Question

A platinum sphere floats in mercury. Find the percentage change in the fraction of volume of sphere immersed in mercury when the temperature is raised by ${80}^{0}C$ ( volume expansivity of mercury is $182\times {10}^{-6}{/}^{0}C$ and linear expansivity of platinum is $9\times {10}^{-6}{/}^{0}C$ respectively)

• A. 1.24%
• B. 1.10 %
• C. 2.48 %
• D. 2.76 %

Answer 1

The correct option is A 1.24%

Let $f_0$ be the initial fraction of platinum sphere inside mercury.
Thus, $V{d}_{0}g=f_{0}V{\rho }_{0}g$

Let $f_1$ be the fraction of platinum sphere inside mercury after raising temperature.

Thus $V{d}_{1}g=f_{1}V{\rho }_{1}g$

$⟹V{f}_0\frac{d_0}{1+{\gamma }_{Pt}\Delta T}g=f_{1}V\frac{{\rho }_0}{1+{\gamma }_{Hg}\Delta T}g$

Thus $\frac{f_1}{f_0}=\frac{1+{\gamma }_{Hg}\Delta T}{1+{\gamma }_{Pt}\Delta T}$

Percentage change in fraction inside mercury=$\frac{f_1-f_0}{f_0}\times 100$%

$=(\frac{f_1}{f_0}-1)\times 100$%

$\approx 1.24$%