Question

A light cylindrical tube T of length $\iota$ and radius r containing air is inverted in water (density d). One end of the tube is open and the other is closed. A block B of density 2d is kept on the tube as shown in the figure. The tube stays in equilibrium in the position shown. (Assume the atmospheric pressure is to be $P_0$.). Pick up the correct statement(s).

• A. the volume of block B is $\frac{\pi r^2\iota }{3}$
• B. the volume of block B is $\frac{2\pi r^2\iota }{3}$
• C. the pressure of air trapped in the tube is $P_0+dg(h+\frac{\iota }{3})$
• D. the pressure of air trapped in the tube is $P_0+dg(h+\frac{2\iota }{3})$

Answer 1

Answer: (A), (C)

The correct options are
A the volume of block B is $\frac{\pi r^2\iota }{3}$
C the pressure of air trapped in the tube is $P_0+dg(h+\frac{\iota }{3})$

The free body diagram of block + tube inclusive of water is as shown below
Since the block + tube system shown in figure is in equilibrium
$\therefore$ Net weight of system = buoyant force
$2dVg+\pi r^2\frac{2\iota }{3}dg=(V+\pi r^2\iota )dg$
where V is the volume of block B
$\Rightarrow (2d-d)V=\frac{\pi r^2\iota d}{3}$ or $V=\frac{\pi r^2\iota }{3}$
The pressure of the air trapped inside the tube is same as pressure at point A in the water as shown in figure.
$\therefore P=P_0+(h+\frac{\iota }{3})dg$ Ans: $\therefore P_a=P_0+dg(h+\frac{\iota }{3})$