Plz Plz solve it
21. The diameter of a cylindrical beaker is 12 cm. The beaker is filled with water to a height such that
when some lead balls of diameter 3 cm are dropped in the beaker, the balls fully submerge. How
many balls should be dropped in order to raise the height of water by 2 cm?
Answers
Answered by
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Step-by-step explanation:
ANSWER
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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