First slide

Thermal expansion

Question

45 gm of alcohol are needed to completely fill up a weight thermometer at 15°.C. The weight of alcohol which will overflow when the weight thermometer is heated to 33°C is (Take $γ_{a} = 121 \times 10^{- 5} /^{\circ} C$ )

Moderate

A

0.9 gm
B

0.96 gm
C

0.98 gm
D

none of these

Solution

$γ_{a} = \frac{m_{15} - m_{33}}{m_{33} (T_{2} - T_{1})} = \frac{45 - m_{33}}{m_{33} (33 - 15)}$
or $121 \times 10^{- 5} = \frac{45 - m_{33}}{m_{33} \times 18}$
Solving, we get $m_{33} = 45 / (1 \cdot 02178) gm$
Weight of alcohol overflowing $= m_{15} - m_{33}$
$\begin{array}{l} = 45 - \frac{45}{1 \cdot 02178} = \frac{45 (1 \cdot 02178 - 1)}{1 \cdot 02178} \\ = \frac{45 \times 0 \cdot 02178}{1 \cdot 02178} = 0 \cdot 96 gm \end{array}$
$∴$ Weight of alcohol overflowing = 0.96 gm

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