Question

Find the minimum value of height $h_1$ (in situation 1) for which the block just starts to move up?

• A. $2h/3$
• B. $5h/4$
• C. $5h/3$
• D. $5h/2$

Answer 1

Answer: (C)

$5h/3$

Consider the equilibrium of wooden block. Force acting in the downward direction are as follows.

weight of wooden cylinder

$=\pi (4r{)}^2\times h\times \frac{\rho }{3}\times \rho =\pi \times 16r^2\frac{h\rho }{3}g$

Force due to pressure $\left(P_1\right)$ created by liquid of height $\left(h_1\right)$ above the wooden block's

$=P_1\times \pi (4r{)}^2=[P_0+h_1\rho g]\times \pi (4r{)}^2$

$=(P_0+h_1\rho g)\pi \times 16r^2$

Four acting on the upward direction due to pressure $P_0$ exerted from below the wooden block and atmospheric is

$=P_2\times \pi \left[\right(4r{)}^2-\left(2r{)}^2\right]+P_0\times (2r{)}^2$

$=[P_0+(h_1+h\left)\rho g\right]\times \pi \times 12r^2+4r^2{P}_0$

At verge of rising

$[P_0+(h_1+h\left)\rho g\right]\pi \times 12r^2+4r^3{P}_0=\pi \times 10r^{2}h\times \frac{\rho }{3}g+[P_0+h_1\rho g]\times \pi 16r^2$

$12h_1+12h=\frac{16h}{3}+16h$

$h_1=\frac{5h}{3}$