the volume of the ball, and thus the amount of water displaced, is about 0.004851 cubic meters.
Q3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3?
Solution:
Radius of the ball is half the diameter, so \(r=\dfrac{4.2}{2}=2.1\ \text{cm}.\)
The density is \(\rho=8.9\ \text{g/cm}^3\).
The volume \(V\) of a sphere is given by
\[
V=\frac{4}{3}\pi r^{3}.
\]
Substitute \(r=2.1\ \text{cm}\).
First compute \(r^3\):
\[\begin{aligned}
r^3&=(2.1)^3\\&=2.1\times2.1\times2.1\\&=9.261\ \text{cm}^3.
\end{aligned}\]
Now
\[\begin{aligned}
V&=\frac{4}{3}\pi\times 9.261 \\\\&\approx \frac{4}{3}\times 3.142\times 9.261 \\\\&\approx 38.7924\
\text{cm}^3.
\end{aligned}\]
Mass \(m\) = (density) × (volume): \[\begin{aligned} m&=\rho V\\&=8.9\times 38.7924 \\&\approx 345.252\ \text{g}. \end{aligned}\]
The mass of the ball is approximately \(345.25\ \text{g}\).
Q4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Solution:
Assume both Earth and Moon are spheres.
Let \(r_e\) be the radius of the Earth and \(r_m\) the radius of
the Moon.
The problem states the diameter of the Moon is one-fourth the diameter of the Earth, so their
radii satisfy \[r_m=\tfrac{1}{4}r_e\] equivalently \[r_e=4r_m\]
The volume of a sphere is \[V=\dfrac{4}{3}\pi r^{3}\] Thus the volumes of Earth and Moon are \[\begin{aligned} V_e&=\frac{4}{3}\pi r_e^{3}\quad\text{and}\\\\ V_m&=\frac{4}{3}\pi r_m^{3} \end{aligned}\]
The ratio of their volumes is \[\begin{aligned} \frac{V_e}{V_m}&=\frac{\dfrac{4}{3}\pi r_e^{3}}{\dfrac{4}{3}\pi r_m^{3}}\\\\&=\left(\frac{r_e}{r_m}\right)^{3} \end{aligned}\] Since \(r_e=4r_m\) we get \[\begin{aligned} \frac{V_e}{V_m}&=(4)^{3}\\\\&=64 \end{aligned}\]
Therefore the fraction of the Earth's volume that the Moon occupies is \[ \frac{V_m}{V_e}=\frac{1}{64} \] As a percentage this is \(\dfrac{1}{64}\times 100\% \approx 1.5625\%\).
Q5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Solution:
The diameter of the hemispherical bowl is \(10.5\,\text{cm}\),
so its radius is
\[
r=\frac{10.5}{2}=5.25\ \text{cm}
\]
The volume of a hemisphere is \[ V=\frac{2}{3}\pi r^{3} \] Substituting \(r=5.25\): \[\begin{aligned} r^{3}&=5.25\times 5.25\times 5.25\\&=144.703125\ \text{cm}^3 \end{aligned}\] Thus, \[ V=\frac{2}{3}\pi \times 144.703125 \]
Using \(\pi=\frac{22}{7}\), \[\begin{aligned} V&=\frac{2}{3}\times \frac{22}{7}\times 144.703125 \\&\approx 303.0\ \text{cm}^3 \end{aligned}\]
Since \(1000\ \text{cm}^3 = 1\ \text{litre}\), \[\begin{aligned} \text{Capacity}&=\frac{303.0}{1000}\\&\approx 0.303\ \text{litres} \end{aligned}\]
Therefore, the hemispherical bowl can hold approximately \(0.303\ \text{L}\) of milk.
Q6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Solution:
The inner radius of the hemispherical tank is given as \(r_i = 1\ \text{m}\). The iron sheet is \(1\ \text{cm} = 0.01\ \text{m}\) thick, so the outer radius becomes \[ r_o = r_i + 0.01 = 1.01\ \text{m}. \]
The volume of iron used is the difference between the volumes of the outer and inner hemispheres: \[\begin{aligned} V &= \frac{2}{3}\pi r_o^{3} - \frac{2}{3}\pi r_i^{3} \\\\&= \frac{2}{3}\pi \left(r_o^{3}-r_i^{3}\right). \end{aligned}\] Substituting values, \[ \begin{aligned} V=\frac{2}{3}\times \frac{22}{7}\left[(1.01)^{3}-1^{3}\right] \end{aligned} \] Since \((1.01)^{3}=1.030301\), \[ \begin{aligned} V&=\frac{2}{3}\times \frac{22}{7}\times 0.030301 \\\\&\approx 0.06348\ \text{m}^3. \end{aligned}\]
Thus, the volume of iron used to make the hemispherical tank is approximately \(0.06348\ \text{m}^3\).
Q7. Find the volume of a sphere whose surface area is 154 \(cm^2\).
Solution:
The surface area of a sphere is given by
\[
S = 4\pi r^{2}
\]
Here, \(S = 154\ \text{cm}^2\)
Let the radius be \(r\). Substituting the value,
\[
154 = 4\pi r^{2}
\]
Solving for \(r^{2}\),
\[
\begin{aligned}
r^{2}&=\frac{154}{4\pi}
\\\\&=\frac{154}{4\times \frac{22}{7}}
\\\\&=\frac{154\times 7}{88}
\\\\&=12.25.
\end{aligned}\]
Thus,
\[
r=\sqrt{12.25}=3.5\ \text{cm}
\]
Now the volume of the sphere is \[ V=\frac{4}{3}\pi r^{3} \] Substituting \(r=3.5\), \[ V=\frac{4}{3}\times \frac{22}{7}\times 3.5\times 3.5\times 3.5 \] Evaluating, \[ V\approx 179.67\ \text{cm}^3 \]
Therefore, the volume of the sphere is approximately \(179.67\ \text{cm}^3\).
Q8. A dome of a building is in the form of a hemisphere. From inside, it was
white-washed at the cost of ₹ 4989.60. If the cost of white-washing is ₹ 20 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
Solution:
The total cost of white-washing the inside surface of the hemispherical dome is ₹ 4989.60 and the
rate is ₹ 20 per square metre.
The inside curved surface area is therefore
\[\begin{aligned}
\text{Surface Area}&=\frac{4989.60}{20}\\\\&=249.48\ \text{m}^2
\end{aligned}\]
Let the radius of the dome be \(r\). For a hemisphere, the curved surface area is \[ \text{CSA}=2\pi r^{2} \] So, \[ 249.48 = 2\pi r^{2} \] Solving for \(r^{2}\), \[\begin{aligned} r^{2}&=\frac{249.48}{2\pi}\\\\ &=\frac{249.48}{2\times \frac{22}{7}}\\\\ &=\frac{249.48\times 7}{44}\\\\ &=39.69 \end{aligned}\] Thus, \[ r=\sqrt{39.69}=6.3\ \text{m} \]
The volume of the air inside the dome is the volume of the hemisphere: \[ V=\frac{2}{3}\pi r^{3} \] Substituting \(r=6.3\), \[\begin{aligned} V&=\frac{2}{3}\times \frac{22}{7}\times 6.3\times 6.3\times 6.3 \\\\&\approx 523.908\ \text{m}^3 \end{aligned}\]
Therefore, the inside curved surface area of the dome is \(249.48\ \text{m}^2\) and the volume of air it contains is approximately \(523.908\ \text{m}^3\).
Q9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to
form a sphere with surface area S′. Find the
(i) radius r′ of the new sphere,
(ii) ratio of S and S′.
Solution:
Each of the twenty-seven small spheres has radius \(r\) and surface area \(S=4\pi r^{2}\). Its volume is \[ V=\frac{4}{3}\pi r^{3} \] When 27 such spheres are melted, the total volume becomes \[\begin{aligned} V_{\text{total}}&=27\times \frac{4}{3}\pi r^{3}\\\\&=36\pi r^{3} \end{aligned}\]
Let the radius of the new sphere be \(r'\). Its volume is \[ \frac{4}{3}\pi r'^{3} \] Since the volume is preserved during melting, \[ 36\pi r^{3}=\frac{4}{3}\pi r'^{3} \] Solving for \(r'\), \[ r'^{3}=\frac{36\times 3}{4}\,r^{3}=27r^{3}, \] \[ r'=(27r^{3})^{1/3}=3r \]
The surface area of the new sphere is \[\begin{aligned} S'&=4\pi (r')^{2}\\&=4\pi (3r)^{2}\\&=36\pi r^{2} \end{aligned}\]
The required ratio of the original and new surface areas is \[\begin{aligned} S:S' &= 4\pi r^{2} : 36\pi r^{2} \\&= 1:9 \end{aligned}\]
Thus, the new radius is \[r' = 3r\] The ratio of surface areas is \[1 : 9\]
Q10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?
Solution:
The diameter of the spherical capsule is \(3.5\,\text{mm}\),
so its radius becomes
\[
r=\frac{3.5}{2}=1.75\,\text{mm}
\]
The volume of a sphere is given by \[ V=\frac{4}{3}\pi r^{3} \] Substituting \(r=1.75\), \[ V=\frac{4}{3}\times \frac{22}{7}\times 1.75\times 1.75\times 1.75 \] Evaluating the expression, \[ V\approx 22.46\ \text{mm}^{3} \]
Therefore, approximately \(22.46\ \text{mm}^3\) of medicine is required to fill the capsule.
