Question

A liquid of density ρ is filled in a vessel of cross sectional area A upto height h. A block of mass M and cross sectional area a is made to float in it as shown in figure. The pressure at the bottom of the vessel will be [P0 : Atmospheric pressure]​

Answer 1

The pressure at the bottom of the vessel will be

$\boxed{P_0+h\rho g+\frac{Mg}{A}}$

Therefore, option (3) is correct.

Explanation:

Initially the height of the liquid in the vessel was h

If a block of mass M is floating on it and due to this if the increase in height of the liquid is h' then

The pressure at the bottom of the vessel will be

$P_0+\rho g(h+h')$

Now we need to find the value of h'

In equilibrium condition

Weight of block = buoyant force = Weight of the liquid displaced

or, Weight of the block = Cross sectional area of the vessel x increase in height x density x g

or, $Mg=Ah'\rho g$

or, $h'=\frac{M}{A\rho}$

Therefore, the pressure at the bottom

$=P_0+\rho g(h+\frac{M}{A\rho}$

$=P_0+h\rho g+\frac{Mg}{A}$

Therefore, option (3) is correct.

Hope this answer is helpful.

Know More:

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