Question

# The loss of weight of a solid when immersed in a liquid at 0∘C is W0 then the loss of weight W at t∘C is - where α and β are the cubical expansion coefficients of solid and liquid respectively.

Solution

## The correct option is A Let Ws be the weight of the solid and ds be the density at 0∘C. Let dL be the density of the liquid at that temperature. The loss of weight of solid when immersed in the liquid at 0∘C. =(Wsds)d1=W0  (given)……(1) Similarly, the loss of weight of the solid when immersed in the liquid at t∘C. =(Wsd′s)×d′L=W (given)……(2) where $$d^{'}_{s}$$ = density of solid at t∘C. and $$d^{'}_{L}$$ = density of liquid at t∘C. Now ds and $$d^{'}_{s}$$ are related as d′s=ds1+αtSimilarly, d′L=dL(1+βt) where α and β are the coefficients of cubical expansion of the solid and the liquid respectively. Now equation (2) can be written as Ws×1+αtds×dL1+βt=Wor (Wsds×dL)(1+αt1+βt)=Wor, W01+αt1+βt=W   [from (1)]or, W0(1+αt)(1+βt)−1=Wor, W0(1+αt)(1−βt)=W    (approx)or, W0[1+(α−β)t−αβt2]=Wor, W0[1+(α−β)t]=W (approx)

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