Question

A sphere of diameter 7.0 cm and mass 266.5 g floats in a bath of a liquid. As the temperature is raised, the sphere begins to sink at a temperature of ${35}^{\circ }C$. If the density of the liquid is $1.527g/c{m}^3$ at $0^{\circ }C$, find the coefficient of cubical expansion of the liquid. Neglect the expansion of the sphere.

• A. 2.32 × 10^-4/°C
• B. 1.22 × 10^-4/°C
• C. 8.28 × 10^-4/°C
• D. 6.36 × 10^-4/°C

Answer 1

The correct option is C

8.28 × 10^-4/°C

It is given that the expansion of the sphere is negligible as compared to the expansion of the liquid. At $0^{\circ }C$, the density of the liquid is ${\rho }_0$ = $1.527g/c{m}^3$. At ${35}^{\circ }C$, the density of the liquid equals the density of the sphere. Thus,

${\rho }_{35}$ = $\frac{266.5g}{\frac{4}{3}\pi (3.5cm{)}^3}$

= $1.484g/c{m}^3$.

We have $\frac{{\rho }_{\theta }}{{\rho }_0}$ = $\frac{v_0}{v_{\theta }}$ = $\frac{1}{(1+{\gamma }_{\theta })}$

or, ${\rho }_{\theta }$ = $\frac{{\rho }_0}{1+{\gamma }_{\theta }}$

Thus, $\gamma$ = $\frac{{\rho }_0-{\rho }_{\theta }}{{\rho }_{35}\left({35}^{\circ }C\right)}$

= $\frac{(1.527-1.484)g/c{m}^3}{\left(1.484g/c{m}^3\right)\left({35}^{\circ }C\right)}$

= $8.28\times {10}^{-4}{/}^{\circ }C$.