In the figure, the cross-sectional area of the smaller tube is a and that of the larger tube is $2$ a. 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

\frac{p_{0}}{ρ g} + \frac{m}{a ρ}

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\frac{p_{0}}{ρ g} + \frac{m}{2 a ρ}

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\frac{m}{a ρ}

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\frac{m}{2 ρ a}

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Solution

The correct option is C $\frac{m}{a ρ}$
Pressure at Q
$h ρ g = \frac{m g}{a} \Rightarrow h = \frac{m}{a ρ}$
m=mass till point Q

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