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Asked in JEE Main Online 2013 · Series and parallel wires
Given: the steel wire hangs from the ceiling and carries the block M together with the block 2M below it, so Tₛₜₑₑₗ=3Mg, while the brass wire carries only the lower block, T_brass=2Mg.
Δℓ=(TL)/(π r²Y).
(Δℓₛₜₑₑₗ)/(Δℓ_brass)=(Tₛₜₑₑₗ)/(T_brass)·(Lₛₜₑₑₗ)/(L_brass)·((r_brass)/(rₛₜₑₑₗ))²·(Y_brass)/(Yₛₜₑₑₗ).
With a=(Lₛₜₑₑₗ)/(L_brass), b=(rₛₜₑₑₗ)/(r_brass), c=(Yₛₜₑₑₗ)/(Y_brass), this is 3/2· a·1/(b²)·1/c.
(Δℓₛₜₑₑₗ)/(Δℓ_brass)=(3a)/(2b²c).
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