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An object suspended by a wire stretches it by 10 mm. When object is immersed in a liquid the elongation in wire reduces by 5/3mm. The ratio of relative density of the object and liquid is

a
1/3
b
5/6
c
1/6
d
6/5

detailed solution

Correct option is C

Given elongation e1=10 mm e2=(10−53) mm Elongation e2=253mm We know Young's modulus Y=FlAe...(1) ⇒e1∝weight...(2) ⇒e2∝ Apparent weight...(3) equations(2)(3)⇒1010−53=MgMg−Mgρσ ⇒55(22−13)=11−ρ6 ⇒65=11−ρ6 ⇒1−ρ6=56 ⇒ρσ=1−56=16

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