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(B) Let the initial volume of the liquid be $V_0 = A_0 l_0$,where $A_0$ is the cross-sectional area and $l_0$ is the length of the liquid column.When

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$\gamma = 2\alpha$

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(B) Let the initial volume of the liquid be $V_0 = A_0 l_0$,where $A_0$ is the cross-sectional area and $l_0$ is the length of the liquid column.When the system is heated by a temperature change $\Delta T$,the new volume of the liquid is $V_L = V_0(1 + \gamma \Delta T)$.The new cross-sectional area of the glass tube is $A = A_0(1 + 2\alpha \Delta T)$,where $2\alpha$ is the coefficient of area expansion.Since the length of the liquid column remains constant at $l_0$,the new volume of the liquid must be equal to the new volume of the tube up to height $l_0$,which is $V_{tube} = A 	imes l_0 = A_0(1 + 2\alpha \Delta T)l_0 = V_0(1 + 2\alpha \Delta T)$.Equating the two volumes: $V_0(1 + \gamma \Delta T) = V_0(1 + 2\alpha \Delta T)$.This simplifies to $1 + \gamma \Delta T = 1 + 2\alpha \Delta T$,which gives $\gamma = 2\alpha$.

The figure shows a glass tube (linear coefficient of expansion is $\alpha$) completely filled with a liquid of volume expansion coefficient $\gamma$. On heating,the length of the liquid column does not change. Choose the correct relation between $\gamma$ and $\alpha$.

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