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A small sphere of radius r is falling through a viscous liquid when the sphere has attained terminal velocity, its kinetic energy is 15 μJ. If radius of the sphere is doubled, then its kinetic energy at terminal velocity will be 

a
480 μJ
b
1920 μJ
c
30 μJ
d
960 μJ

detailed solution

Correct option is B

At terminal velocity, V ∞ r2Also, mass of sphere m ∞ r3Since kinetic energy K=12mV2we can write, K ∞ (r)7∴ K1K=(2rr)7⇒K1=15×27μJ=1920 μJ

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