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In rising from the bottom of a lake to the top, the temperature of an air bubble remains unchanged, but its diameter is doubled. If h is the barometric height (expressed in meters of mercury of relative density \rho) at the surface of the lake, the depth of the lake is (in meters)

Option: 1

\mathrm{8 \rho h}


Option: 2

\mathrm{4 \rho h}


Option: 3

\mathrm{7 \rho h}


Option: 4

\mathrm{2 \rho h}


Answers (1)

best_answer

\mathrm{\text { Volume } \propto(\text { diameter })^3}

Since the diameter of the bubble is doubled in rising from the
bottom to the top of the lake, its volume becomes 8 times. Now PV = constant.
Therefore, the pressure at the bottom of lake = 8 times that at the top. Let H be the
depth of the lake

\mathrm{\begin{aligned} & H \rho_w \mathrm{~g}=(8 h-h) \rho_m \mathrm{~g} \\ & \text { Or } H=7 h \frac{\rho_m}{\rho_w}=7 h \rho\left(\because \rho=\frac{\rho_m}{\rho_w}\right) \end{aligned}}

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manish painkra

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