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

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(A) Let $V$ be the volume of the body and $\rho_L$ be the density of the liquid.Since the body floats, the weight of the body equals the weight of the

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $V$ be the volume of the body and $ho_L$ be the density of the liquid.Since the body floats, the weight of the body equals the weight of the displaced liquid: $mg = V_{imm} ho_L g$.Thus, $V_{imm} ho_L = 	ext{constant}$.At $t_1 = 20^{\circ}C$, $V_{imm,1} = \frac{2}{3}V$, so $\frac{2}{3}V ho_1 = m$.At $t_2 = 100^{\circ}C$, $V_{imm,2} = \frac{3}{4}V$, so $\frac{3}{4}V ho_2 = m$.Equating the two: $\frac{2}{3} ho_1 = \frac{3}{4} ho_2$, which gives $ho_2 = \frac{8}{9} ho_1$.Using the relation $ho_2 = \frac{ho_1}{1 + \gamma_R \Delta t}$, we have $1 + \gamma_R \Delta t = \frac{ho_1}{ho_2} = \frac{9}{8}$.$\gamma_R \Delta t = \frac{9}{8} - 1 = \frac{1}{8}$.Given $\Delta t = 100 - 20 = 80^{\circ}C$, so $\gamma_R = \frac{1}{8 	imes 80} = \frac{1}{640} = 0.0015625 = 15.6 	imes 10^{-4} \, ^{\circ}C^{-1}$.

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