Question

Two immiscible liquids of density $\rho$ and $2\rho$ filled in a container to each height h. Base length is a and width of container is b. Then : (Neglect atmospheric pressure)

• A. Force applied by liquids on the bottom is equals to the sum of their weight.
• B. Net force by liquids on left vertical face is $\frac{5\rho gbh^2}{2}$.
• C. Net force by liquids on right face is $\frac{5\rho gbh^2}{3}$.
• D. Pressure at bottom is $3\rho gh$.

Answer 1

Answer: (A), (B), (D)

A, B, D

Answer 2

Pressure at the bottom due to the upper liquid is $\rho gh$. Pressure at the bottom due to the lower liquid is $2\rho gh$. Total pressure at the bottom is $P_{bottom} = \rho gh + 2\rho gh = 3\rho gh$. Force on the bottom is $F_{bottom} = P_{bottom} \times Area = (3\rho gh) \times (ab) = 3\rho gabh$. Weight of the upper liquid is $W_1 = (\rho \times abh) \times g = \rho gabh$. Weight of the lower liquid is $W_2 = (2\rho \times abh) \times g = 2\rho gabh$. Total weight of liquids is $W_{total} = W_1 + W_2 = 3\rho gabh$. Thus, $F_{bottom} = W_{total}$. (Option A is correct)

For the vertical face, consider a strip of height $dz$ at depth $z$. For $0 \le z \le h$, pressure $P(z) = \rho g z$. For $h \le z \le 2h$, pressure $P(z) = \rho gh + 2\rho g(z-h)$. Force on the vertical face is $F = \int_0^{2h} P(z) b dz$. $F = b \left( \int_0^h \rho g z dz + \int_h^{2h} (\rho gh + 2\rho g(z-h)) dz \right)$ $F = b \left( [\frac{1}{2}\rho gz^2]_0^h + [\rho ghz + \rho g(z-h)^2]_h^{2h} \right)$ $F = b \left( \frac{1}{2}\rho gh^2 + (\rho gh(2h) + \rho g(2h-h)^2) - (\rho gh^2 + \rho g(h-h)^2) \right)$ $F = b \left( \frac{1}{2}\rho gh^2 + 2\rho gh^2 + \rho gh^2 - \rho gh^2 \right)$ $F = b \left( \frac{1}{2}\rho gh^2 + 2\rho gh^2 \right) = \frac{5}{2}\rho gbh^2$. (Option B is correct)

Option C is incorrect as the force on the right face would be the same as the left face if the container is a rectangular prism.