A non-conducting body floats in a liquid at 2 | vedclass

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Question

(A) Let $V$ be the volume of the body and $\rho_b$ be its density. Let $\rho_1$ and $\rho_2$ be the densities of the liquid at $20^{\circ} C$ and $100

Vedclass · Accepted Answer

$15.6 	imes 10^{-4} {}^{\circ} C^{-1}$

Vedclass · Answer

(A) Let $V$ be the volume of the body and $ho_b$ be its density. Let $ho_1$ and $ho_2$ be the densities of the liquid at $20^{\circ} C$ and $100^{\circ} C$ respectively.By the law of flotation,the weight of the body equals the weight of the displaced liquid.At $20^{\circ} C$: $V ho_b g = (\frac{2}{3} V) ho_1 g \implies ho_b = \frac{2}{3} ho_1$.At $100^{\circ} C$: $V ho_b g = (\frac{3}{4} V) ho_2 g \implies ho_b = \frac{3}{4} ho_2$.Equating the two expressions for $ho_b$: $\frac{2}{3} ho_1 = \frac{3}{4} ho_2 \implies \frac{ho_1}{ho_2} = \frac{9}{8}$.Since density $ho = \frac{ho_0}{1 + \gamma \Delta t}$,we have $\frac{ho_1}{ho_2} = \frac{1 + \gamma \Delta t_2}{1 + \gamma \Delta t_1} = 1 + \gamma \Delta T$ (where $\Delta T = 80^{\circ} C$).$1 + \gamma (80) = \frac{9}{8} \implies 80 \gamma = \frac{1}{8} \implies \gamma = \frac{1}{640} = 0.0015625 {}^{\circ} C^{-1}$.$\gamma = 15.6 	imes 10^{-4} {}^{\circ} C^{-1}$.

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