Question

Two identical containers contains liquid of same mass m but different densities as shown in the figure given $\rho_1 > \rho_2$. $P_1$ and $P_2$ are pressures at the bottoms of the container1 and container -2 respectively and $F_1$ and $F_2$ are the forces exerted by the liquid on the walls of container-1 and container -2 respectively. (Neglect the atmospheric pressure)

Answer 1

P_1 > P_2 and F_1 < F_2

Answer 2

1. Volume and Density: Since the mass $m$ is the same for both liquids and $\rho_1 > \rho_2$, we have $V_1 < V_2$ (because $m = \rho V$).

2. Height and Volume Relationship: The containers are identical and wider at the top. For this shape, the volume $V(H)$ increases with height $H$, and the ratio $V(H)/H$ also increases with $H$. Since $V_1 < V_2$, this implies $H_1 < H_2$.

3. Pressure Comparison: The pressure at the bottom is $P = \rho g H$. Thus, $P_1 = \rho_1 g H_1$ and $P_2 = \rho_2 g H_2$. We can rewrite these as $P_1 = mg \frac{H_1}{V_1}$ and $P_2 = mg \frac{H_2}{V_2}$. Since $V(H)/H$ increases with $H$, $V_1/H_1 < V_2/H_2$. Taking the reciprocal, $H_1/V_1 > H_2/V_2$. Therefore, $P_1 > P_2$.

4. Vertical Force Balance: The vertical equilibrium of the liquid gives $P_{bottom} A_{bottom} + F_{walls, vertical} = mg$. Since $P_1 > P_2$, $P_1 A_{bottom} > P_2 A_{bottom}$. As $mg$ and $A_{bottom}$ are the same, $F_{1, vertical} = mg - P_1 A_{bottom} < mg - P_2 A_{bottom} = F_{2, vertical}$.

5. Total Force on Walls: For the given container shape (frustum of a cone), the angle of the wall with the vertical is constant. The total force on the walls $F$ is related to its vertical component $F_{vertical}$ by $F = F_{vertical} / \cos\theta$, where $\theta$ is the angle with the vertical. Since $\theta$ is the same for both containers, comparing $F_1$ and $F_2$ is equivalent to comparing $F_{1, vertical}$ and $F_{2, vertical}$. Thus, $F_1 < F_2$.

Therefore, $P_1 > P_2$ and $F_1 < F_2$.