Question

Two containers A and B contain the same liquid upto height H and $\frac{3H}{2}$respectively. P and Q are points inside containers A and B respectively, where pressure exerted due to the liquid are same. If point P lies at height $\frac{H}{3}$ from the bottom of container A, then point Q lies

• A.
  $\frac{H}{6}$above bottom of container B
• B.
  $\frac{H}{3}$above bottom of container B
• C.
  $\frac{5H}{6}$above bottom of container B
• D.
  $\frac{H}{3}$above bottom of container B

Answer 1

The correct option is C

$\frac{5H}{6}$above bottom of container B
Pressure exerted due to fluid at point
P = ρgd

Here, $\rho$ = density of fluid,

g = acceleration due to gravity and
d = depth from the surface of the fluid



From the figure, $d=H-\frac{H}{3}=\frac{2H}{3}$

Pressure at point P in container A $P_A=\rho g\left(\frac{2H}{3}\right)=\frac{2}{3}\rho gH$

Consider point Q is at height x from the bottom of container B. Pressure at point Q,$P_B=\rho g{d}^{\prime }{P}_B=\rho g(\frac{3H}{2}-x)(Where,d^{\prime }=\frac{3H}{2}-x)$

According to the question:

$P_A=P_B\Rightarrow \rho g\frac{2H}{3}=\rho g(\frac{3H}{2}-x)\Rightarrow \frac{2H}{3}=\frac{3H}{2}-x\Rightarrow x=\frac{5H}{6}$

Thus, Q lies at height$\frac{5H}{6}$ from the bottom.

Hence, the correct answer is option (c).