solid spheres of d 6cm are dropped into a cylindrical beaker containing some water and are fully submerged. if the diameter of the beaker is 18 cm qnd the water rises by 40 cm find the number of solid spheres dropped in the water
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Given,
Diameter of lead spheres d=6 cm
The diameter of cylindrical beaker D=18 cm
Hight of rises water in the beaker h=40 cm
Hence,
Radius of lead spheres r=
2
d
=3 cm
The radius of cylindrical beaker R=
2
D
=9 cm
The volume of rises water in the beaker V=πR
2
h
The volume of one sphere v=
3
4
πr
3
Hence, Volume of n sphere V=nv=
3
4
nπr
3
Let the volume of water in the beaker rises by 40 cm by dropping n lead spheres of diameter 6 cm
→Volume of n spheres = increased volume of water in the beaker
→
3
4
nπr
3
=πR
2
h
→n=
4πr
3
3πR
2
h
→n=
4r
3
3R
2
h
→n=
4×3
3
3×9
2
×40
→n=
4×27
3×81×40
→n=
108
9720
→n=90
Hence, for rising the water by 40 cm , 90 lead spheres dropped in the water.
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