A table tennis ball has radius (3/2) × 10⁻² m and mass (22/7) × 10⁻³ kg. It is slowly pushed down into a swimming pool to a depth of 𝑑 = 0.7 m below the water surface and then released from rest. It emerges from the water surface at speed 𝑣, without getting wet, and rises up to a height 𝐻. Which of the following option(s) is(are) correct ?

[Given: 𝜋 = 22/7, 𝑔 = 10 ms⁻² , density of water = 1 × 10³ kg m⁻³ , viscosity of water = 1 × 10⁻³ Pa-s.]

The correct options are:
1. The work done in pushing the ball to the depth d is 0.077 J.
2. If we neglect the viscous force in water, then the speed v = 7 m/s.
3. The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500/9.

1. Calculation of Volume and Forces:
Given parameters:
Radius of the ball, R = 3/2 × 10^-2 m
Mass of the ball, m = 22/7 × 10^-3 kg = 𝜋 × 10^-3 kg (since 𝜋 = 22/7)
Depth, d = 0.7 m
Density of water, 𝜌_w = 10³ kg/m³
Acceleration due to gravity, g = 10 m/s²
Viscosity of water, 𝜂 = 10^-3 Pa-s

The volume of the ball V is given by:

$V\; =\; 4/3\; 𝜋\; R3=\; 4/3\; \times \; (22/7)\; \times \; (3/2\; \times \; 10-2)3=\; 99/7\; \times \; 10-6m3$

The buoyant force F_B acting on the fully submerged ball is:

$F$_B = V 𝜌_w g = (99/7 × 10^-6) × 10³ × 10 = 0.1414 N = 99/7 × 10^-2 N

The gravitational force F_g acting on the ball is:

$F$_g = m g = (22/7 × 10^-3) × 10 = 0.0314 N = 22/7 × 10^-2 N

The net upward force excluding viscous force is:

$F$_net = F_B - F_g = (99/7 - 22/7) × 10^-2 = 77/7 × 10^-2 = 0.11 N

2. Work done in pushing the ball to depth d:
The work done by the external force to slowly push the ball down by a distance d (once fully submerged) is:

$W\; =\; F$_net × d = 0.11 × 0.7 = 0.077 J

Thus, the first option is correct.

3. Speed of emergence v when viscous force is neglected:
Using the work-energy theorem, the kinetic energy of the ball as it reaches the surface is equal to the work done by the net upward force:

$1/2\; m\; v2=\; F$_net d

$1/2\; \times \; (22/7\; \times \; 10-3)\; \times \; v2=\; 0.077$

$11/7\; \times \; 10-3\times \; v2=\; 77\; \times \; 10-3$

$v2=\; 49\; \Rightarrow \; v\; =\; 7\; m/s$

Thus, the second option is correct.

4. Maximum height H reached in air (excluding viscous force):
Once the ball leaves the water, it rises under the influence of gravity alone:

$H\; =\; v2/\; (2g)\; =\; 49\; /\; (2\; \times \; 10)\; =\; 2.45\; m$

Thus, the third option is incorrect.

5. Ratio of net force to maximum viscous force:
The maximum viscous force occurs when the speed is maximum (i.e., at the surface, v = 7 m/s). Using Stokes' Law:

$F$_v,max = 6 𝜋 𝜂 R v

$F$_v,max = 6 × (22/7) × 10^-3 × (3/2 × 10^-2) × 7

$F$_v,max = 6 × 22 × 10^-3 × 3/2 × 10^-2 = 198 × 10^-5 N

Now, the ratio of the net force excluding viscous force to the maximum viscous force is:

$F$_net / F_v,max = 0.11 / (198 × 10^-5) = (11 × 10^-2) / (1.98 × 10^-3) = 110 / 1.98 = 500/9

Thus, the fourth option is correct.