Question

In the arrangement shown below, a block of $2800kg$ is in equilibrium on applying a force of $45N$. Find the density of the liquid?

Answer 1

Pressure at horizontal line will be same

Lets consider the bottom point of system .

Let the height of 2800 kg side be H and area is $500\times {10}^{-4}{m}^2$

let the density of liquid be $\rho$

$P=\frac{weight}{area}+\rho gh$

So the pressure $P_1=\frac{2800g}{500\times {10}^{-4}}+\rho gh$

pressure at other end is

$P_2=\frac{45}{9\times {10}^{-4}}+\rho g(h+10)$

now we know that $P_1=P_2$

$\frac{2800g}{500\times {10}^{-4}}+\rho gh=\frac{45}{9\times {10}^{-4}}+\rho g(h+10)$

$\rho =5100kg{m}^{-3}$