In this chapter, we will discuss questions related to volume of sphere and its solution.
The formula for volume of sphere is given as;
Volume = \mathtt{\ \frac{4}{3} .\pi .r^{3}}
Where, r is the radius of sphere.
Questions on volume of sphere
Question 01
The radius of sphere is given as 21 cm. Find the volume.
Solution
radius (r) = 21 cm
Volume = \mathtt{\ \frac{4}{3} .\pi .r^{3}}
Putting the values;
\mathtt{Volume\ =\ \frac{4}{3} \times \frac{22}{7} \times ( 21)^{3}}\\\ \\ \mathtt{Volume\ =\ 38808\ \ cm^{3}}
Hence, 38808 cu cm is the volume of sphere.
Question 02
Find the volume of sphere with diameter 14 cm.
Solution
Diameter (d) = 14 cm
Radius (r) = 14 / 2 = 7 cm
Volume of sphere is given by following formula;
Volume = \mathtt{\ \frac{4}{3} .\pi .r^{3}}
Putting the values in formula;
\mathtt{Volume\ =\ \frac{4}{3} \times \frac{22}{7} \times ( 7)^{3}}\\\ \\ \mathtt{Volume\ =\ 1437.33\ \ cm^{3}}
Hence, 1437.33 cu cm is the volume of cube.
Question 03
A hemisphere tank of 3 meter is given. Find the volume of water (in liters) that can be stored in the tank.
Solution
Radius (r) = 3 meters
We know that when sphere is cut into half, we get two hemisphere. Hence, the volume of hemisphere is half of the sphere.
Volume of hemisphere = \mathtt{\frac{1}{2} .\ \frac{4}{3} .\pi .r^{3}
Putting the values, we get;
\mathtt{Volume\ =\frac{1}{2} \times \frac{4}{3} \times \frac{22}{7} \times ( 3)^{3}}\\\ \\ \mathtt{Volume\ =\ 56.57\ \ m^{3}}
Hence, 56.57 cu meter of water can be stored in the tank.
Now converting the volume into liters.
1 cu m = 1000 liters
56.57 cu. m = 56570 liters
So, total of 56570 liters of water is stored in the hemi sphere.
Question 04
A cube of lead of size 21 cm is given. How many spherical bullets of radius 3 cm can be made from the given lead cube.
Solution
First calculate the volume of lead into the given cube.
Volume of cube = side x side x side
Volume of cube = 21 x 21 x 21 = 9261 cu. cm
So from 9261 cu cm of the cube, the spherical bullets will be made.
Let ” n ” number of spherical bullets (radius 3 cm) are made from the above cube.
Volume of all bullets = volume of cube
\mathtt{n\times \frac{4}{3} .\pi .r^{3} \ =\ 9261}\\\ \\ \mathtt{n\times \frac{4}{3} \times \frac{22}{7} \times ( 3)^{3} =\ 9261}\\\ \\ \mathtt{n\ =\ \frac{9261\times 21}{4\times 22\times 27}}\\\ \\ \mathtt{n\ =\ \frac{194481}{2376} =81.8\ bullets}
Hence, total of 81 bullets are made from cube.
Question 05
A shopkeeper has spherical chocolate of radius 5 cm. He breaks the chocolate and with the same material he started making chocolate sphere of radius 2.5 cm. Find how many small spheres can be made ?
Solution
Let ” n ” number of small spheres can be manufactured.
Volume of big sphere = n x volume of small sphere
\mathtt{\frac{4}{3} .\pi .( 5)^{3} \ =\ n\times \frac{4}{3} .\pi .( 2.5)^{3}}\\\ \\ \mathtt{n\ =\ \frac{5^{3}}{2.5^{3}} \ }\\\ \\ \mathtt{n=\ 8}
Hence, total of 8 small spheres can be made from the large one.
Question 06
A cylindrical vessel half filled with water is given. A spherical ball of radius 5 cm is immersed in the vessel which results in rise of water level by 5/3 cm. Find the radius of cylindrical vessel.
Solution
Radius of sphere (r) = 5 cm
Radius of cylinder = R
Here the volume of rise in water level is equal to the volume of sphere.
Volume of rise in water level = volume of sphere
\mathtt{\pi .R^{2} .\left(\frac{5}{3}\right) =\frac{4}{3} .\pi .( 5)^{3}}\\\ \\ \mathtt{R^{2} =4 \times 5^{2}}\\\ \\ \mathtt{R\ =\ 2\times 5=\ 10\ cm\ }
Hence, the radius of cylindrical vessel is 10 cm.
Question 07
A sphere of radius 3 cm is melted and drawn into cylindrical wire of radius 0.1 cm. Find the length of wire formed?
Solution
Sphere radius (r) = 3 cm
Wire radius (R) = 0.1 cm
Here the material used in the sphere is converted into cylindrical wire.
Volume of wire = volume of sphere
\mathtt{\pi .R^{2} .h=\frac{4}{3} .\pi .( r)^{3}}\\\ \\ \mathtt{\pi .( 0.1)^{2} .h=\frac{4}{3} .\pi .( 3)^{3}}\\\ \\ \mathtt{h\ =\ \frac{4}{3} \times \frac{3^{3}}{0.1^{2}}}\\\ \\ \mathtt{h\ =\ 3600\ cm\ }
Hence, the length of wire is 3600 cm.