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Question

A ball rises to surface at a constant velocity in liquid whose density is 4 times greater than that of the material of the ball. The ratio of the force of friction acting on the rising ball and its weight is :

A
3:1
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B
4:1
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C
1:3
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D
1:4
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Solution

The correct option is A 3:1
Frictional Force=4πr33(d1d2)g
Weight=4πr33(d1g)
where d1 is density of ball
d2 is density of liquid
Given:d2=4d1
Frictional Force=4πr33(d14d1)g=4πr33(3d1)g
Considering only magnitude of frictional force because its already known that both forces act in opposite directions.
Frictional Force=34πr33(d1g)=3× Weight
Therefore the correct opton is (A).



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