Question

A sphere of mass 266.5g and diameter 7 cm floats on the surface of a liquid. When the liquid is heated to 35${}^0$ C, the sphere sinks in the liquid. If the density of liquid at 0${}^0$ C is 1.527 gcm${}^{-3}$. The coefficient of real expansion of liquid is

• A. $8.48\times {10}^{-4}{(}^{0}C{)}^{-1}$
• B. $7.88\times {10}^{-4}{(}^{0}C{)}^{-1}$
• C. $6.48\times {10}^{-4}{(}^{0}C{)}^{-1}$
• D. $5.48\times {10}^{-4}{(}^{0}C{)}^{-1}$

Answer 1

Answer: (A)

$8.48\times {10}^{-4}{(}^{0}C{)}^{-1}$

For the sphere to sink in the liquid at ${35}^{\circ }C$, the density of sphere and that of liquid at the temperature should be same.

Hence, $\frac{m}{\frac{4}{3}\pi r^3}=\frac{{\rho }_0}{1+\gamma \Delta \theta }$

$⟹\frac{266.5}{\frac{4}{3}\pi (3.5{)}^3}=\frac{1.527}{1+35\gamma }$

$⟹\gamma \approx 8.48\times {10}^{-4}{/}^{\circ }C$