Question

In the arrangement shown below a block of mass $2700kg$ is in equilibrium on applying a force F. The value of force F is $d_{liquid}=0.75gc{m}^{-3}$ is

• A. $147N$
• B. $300N$
• C. $153N$
• D. $1755N$

Answer 1

Answer: (D)

$1755N$

We know that pressure in a fluid changes by depth as $\Delta P=\rho g\Delta h$ and also, Pressure is constant on a horizontal plane.

In the given system, $\rho =0.75{\text{g cm}}^{-3}=750{\text{kg m}}^{-3},\Delta h=2\text{m}$ and $m=2700\text{kg}$

Let the pressure on the horizontal line be $P$

Equilibrium on the left hand side is given by

$P{A}_2=mg$

Equilibrium on the right hand side is given by

$(P-\rho g\Delta h)A_1=F$

Combining both the equations, $F=(\frac{mg}{A_2}-\rho g\Delta h)A_1=(2700\times 9.8\times \frac{6}{90})-750\times 9.8\times 2\times 6\times {10}^{-4}=1755.18\text{N}$