Surface Areas and Volumes: 9 Math
NCERT Exercise 13.8: 9th math
Important Formula
Volume of a Sphere `=4/3\pi\ r^3`
Volume of a Hemisphere `=2/3\ pi\ r^3`
NCERT Exercise 13.8 Questions and Answers
Assume ℼ = 22/7, unless stated otherwise.
Surface Areas And Volumes Class nine Math NCERT Exercise 13.8 Question (1) Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m
solution
(i) 7 cm
Given, Radius of a sphere = 7 cm
Therefore, volume of the given sphere = ?
We know that Volume of a Shpere = 4/3 ℼ r3
Therefore, Volume of the given sphere
= `4/3xx22/7` × 7 cm × 7 cm × 7 cm
= `4/3` xx 22 × 49 cm3
`=4312/3` cm3
= 1437.3 cm3
Thus, volume of the given sphere = 1437.3 cm3 Answer
(ii) Radius = 0.63 m
Given, Radius of the sphere = 0.63 m
Thus, volume of the given sphere = ?
We know that Volume of a Shpere = 4/3 ℼ r3
Therefore, Volume of the given sphere
= `4/3xx22/7` × 0.63 m × 0.63 m × 0.63 m
= 4 × 22 × 0.09 m × 0.21 m × 0.63 m
= 88 × 0.011907 m3
= 1.047816 m3
≃ 1.05 cm3
Thus, volume of the given sphere ≃ 1.05 cm3 Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.8 Question (2) Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m
Solution
(i) Volume of water displaced by a sphere having diameter equal to 28 cm
Given, Diameter of a sphere = 28 cm
Thus, radius of the sphere = 28/2 = 14 cm
Thus, volume of water displaced by the given sphere = ?
Here, volume of water displaced by the given sphere = volume of the sphere
We know that Volume of a Shpere = 4/3 ℼ r3
Thus, volume of the given sphere
= `4/3xx22/7` × 14 cm × 14 cm × 14 cm
= `4/3xx22` × 2 cm × 14 cm × 14 cm
= 11498.66 cm3
Thus, volume of the water displace by the given sphere = = 11498.66 cm3 Answer
(ii) Diameter = 0.21 m
Thus, radius of the sphere = 0.21/2 = 0.105 m
Thus, volume of water displaced by the given sphere = ?
Here, volume of water displaced by the given sphere = volume of the sphere
We know that Volume of a Shpere = 4/3 ℼ r3
Thus, volume of the given sphere
= `4/3xx22/7` × 0.105 m × 0.105 m × 0.105 m
= 4 × 22 × 0.035 m × 0.015 m × 0.105 m
= 88 × 0.000055125 m3
= 0.004851 m3
Thus, volume of water displaced by the given sphere = = 0.004851 m3 Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.8 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 per cm3?
Solution
Given, Diameter of a metallic ball = 4.2 cm
Therefore, Radius of the given ball = 4.2/2 = 2.1 cm
And, Density of the metal = 8.9 g/cm3
Thus, mass of the ball = ?
We know that Volume of a Shpere = 4/3 ℼ r3
Thus, volume of the given sphere
= `4/3xx22/7` × 2.1 cm × 2.1 cm × 2.1 cm
= 4 × 22 × 0.7 cm × 0.3 cm × 2.1 cm
= 88 × 0.441 cm3
= 38.808 cm3
Now, we know that Mass = Volume × Density
Thus, mass of the given metallic ball
= 38.808 cm3 × 8.9 g/cm3
= 345.3912 g ≃ 345.39 g
Thus, mass of the given metallic ball & 345.39 g Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.8 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?
Solution
Given, Diameter of the moon = 1/4 th of the diameter of the earth.
Therefore, what fraction of the volume of the earth is the volume of the moon?
Let diameter of the moon = 2r
Therefore, radius of the moon = r
Therefore, according to question, diameter of the moon is ¼ th of the diameter of the earth
This means that diameter of the earth = 4 times of the moon
Thus, diameter of the earth = 4 × 2r = 8r
Thus, radius of the earth = 8r/2 = 4r
We know that Volume of a Shpere = 4/3 ℼ r3
Thus, volume of the moon
= `4/3` ℼ r3
And, volume of the earth
= `4/3` ℼ (4r)3
= `4/3` ℼ 64 r3
Now, fraction of the volume of moon to the volume of the earth
= Volume of the moon/Volume of the earth
`= (4/3xxpi\ r^3)/(4/3xxpi\ 64\ r^3)`
= 1/64
Thus, volume of the moon is 1/64 of the volume of the earth Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.8 Question (5) How many liters of milk can a hemispherical bowl of diameter 10.5 cm hold?
Solution
Given, Diameter of a hemispherical bowl = 10.5 cm
Thus, Radius of the given hemispherical bowl = 10.5/2 = 5.25 cm
Thus, Litres of milk can hold by given hemispherical ball = ?
We know that, Volume of a hemisphere `=2/3\ pi\ r^3`
Thus, volume of the given hemispherical ball
= `2/3xx22/7` × 5.25 cm × 5.25 cm × 5.25 cm
= 2 × 22 × 1.75 cm × 0.75 × 5.25 cm
= 44 × 6.890625 cm3
= 303.1875 cm3
Thus, volume of the given bowl = 303.1875 cm3
Thus, the given bowl can hold = 303.1875 cm3 of milk
Now, since 1000 cm3 = 1`l`
Therefore, 1 cm3 = 1/1000 `l`
Therefore, 303.1875 cm3 of milk
`= 303.1875/1000\ l`
= 0.3031875 `l`
≃ 0.303 `l`
Thus, given hemispherical bowl can hold approximately 0.303 Litre of milk. Answer
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