Question

The loss of weight of a solid when immersed in a liquid at $0^{\circ }C$ is $W_0$ then the loss of weight W at $t^{\circ }C$ is - where $\alpha$ and $\beta$ are the cubical expansion coefficients of solid and liquid respectively.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let $W_s$ be the weight of the solid and $d_s$ be the density at $0^{\circ }C$. Let $d_L$ be the density of the liquid at that temperature.
The loss of weight of solid when immersed in the liquid at $0^{\circ }C$.
$=\left(\frac{W_s}{d_s}\right)d_1=W_0\left(given\right)\dots \dots \left(1\right)$
Similarly, the loss of weight of the solid when immersed in the liquid at $t^{\circ }C$.
$=\left(\frac{W_s}{d_s^{{}^{\prime }}}\right)\times d_L^{{}^{\prime }}=W\left(given\right)\dots \dots \left(2\right)$
where \(d^{'}_{s}\) = density of solid at $t^{\circ }C$.
and \(d^{'}_{L}\) = density of liquid at $t^{\circ }C$.
Now $d_s$ and \(d^{'}_{s}\) are related as
$d_s^{{}^{\prime }}=\frac{d_s}{1+\alpha t}Similarly,d_L^{{}^{\prime }}=\frac{d_L}{(1+\beta t)}$
where $\alpha$ and $\beta$ are the coefficients of cubical expansion of the solid and the liquid respectively.
Now equation (2) can be written as
$W_s\times \frac{1+\alpha t}{d_s}\times \frac{d_L}{1+\beta t}=Wor(\frac{W_s}{d_s}\times d_L)\left(\frac{1+\alpha t}{1+\beta t}\right)=Wor,W_0\frac{1+\alpha t}{1+\beta t}=W\left[from\right(1\left)\right]or,W_0(1+\alpha t)(1+\beta t{)}^{-1}=Wor,W_0(1+\alpha t\left)\right(1-\beta t)=W(approx)or,W_0[1+(\alpha -\beta )t-\alpha \beta t^2]=Wor,W_0[1+(\alpha -\beta )t]=W(approx)$