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When a body of mass ‘m’, density dB is suspended from a wire, its elongation is ‘e’ when the body is in air. If the body is completely immersed in a non – viscous liquid of density dℓ then its elongation is

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a

e(1−dℓdB)

b

(1−dℓdB)

c

e(dℓdB)

d

(dℓdB)

answer is A.

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Detailed Solution

In air, F=Wa=mg In liquid, F1=Wℓ=mg(1−dℓdB)As e=FℓAY eαF (as ℓ,A,Y are same in both cases)e1e=F1F=mg(1−dℓdB)mg⇒e1=e(1−dℓdB)

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