EXERCISE 13.8
Q 1, Ex. 13.8, Page No 236, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Page No 236:
Question 1:
Find the volume of a sphere whose radius is(i) 7 cm (ii) 0.63 m
$\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
(i) Radius of sphere = 7 cmVolume of sphere $=\frac{4}{3} \pi r^{3}$
$=\left[\frac{4}{3} \times \frac{22}{7} \times(7)^{3}\right] \mathrm{cm}^{3}$
$=\left(\frac{4312}{3}\right) \mathrm{cm}^{3}$
$=1437 \frac{1}{3} \mathrm{cm}^{3}$
Therefore, the volume of the sphere is 1437$\frac{1}{3}$ cm3.
(ii) Radius of sphere = 0.63 m
Volume of sphere $=\frac{4}{3} \pi r^{3}$
$=\left[\frac{4}{3} \times \frac{22}{7} \times(0.63)^{3}\right] \mathrm{m}^{3}$
$=1.0478 \mathrm{m}^{3}$
Therefore, the volume of the sphere is 1.05 m3 (approximately).
Q 2, Ex. 13.8, Page No 236, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 2:
Find the amount of water displaced by a solid spherical ball of diameter(i) 28 cm (ii) 0.21 m
$\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
(i) Radius (r) of ball $=\left(\frac{28}{2}\right) \mathrm{cm}$=14cm
Volume of ball $=\frac{4}{3} \pi r^{3}$
$=\left[\frac{4}{3} \times \frac{22}{7} \times(14)^{3}\right] \mathrm{cm}^{3}$
$=11498 \frac{2}{3} \mathrm{cm}^{3}$
Therefore, the volume of the sphere is $11498 \frac{2}{3}$cm3.
(ii)Radius (r) of ball $=\left(\frac{0.21}{2}\right) \mathrm{m}$= 0.105 m
Volume of ball $=\frac{4}{3} \pi r^{3}$
$=\left[\frac{4}{3} \times \frac{22}{7} \times(0.105)^{3}\right] \mathrm{m}^{3}$
=0.004851 m3
Therefore, the volume of the sphere is 0.004851 m3.
Q 3, Ex. 13.8, Page No 236, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 3:
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? $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Radius (r) of metallic ball $=\left(\frac{4.2}{2}\right) \mathrm{cm}$=2.1cm
Volume of metallic ball $=\frac{4}{3} \pi r^{3}$
$=\left[\frac{4}{3} \times \frac{22}{7} \times(2.1)^{3}\right] \mathrm{cm}^{3}$
$=38.808 \mathrm{cm}^{3}$
Density $=\frac{\text { Mass }}{\text { Volume }}$
Mass = Density × Volume
= (8.9 × 38.808) g
= 345.3912 g
Hence, the mass of the ball is 345.39 g (approximately)
Q 4, Ex. 13.8, Page No 236, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 4:
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?Answer:
Let the diameter of earth be d. Therefore, the radius of earth will be $\frac{d}{2}$.Diameter of moon will be $\frac{d}{4}$ and the radius of moon will be $\frac{d}{8}$.
Volume of moon $=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \pi\left(\frac{d}{8}\right)^{3}$
$=\frac{1}{512} \times \frac{4}{3} \pi d^{3}$
Volume of earth $=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \pi\left(\frac{d}{2}\right)^{3}$
$=\frac{1}{8} \times \frac{4}{3} \pi d^{3}$
$\frac{\text { Volume of moon }}{\text { Volume of earth }}=\frac{\frac{1}{512} \times \frac{4}{3} \pi d^{3}}{\frac{1}{8} \times \frac{4}{3} \pi d^{3}}$
$=\frac{1}{64}$
Volume of moon$=\frac{1}{64}$ Volume of earth
Therefore, the volume of moon is $\frac{1}{64}$ of the volume of earth
Q 5, Ex. 13.8, Page No 236, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 5:
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?$\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Radius (r) of hemispherical bowl $=\left(\frac{10.5}{2}\right) \mathrm{cm}$= 5.25 cm
Volume of hemispherical bowl $=\frac{2}{3} \pi r^{3}$
$=\left[\frac{2}{3} \times \frac{22}{7} \times(5.25)^{3}\right] \mathrm{cm}^{3}$
= 303.1875 cm3
Capacity of the bowl $=\left(\frac{303.1875}{1000}\right)$ litre
=0.3031875 litre
=0.303 litre(approximately)
Therefore, the volume of the hemispherical bowl is 0.303 litre
Q 6, Ex. 13.8, Page No 236, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 6:
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. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Inner radius (r1) of hemispherical tank = 1 mThickness of hemispherical tank = 1 cm = 0.01 m
Outer radius (r2) of hemispherical tank = (1 + 0.01) m = 1.01 m
Volume of iron used to make such a tank$= \frac{2}{3}\left(r_{2}^{3}-r_{1}^{3}\right)$
$=\left[\frac{2}{3} \times \frac{22}{7} \times\left\{(1.01)^{3}-(1)^{3}\right\}\right] \mathrm{m}^{3}$
$=\left[\frac{44}{21} \times(1.030301-1)\right] \mathrm{m}^{3}$
$=0.06348 \mathrm{m}^{3}$ (approximatley)
Q 7, Ex. 13.8, Page No 236 - Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 7:
Find the volume of a sphere whose surface area is 154 cm2. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Let radius of sphere be r.Surface area of sphere = 154 cm2
⇒ 4πr2 = 154 cm2
$\Rightarrow r^{2}=\left(\frac{154 \times 7}{4 \times 22}\right) \mathrm{cm}^{2}$
$\Rightarrow r=\left(\frac{7}{2}\right) \mathrm{cm}$
=3.5cm
Volume of sphere $=\frac{4}{3} \pi r^{3}$
$=\left[\frac{4}{3} \times \frac{22}{7} \times(3.5)^{3}\right] \mathrm{cm}^{3}$
$=179 \frac{2}{3} \mathrm{cm}^{3}$
Therefore, the volume of the sphere is $179 \frac{2}{3}$cm3.
Q 8, Ex. 13.8, Page No 236, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 8:
A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square meter, find the(i) inside surface area of the dome,
(ii) volume of the air inside the dome. $\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
(i) Cost of white-washing the dome from inside = Rs 498.96Cost of white-washing 1 m2 area = Rs 2
Therefore, CSA of the inner side of dome $=\left(\frac{498.96}{2}\right) \mathrm{m}^{2}$
= 249.48 m2
(ii) Let the inner radius of the hemispherical dome be r.
CSA of inner side of dome = 249.48 m2
2πr2 = 249.48 m2
$\Rightarrow 2 \times \frac{22}{7} \times r^{2}=249.48 \mathrm{m}^{2}$
$\Rightarrow r^{2}=\left(\frac{249.48 \times 7}{2 \times 22}\right) \mathrm{m}^{2}$
=39.69m2
⇒ r = 6.3 m
Volume of air inside the dome = Volume of hemispherical dome
$=\frac{2}{3} \pi r^{3}$
$=\left[\frac{2}{3} \times \frac{22}{7} \times(6.3)^{3}\right] \mathrm{m}^{3}$
= 523.908 m3
= 523.9 m3 (approximately)
Therefore, the volume of air inside the dome is 523.9 m3.
Q 9, Ex. 13.8, Page No 236, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 9:
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’.
Answer:
(i)Radius of 1 solid iron sphere = rVolume of 1 solid iron sphere $=\frac{4}{3} \pi r^{3}$
Volume of 27 solid iron spheres $=27 \times \frac{4}{3} \pi r^{3}$
27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres. Let the radius of this new sphere be r‘.
Volume of new solid iron sphere $=\frac{4}{3} \pi {r'}^{3}$
$\frac{4}{3} \pi {r'}^{3}=27 \times \frac{4}{3} \pi r^{3}$
$r^{\prime 3}=27 r^{3}$
r'=3r
(ii) Surface area of 1 solid iron sphere of radius r = 4πr2
Surface area of iron sphere of radius r‘ = 4π (r‘)2
= 4 π (3r)2 = 36 πr2
$\frac{S}{S^{\prime}}=\frac{4 \pi r^{2}}{36 \pi r^{2}}$
$=\frac{1}{9}$
=1:9
Q 10, Ex 13.8, Page No 236, Surface Areas & Volumes, Maths Class 9th
Question 10:
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?$\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Radius (r) of capsule $=\left(\frac{3.5}{2}\right) \mathrm{mn}$=1.75mm
Volume of spherical capsule $=\frac{4}{3} \pi r^{3}$
$=\left[\frac{4}{3} \times \frac{22}{7} \times(1.75)^{3}\right] \mathrm{mm}^{3}$
= 22.458 mm3
= 22.46 mm3 (approximately)
Therefore, the volume of the spherical capsule is 22.46 mm3
0 comments:
Post a Comment