A hollow aluminium sphere of mass 150kg floats on water. It is observed that an additional mass 30kg is required to just submerge it at a temperature of 15⁰C. How much mass is required to submerge the sphere if the temperature is at 35⁰C.
$(γ_{water} = 150 \times 10^{- 6} /^{\circ} C, α_{Al} = 23 \times 10^{- 6} /^{\circ} C)$

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291.6 gm

201.6 gm

191.6 gm

141.2 gm

answer is A.

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Detailed Solution

$\begin{array}{l} 180 = V_{0} d_{0} (at 15^{\circ} C) \\ 150 + Δm = V^{1} d^{1} (at 35^{\circ} C) \\ \frac{180}{150 + Δm} = \frac{V_{0}}{V_{0} (1 + γ_{Al} t)} \times (1 + γ_{w} Δ θ) \\ \frac{180}{150 + Δm} = \frac{[1 + 150 \times 10^{- 6} \times 20]}{1 + 23 \times 10^{- 6} \times 20]} \\ Δm = 291.6 gm \end{array}$