Question

a cuboid (a*a*2a) is filled with 2 immiscible liquids of density 2p & p. neglecting atmospheric pressure, ratio of force on base & side wall of cuboid

Answer 1

Answer: The ratio of force is $\dfrac{F_{b}}{F_{w}} = \dfrac{6}{5}$

Explanation:

Given that,

Density of first liquid $= \rho$

Density of second liquid $= 2\rho$

We know that,

The pressure is

$P = \dfrac{F}{A}$

Where, p = pressure

A = area of cuboid

F = force

$\rho gh = \dfrac{F}{a^{2}}$

The force on base of cuboid is

$F_{b} = (2\rho gh+\rho gh)a^{2}$

$F_{b} = 3\rho gha^{a}$

The force on side wall of cuboid is

$F_{w} = (\dfrac{\rho gh}{2})ah + (2\rho gh)ajh$

$F_{w} = \dfrac{5}{2}\rho gha^{2}$

Now, the ratio of force on base and side wall of cuboid

$\dfrac{F_{b}}{F_{w}} = \dfrac{3\rho gh a^{2}}{\dfrac{5}{2}\rho gha^{2}}$

$\dfrac{F_{b}}{F_{w}} = \dfrac{6}{5}$

Hence, the ratio of force is $\dfrac{F_{b}}{F_{w}} = \dfrac{6}{5}$

Answer 2

Explanation:

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