Q: 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 tube is

(a) $ \displaystyle \frac{P_0}{\rho g}+\frac{m}{a \rho} $

(b) $ \displaystyle \frac{P_0}{\rho g}+\frac{m}{2 a \rho} $

(c) $ \displaystyle \frac{ m}{a \rho } $

(d) $ \displaystyle \frac{ m}{2 a \rho } $

Ans: (c)

Sol: $\large P_0 + \frac{m g}{a} = P_0 + \rho g h $

$\large \frac{m g}{a} = \rho g h $

$ \displaystyle h = \frac{ m}{a \rho } $