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Question

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

Vedclass · Accepted Answer

$\frac{f_1 - f_2}{f_1 t_1 - f_2 t_2}$

Vedclass · Answer

(B) Let $V$ be the volume of the sphere and $\rho_L$ be the density of the liquid. Since the sphere floats,the weight of the sphere equals the weight of the displaced liquid.$Mg = V_{submerged} \rho_L g$$V_{sphere}(t) \rho_S(t) g = f V_{sphere}(t) \rho_L g$Since the liquid density $\rho_L$ is constant,$\rho_S(t) = f \rho_L$. As the sphere expands,its density changes as $\rho_S(t) = \frac{\rho_0}{1 + \gamma t}$,where $\gamma$ is the coefficient of volume expansion.Thus,$f_1 = \frac{\rho_S(t_1)}{\rho_L} = \frac{\rho_0}{\rho_L(1 + \gamma t_1)}$ and $f_2 = \frac{\rho_S(t_2)}{\rho_L} = \frac{\rho_0}{\rho_L(1 + \gamma t_2)}$.Taking the ratio: $\frac{f_1}{f_2} = \frac{1 + \gamma t_2}{1 + \gamma t_1}$.$f_1(1 + \gamma t_1) = f_2(1 + \gamma t_2)$$f_1 + f_1 \gamma t_1 = f_2 + f_2 \gamma t_2$$\gamma (f_1 t_1 - f_2 t_2) = f_2 - f_1$$\gamma = \frac{f_2 - f_1}{f_1 t_1 - f_2 t_2} = \frac{f_1 - f_2}{f_2 t_2 - f_1 t_1}$.Wait,re-evaluating the ratio: $f_1(1 + \gamma t_2) = f_2(1 + \gamma t_1)$ is incorrect. The correct relation is $f_1(1 + \gamma t_1) = C$ (constant). Actually,$V(t) = V_0(1 + \gamma t)$. The submerged volume $V_{sub} = f V(t)$. Since $V_{sub} \rho_L = M$ (constant),$f V_0(1 + \gamma t) \rho_L = M$.So $f_1(1 + \gamma t_1) = f_2(1 + \gamma t_2)$.$f_1 + f_1 \gamma t_1 = f_2 + f_2 \gamma t_2$$\gamma (f_1 t_1 - f_2 t_2) = f_2 - f_1$$\gamma = \frac{f_2 - f_1}{f_1 t_1 - f_2 t_2} = \frac{f_1 - f_2}{f_2 t_2 - f_1 t_1}$.

$A$ sphere floats on a liquid whose volume does not change with temperature. At temperatures $t_1$ and $t_2$,the fractions $f_1$ and $f_2$ of the sphere are submerged in the liquid,respectively. What is the coefficient of volume expansion of the sphere?

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