First slide

Thermal expansion

Question

A metal ball immersed in alcohol weighs W₁ at 0°C and W₂at 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

Easy

A

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

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

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

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

Solution

The upthrust on the ball is given by $\frac{4}{3} {πR}_{t}^{3} ρ_{t} g$
where $R_{t}^{3} = R_{0}^{3} (1 + γt) and ρ_{t} = ρ_{0} / (1 + γt)$
Upthrust at t °C
$\begin{array}{l} = \frac{4}{3} {πR}_{0}^{3} (1 + γt) \times ρ_{0} / (1 + γt) \times g \\ = \frac{4}{3} {πR}_{0}^{3} ρ_{0} g \end{array}$
It is same as that at 0°C. So, decrease in weight remains unchanged
$∴$ $W_{1} = W_{2}$

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