Question

A small solid ball of density $\rho$ is held inside at point A in a closed cubical container of side L, filled with an ideal liquid of density $4\rho$ as shown in the figure. Now, if the container starts moving with constant acceleration a horizontally and the ball is released from point A simultaneously, then

• A. For ball to hit the top of container at end Q, a = 3g
• B. For ball to hit the top of container at end Q, a = 2g
• C. Ball hits the top of container at end Q after a time $t = \sqrt{L/3g}$
• D. Ball hits the top of container at end Q after a time $t = \sqrt{2L/3g}$

Answer 1

Answer: (B), (C)

B, C

Answer 2

The problem involves analyzing the motion of a ball in an accelerating liquid-filled container. We will use a non-inertial frame of reference, which is the frame of the container accelerating horizontally.

1. Determine the forces acting on the ball in the non-inertial frame:

Let the mass of the ball be $m$. Density of the ball, $\rho_{ball} = \rho$. Density of the liquid, $\rho_{liquid} = 4\rho$. Volume of the ball, $V = \frac{m}{\rho_{ball}} = \frac{m}{\rho}$.

The container accelerates horizontally with $\vec{a}_{container} = a \hat{i}$. The acceleration due to gravity is $\vec{g} = -g \hat{j}$.

The forces acting on the ball are:

• Gravitational force: $\vec{F}_g = m\vec{g} = -mg \hat{j}$.
• Pseudo force: This force acts opposite to the container's acceleration: $\vec{F}_{pseudo} = -m\vec{a}_{container} = -ma \hat{i}$.
• Buoyant force: In an accelerating fluid, the effective acceleration due to gravity is $\vec{g}_{eff} = \vec{g} - \vec{a}_{container}$. The buoyant force is given by $\vec{F}_B = -\rho_{liquid} V \vec{g}_{eff}$. $\vec{F}_B = -(4\rho) \left(\frac{m}{\rho}\right) (\vec{g} - \vec{a}_{container})$ $\vec{F}_B = -4m (-g \hat{j} - a \hat{i})$ $\vec{F}_B = 4mg \hat{j} + 4ma \hat{i}$.

2. Calculate the net force and acceleration of the ball relative to the container:

The net force on the ball in the non-inertial frame is the vector sum of these forces: $\vec{F}_{net} = \vec{F}_g + \vec{F}_{pseudo} + \vec{F}_B$ $\vec{F}_{net} = (-mg \hat{j}) + (-ma \hat{i}) + (4mg \hat{j} + 4ma \hat{i})$ $\vec{F}_{net} = (-ma + 4ma) \hat{i} + (-mg + 4mg) \hat{j}$ $\vec{F}_{net} = 3ma \hat{i} + 3mg \hat{j}$.

The acceleration of the ball relative to the container, $\vec{a}_{ball/container} = \frac{\vec{F}_{net}}{m}$: $\vec{a}_{ball/container} = \frac{3ma \hat{i} + 3mg \hat{j}}{m} = 3a \hat{i} + 3g \hat{j}$. So, the horizontal acceleration relative to the container is $a_x = 3a$, and the vertical acceleration relative to the container is $a_y = 3g$.

3. Apply kinematic equations for the ball to hit point Q:

The container has side length $L$. Point A is located at $(0, L/2)$ if we set the origin at point R (bottom-left corner). Point Q (top-right corner) is located at $(L, L)$. The ball is released from rest relative to the container, so initial velocity $v_{0x} = 0$ and $v_{0y} = 0$.

For the ball to hit point Q, its horizontal displacement must be $L$ (from $x=0$ to $x=L$), and its vertical displacement must be $L/2$ (from $y=L/2$ to $y=L$).

Using the kinematic equation $s = ut + \frac{1}{2}at^2$:

Horizontal motion: $L = 0 \cdot t + \frac{1}{2} a_x t^2$ $L = \frac{1}{2} (3a) t^2$ $L = \frac{3}{2} a t^2 \implies t^2 = \frac{2L}{3a}$ (Equation 1)

Vertical motion: $L/2 = 0 \cdot t + \frac{1}{2} a_y t^2$ $L/2 = \frac{1}{2} (3g) t^2$ $L/2 = \frac{3}{2} g t^2 \implies t^2 = \frac{L}{3g}$ (Equation 2)

4. Solve for 'a' and 't':

Equating Equation 1 and Equation 2 (since the time 't' is the same for both motions to reach point Q): $\frac{2L}{3a} = \frac{L}{3g}$ Cancel $L/3$ from both sides: $\frac{2}{a} = \frac{1}{g}$ $a = 2g$.

Now, substitute $a=2g$ back into Equation 2 to find the time $t$: $t^2 = \frac{L}{3g} \implies t = \sqrt{\frac{L}{3g}}$.

5. Compare with the given options:

(A) For ball to hit the top of container at end Q, a = 3g. (Incorrect, we found $a = 2g$) (B) For ball to hit the top of container at end Q, a = 2g. (Correct) (C) Ball hits the top of container at end Q after a time $t = \sqrt{L/3g}$. (Correct, this is the time taken when $a=2g$) (D) Ball hits the top of container at end Q after a time $t = \sqrt{2L/3g}$. (Incorrect)

Both options (B) and (C) are correct statements that describe the conditions for the ball to hit point Q.