A beaker of height H is made up of a material | vedclass

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(D) Let the cross-sectional area of the beaker be $A$. The initial volume of the liquid is $V_0 = A H$.When heated by temperature $T$,the new volume o

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$8 \alpha TH$

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(D) Let the cross-sectional area of the beaker be $A$. The initial volume of the liquid is $V_0 = A H$.When heated by temperature $T$,the new volume of the liquid is $V_L = V_0(1 + \gamma_L T) = AH(1 + \alpha T)$,where $\gamma_L = \alpha$ is the volume expansion coefficient of the liquid.The new cross-sectional area of the beaker is $A' = A(1 + 2\alpha_s T)$,where $\alpha_s = 3\alpha$ is the linear expansion coefficient of the beaker material. Thus,$A' = A(1 + 2(3\alpha)T) = A(1 + 6\alpha T)$.The new height of the beaker is $H' = H(1 + \alpha_s T) = H(1 + 3\alpha T)$.The new height of the liquid $H_L$ in the beaker is $H_L = V_L / A' = AH(1 + \alpha T) / [A(1 + 6\alpha T)]$.Using the binomial approximation $(1 + x)^{-1} \approx 1 - x$ for small $x$,we get $H_L \approx H(1 + \alpha T)(1 - 6\alpha T) \approx H(1 + \alpha T - 6\alpha T) = H(1 - 5\alpha T)$.The reduction in the level of the liquid is $\Delta H = H' - H_L = H(1 + 3\alpha T) - H(1 - 5\alpha T) = H(1 + 3\alpha T - 1 + 5\alpha T) = 8\alpha TH$.

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