An air bubble of volume $V_0$ is released by a fish at a depth $h$ in a lake. The bubble rises to the surface. Assume constant temperature and standard atmospheric pressure $P$ above the lake. The volume of the bubble just before touching the surface will be (density of water is $\rho$):

  • A
    $V_0$
  • B
    $V_0(\rho g h / P)$
  • C
    $\frac{V_0}{1 + \frac{\rho g h}{P}}$
  • D
    $V_0(1 + \frac{\rho g h}{P})$

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The vertical limbs of a $U$ shaped tube are filled with a liquid of density $\rho$ up to a height $h$ on each side. The horizontal portion of the $U$ tube having length $2h$ contains a liquid of density $2\rho$. The $U$ tube is moved horizontally with an acceleration $g/2$ parallel to the horizontal arm. The difference in heights in liquid levels in the two vertical limbs,at steady state,will be:

$A$ fluid container containing a liquid of density $\rho$ is accelerating upward with acceleration $a$ along an inclined plane of inclination $\alpha$ as shown. Then the angle of inclination $\theta$ of the free surface is:

$A$ liquid flows through a horizontal tube. The velocities of the liquid in the two sections,which have areas of cross-section $A_1$ and $A_2$,are $v_1$ and $v_2$ respectively. The difference in the levels of the liquid in the two vertical tubes is $h$.

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