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In the figure, the cross-sectional area of the smaller tube is a and that of the larger tube is 2a. A block of mass m is kept in the smaller tube having the same base area a, as that of the tube. The difference between water levels of the two tubes is $(ρ i s t h e d e n s i t y o f t h e l i q u i d)$

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$\frac{P_{0}}{ρ g} + \frac{m}{a ρ}$

$\frac{m}{a ρ}$

$\frac{P_{0}}{ρ g} + \frac{m}{2 a ρ}$

$\frac{m}{2 ρ a}$

answer is C.

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Detailed Solution

Pressure at the bottom level of the block $P = \frac{m g}{a} + P_{0}$ . The pressure at the same level remains the same. Hence, $\frac{m g}{a} + P_{0} = h ρ g + P_{0}$ $\Rightarrow h = \frac{m}{a ρ}$

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