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(C) Let $h_1 = 120 \, cm$ be the height of the liquid at $0^{\circ}C$ (left tube).Let $h_2 = 120 + 4 = 124 \, cm$ be the height of the liquid at $30^{

Vedclass · Accepted Answer

$11 	imes 10^{-4}/^{\circ}C$

Vedclass · Answer

(C) Let $h_1 = 120 \, cm$ be the height of the liquid at $0^{\circ}C$ (left tube).Let $h_2 = 120 + 4 = 124 \, cm$ be the height of the liquid at $30^{\circ}C$ (right tube).Since the pressure at the bottom of both tubes must be equal:$h_1 ho_1 g = h_2 ho_2 g$$ho_1 h_1 = ho_2 h_2$Since $ho_2 = \frac{ho_1}{1 + \gamma \Delta T}$,where $\Delta T = 30^{\circ}C - 0^{\circ}C = 30^{\circ}C$:$h_1 = h_2 \left( \frac{1}{1 + \gamma \Delta T} ight)$$1 + \gamma \Delta T = \frac{h_2}{h_1} = \frac{124}{120}$$\gamma \Delta T = \frac{124}{120} - 1 = \frac{4}{120} = \frac{1}{30}$$\gamma = \frac{1}{30 	imes 30} = \frac{1}{900} \approx 0.00111 \, /^{\circ}C$$\gamma \approx 11.11 	imes 10^{-4} /^{\circ}C$.

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