First slide

Thermal expansion

Question

A metal ball immersed in alcohol weights W₁ at 0°C and W₂ at 59°C. The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown that

Difficult

A

$W_{1} > W_{2}$
B

$W_{1} = W_{2}$
C

$W_{1} < W_{2}$
D

$W_{2} = (W_{1} / 2)$

Solution

The upthrust is given by $\frac{4}{3} {πR}_{t}^{3} ρ_{t} g$
Here $R_{t}^{3} = R_{0}^{3} (1 + γ_{m} t)$
and $ρ_{t} = ρ_{0} / (1 + γ_{a} t)$
So, the upthrust at t°C is given by
$\begin{aligned} = \frac{4}{3} {πR}_{0}^{3} (1 + γ_{m} t) \times \{ρ_{0} / (1 + γ_{a} t)\} g \\ = \frac{4}{3} {πR}_{0}^{3} ρ_{0} g \{(1 + γ_{m} t) / (1 + γ_{a} t)\} \end{aligned}$
As $γ_{m} < γ_{a}$ , hence upthrust at t°C < upthrust at 0°C. So, the upthrust is decreased. Hence weight in liquid gets increased.

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