Question

A sphere of diameter $7cm$ and mass $266.5gm$ floats in a bath of liquid. As the temperature is raised, the sphere just begins to sink at a temperature ${35}^{\circ }C$. If the density of the liquid at $0^{\circ }C$ is $1.527gm/cc$, then neglecting the expansion of the sphere, the coefficient of cubical expansion of the liquid is $f:$

• A. $8.486\times {10}^{-4}per{}^{\circ }C$
• B. $8.486\times {10}^{-5}per{}^{\circ }C$
• C. $8.486\times {10}^{-6}per{}^{\circ }C$
• D. $8.486\times {10}^{-3}per{}^{\circ }C$

Answer 1

The correct option is A $8.486\times {10}^{-4}per{}^{\circ }C$

Let ${\rho }_t=$ density of liquid at $t^{o}C$ and ${\rho }_0=$ density of liquid at $0^{o}C$
As the sphere just begins to sink at $t={35}^{o}C$, weight of sphere $=$ upward thrust
or $\frac{4}{3}\pi r^3{\rho }_{t}g=266.5g$
or ${\rho }_t=\frac{266.5\times 3}{4\times 3.14\times (3.5{)}^3}=1.485g/cc$
Now ${\rho }_0={\rho }_t[1+\gamma t]$
or $1.527=1.485[1+\gamma \times 35]$
or $\gamma =8.486\times {10}^{-4}{/}^{o}C$