Q1. Find the volume of a sphere whose radius is : (i) 2 cm (ii) 3.5 cm (iii) 10.5 cm.
Sol.
(i) Radius(r) = 2cm
Therefore volume = πr^{3}
= ××(2)^{3}
= 33.52cm^{3}
(ii) Radius(r) = 3.5cm
Therefore volume = πr^{3}
= ××(3.5)^{3} = 179.666cm^{3}
(iii) Radius(r) = 10.5cm
Therefore volume = 4^{3}πr^{3}
= 4851cm^{3}
Q2. Find the volume of a sphere whose diameter is : (i) 14 cm (ii) 3.5 dm (iii) 2.1 m
Sol.
(i) Diameter = 14cm, Radius(r) = = 7cm
Therefore volume = πr^{3}
= 1437.33cm^{3}
(ii) Diameter = 3.5dm, Radius(r) = = 1.75dm
Therefore volume = πr^{3}
= 22.46dm^{3}
(iii) Diameter = 2.1m, Radius(r) = = 1.05m
Therefore volume = πr^{3}
= 4.851m^{3}
Q3. A hemispherical tank has the inner radius of 2.8 m. Find its capacity in liters.
Sol.
Radius of the tank = 2.8m
Therefore Capacity = πr^{3}
= 45.994m^{3}
1m^{3} = 1000l
Therefore capacity in litres = 45994 litres
Q4. A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. Find the volume of steel used in making the bowl.
Sol.
Inner radius = 5cm
Outer radius = 5 + 0.25 = 5.25
Volume of steel used = Outer volume  Inner volume
= 41.282cm^{3}
Q5. How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?
Sol.
Cube edge = 22cm
Therefore volume of the cube = (22)^{3} = 10648cm^{3}
And,
Volume of each bullet = πr^{3 }=
Number of bullets = = 2541
Q6. A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made?
Sol.
Volume of laddoo having radius = 5cm
i.e Volume(V_{1}) = πr^{3}
Also Volume of laddoo having radius 2.5cm
i.e Volume(V_{2}) = πr^{3}
Therefore number of laddoos = = 8
Q7. A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 32cm and 2 cm, find the diameter of the third ball.
Sol.
Volume of lead ball = πr^{3}
Diameter of first balld_{1 } = cm
Radius of first ball r_{1} cm
Diameter of second ball d_{2} = 2cm
Radius of second ball r_{2} = cm = 1cm
Diameter of third balld_{3} = d
Radius of third ball r_{3} = cm
Volume of lead ball =
Volume of lead ball =
d =
d = 2.5cm
Q8. A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises cm. Find the radius of the cylinder.
Sol.
Radius of cylinder = r
Radius of sphere = 5cm
Volume of sphere = πr^{3}
= ×π×(5)^{3}
Height of water rised = cm
Volume of water rised in cylinder = πr^{2}h
Therefore, Volume of water rises in cylinder = Volume of sphere
Let r be the radius of the cylinder
πr^{2}h = πr^{3}
r^{2} = 20×5
r =
r = 10cm
Q9. If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?
Sol.
Let v_{1} and v_{2} be the volumes of the first and second sphere respectively
Radius of the first sphere = r
Radius of the second sphere = 2r
Therefore,
Q10. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
Sol.
Given that
Volume of the cone = Volume of the hemisphere
r^{2}h = 2r^{3}
h = 2r
= 2
Therefore
Ratio of their heights = 2 : 1
Q11. A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.
Sol.
Given that
Volume of water in the hemispherical bowl = Volume of water in the cylinder
Let h be the height to which water rises in the cylinder
Inner radii of the bowl = r_{1 } = 3.5cm
Inner radii of the bowl = r_{2} = 7cm
Q12. A cylinder whose height is two thirds of its diameter has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.
Sol.
Given that
Height of the cylinder = diameter
We know that
Diameter = 2(radius)
Volume of the cylinder = Volume of the sphere
πr^{2}h = πr^{3}
(r)^{3} = (4)^{3}
r = 4cm
Q13. A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder.
Sol.
It is given that
Volume of water in hemispherical bowl = Volume of cylinder
h = 9cm
Q14. A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?
Sol.
Let r be the radius of the iron ball
Radius of the cylinder = 16cm
Then,
Volume of iron ball = Volume of water raised in the hub
= πr^{2}h
= (16)^{2}×9
r^{3 }=
r^{3} = 1728
r = 12cm
Therefore radius of the ball = 12cm.
Q15. A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use = ).
Sol.
Given that :
Radius of the cylinder = 12cm = r_{1}
Raised in raised = 6.75 cm = r_{2}
Volume of water raised = Volume of the sphere
= 12×12×6.75 =
= 729
= r_{2} = 9cm
Radius of the sphere is 9cm
Q16. The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross  section. If the length of the wire is 108 m, find its diameter.
Sol.
Given that diameter of a copper sphere = 18cm
Radius of the sphere = 9cm
Length of the wire = 108m = 10800cm
Volume of cylinder = Volume of sphere
×9×9×9
= 0.009
r_{1} = 0.3cm
Therefore Diameter = 2×0.3 = 0.6cm
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