Question 6

1. Calculation of Volume and Forces

• Volume of the ball ($V$):

$V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{3}{2} \times 10^{-2}\right)^3 = \frac{99}{7} \times 10^{-6}\text{ m}^3$

• Buoyant force ($F_b$):

$F_b = \rho_w V g = 10^3 \times \left(\frac{99}{7} \times 10^{-6}\right) \times 10 = \frac{99}{7} \times 10^{-2}\text{ N} = \frac{99}{700}\text{ N}$

• Gravitational force ($F_g = mg$):

$F_g = \left(\frac{22}{7} \times 10^{-3}\right) \times 10 = \frac{22}{7} \times 10^{-2}\text{ N} = \frac{22}{700}\text{ N}$

• Net Upward Force in Water (excluding viscosity):

$F_{\text{net}} = F_b - F_g = \frac{99}{700} - \frac{22}{700} = \frac{77}{700} = 0.11\text{ N}$

Evaluating the Options

Option (A): Work done in pushing the ball to depth $d$

To push the ball down slowly at a constant speed, the external downward force applied must balance the net upward force: $W = F_{\text{net}} \times d = 0.11\text{ N} \times 0.7\text{ m} = \mathbf{0.077\text{ J}}$

• Option (A) is correct.

Option (B): Speed $v$ neglecting viscous force

When released from rest at depth $d$, the work done by the net upward force converts entirely into kinetic energy at the surface: $W = \frac{1}{2}mv^2 \implies 0.077 = \frac{1}{2} \times \left(\frac{22}{7} \times 10^{-3}\right) \times v^2$ $0.077 = \frac{11}{7} \times 10^{-3} v^2 \implies v^2 = \frac{0.077 \times 7}{11 \times 10^{-3}} = 49$ $v = \mathbf{7\text{ m/s}}$

• Option (B) is correct.

Option (C): Height $H$ neglecting viscous force

After emerging from the water, the ball moves upward against gravity in the air: $H = \frac{v^2}{2g} = \frac{49}{2 \times 10} = \mathbf{2.45\text{ m}}$

• Since the option states $H = 1.4\text{ m}$, Option (C) is incorrect.

Option (D): Ratio of forces

The maximum viscous force occurs when the velocity is highest ($v = 7\text{ m/s}$): $(F_v)_{\text{max}} = 6\pi\eta r v = 6 \times \frac{22}{7} \times 10^{-3} \times \left(\frac{3}{2} \times 10^{-2}\right) \times 7$ $(F_v)_{\text{max}} = 6 \times 22 \times 10^{-3} \times \frac{3}{2} \times 10^{-2} = 198 \times 10^{-5}\text{ N}$ Now, find the ratio of the net force (excluding viscosity) to this maximum viscous force: $\text{Ratio} = \frac{F_{\text{net}}}{(F_v)_{\text{max}}} = \frac{0.11}{198 \times 10^{-5}} = \frac{11 \times 10^{-2}}{198 \times 10^{-5}} = \frac{11}{198} \times 10^3 = \frac{1000}{18} = \mathbf{\frac{500}{9}}$

• Option (D) is correct.