let it is $=\mathrm{q}$
and let current in resistance branch $=\mathrm{i}$
$\therefore \mathrm{V}_{\mathrm{C}_{1}}=\mathrm{i} \mathrm{R}_{1} \Rightarrow \frac{\mathrm{q}}{\mathrm{C}_{1}}=\mathrm{i} \mathrm{R}_{1} \Rightarrow \mathrm{q}=\mathrm{ir}_{1} \mathrm{C}_{1}$
similarly $\mathrm{V}_{\mathrm{C}_{2}}=\mathrm{i} \mathrm{R}_{2} \Rightarrow \frac{\mathrm{q}}{\mathrm{C}_{2}}=\mathrm{i} \mathrm{R}_{2} \Rightarrow \mathrm{q}=\mathrm{i} \mathrm{R}_{2} \mathrm{C}_{2}$
$\therefore \quad \mathrm{iR}_{1} \mathrm{C}_{1}=\mathrm{i} \mathrm{R}_{2} \mathrm{C}_{2}$
$\Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$
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If the whole set up is immersed in water without disturbing the object and the screen positions, what will one observe on the screen?
Statement$-I:$ Acceleration due to gravity is different at different places on the surface of earth.
Statement$-II:$ Acceleration due to gravity increases as we go down below the earth's surface.
In the light of the above statements, choose the correct answer from the options given below