Question

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

• A. $8.048\times {10}^{-4}pe{r}^{0}C$
• B. $8.048\times {10}^{-5}pe{r}^{0}C$
• C. $8.486\times {10}^{-5}pe{r}^{0}C$
• D. $8.486\times {10}^{-3}pe{r}^{0}C$

Answer 1

The correct option is A $8.048\times {10}^{-4}pe{r}^{0}C$

Let coeff of expansion of liquid be $r$

The sphere has mass $=266.5$ and a volume $=\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3$ where $d=7cm$

so density of sphere $=p=\frac{m}{v}=\frac{266.5}{\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3}=1.484$glcc

Due to thermal expansion,the density of liquid decreases which is :-

$P=P_0(\perp -\gamma \Delta T)$

where $\Delta T=35K$

$P_0:$Density of liquid at $0^{o}C$

The sphere begins to ink in liquid when its density equals to density of liquid

$=P_0(1-\gamma \Delta )=1.484$

$=1.527(1-\gamma \times 35)=1.484$

$\Rightarrow 35\gamma =(\perp -\frac{1.484}{1.527})$

$\Rightarrow \gamma =8.05\times {10}^{-4}{K}^{-1}$