A vertical column 50 cm long at 50^{circ}C b | vedclass

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(A) The pressure exerted by a liquid column is given by $P = h \rho g$. Since the two columns balance each other,their pressures are equal: $h_1 \rho_

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$0.005 \ ^{\circ}C^{-1}$

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(A) The pressure exerted by a liquid column is given by $P = h ho g$. Since the two columns balance each other,their pressures are equal: $h_1 ho_1 g = h_2 ho_2 g$,which implies $h_1 ho_1 = h_2 ho_2$.Using the relation for density variation with temperature,$ho = \frac{ho_0}{1 + \gamma 	heta}$,we have $\frac{h_1}{h_2} = \frac{ho_2}{ho_1} = \frac{1 + \gamma 	heta_1}{1 + \gamma 	heta_2}$.Given $h_1 = 50 \ cm$,$	heta_1 = 50^{\circ}C$,$h_2 = 60 \ cm$,and $	heta_2 = 100^{\circ}C$.Substituting these values: $\frac{50}{60} = \frac{1 + 50\gamma}{1 + 100\gamma}$.Cross-multiplying: $50(1 + 100\gamma) = 60(1 + 50\gamma)$.$50 + 5000\gamma = 60 + 3000\gamma$.$2000\gamma = 10$.$\gamma = \frac{10}{2000} = 0.005 \ ^{\circ}C^{-1}$.

$A$ vertical column $50 \ cm$ long at $50^{\circ}C$ balances another column of the same liquid $60 \ cm$ long at $100^{\circ}C$. The coefficient of absolute expansion of the liquid is:

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