A hollow aluminium sphere of mass 150 kg floats on water. It is observed that, an additional mass of 30 kg is required to just submerge it at a temperature of ${15}^{0}C$. How much less mass (in g) is required, in order to submerge the sphere, when the temperature is ${35}^{0}C$? (Coefficient of cubical expansion of water is  $1.5\times {10}^{-4}{/}^{0}C$ and linear expansivity of aluminium is  $23\times {10}^{-6}{/}^{0}C)$

Let  ${\rho }_0$ and  $V_0$ be the density and volume of water displaced by the sphere, at  ${15}^{0}C$, when it just happens to get submerged into water.
     From principle of floatation,  
$V_0{\rho }_0=150+30=180$  …………………(i)
At a temperature of ${35}^{0}C$, the volume of the sphere increases to $V_0(1+{\gamma }_a\times 20)$ and the density of water decreases to ${\rho }_0(1-{\gamma }_w\times 20)$. If m kg be the additional mass required for submergence, then

$V_0(1+{\gamma }_a\times 20){\rho }_0(1-{\gamma }_w\times 20)=(150+m)$

or  $V_0{\rho }_0\left[1-20({\gamma }_w-{\gamma }_a)\right]=(150+m) \dots . \left(ii\right)$
Dividing Eqn. (ii) by (i),  $1-20({\gamma }_w-{\gamma }_a)=(150+m)$
Substituting for ${\gamma }_w$ and ${\gamma }_a$ and, adopting the required difference in mass 
$(30-m) = \Delta m$

$$\begin{gathered} 1-20(150-3\times 23)\times {10}^{-6}=1-\frac{\Delta m}{180} \\ or \Delta m=20\times 81\times 180\times {10}^{-3}g = 291.6g. \end{gathered}$$