Consider two thermometers T_1 and T_2 of equa | vedclass

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(D) The change in volume $\Delta V$ of a liquid is given by $\Delta V = V_0 \gamma \Delta 	heta$,where $V_0$ is the initial volume,$\gamma$ is the co

Vedclass · Accepted Answer

$0.4$

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(D) The change in volume $\Delta V$ of a liquid is given by $\Delta V = V_0 \gamma \Delta 	heta$,where $V_0$ is the initial volume,$\gamma$ is the coefficient of volume expansion,and $\Delta 	heta$ is the change in temperature.Since the increase in length $\Delta l$ is the same for both thermometers,the change in volume is $\Delta V = A \Delta l = \frac{\pi d^2}{4} \Delta l$.Given that the initial volumes $(V_0)_{Hg} = (V_0)_{Br}$ and the change in length $\Delta l$ is the same for both,we have:$\Delta V_{Hg} = \frac{\pi d_1^2}{4} \Delta l = V_0 \gamma_{Hg} \Delta 	heta$$\Delta V_{Br} = \frac{\pi d_2^2}{4} \Delta l = V_0 \gamma_{Br} \Delta 	heta$Dividing the two equations:$\frac{d_1^2}{d_2^2} = \frac{\gamma_{Hg}}{\gamma_{Br}}$$\frac{d_1}{d_2} = \sqrt{\frac{\gamma_{Hg}}{\gamma_{Br}}}$Substituting the given values:$\frac{d_1}{d_2} = \sqrt{\frac{18 	imes 10^{-5}}{108 	imes 10^{-5}}} = \sqrt{\frac{1}{6}} \approx \sqrt{0.166} \approx 0.408$Thus,the ratio $d_1: d_2$ is closest to $0.4$.

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