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 m s −2 , density of water = 1 × 103 kg m−3 , viscosity of water = 1 × 10−3 Pa-s.] (A) The work done in pushing the ball to the depth 𝑑 is 0.077 J. (B) If we neglect the viscous force in water, then the speed 𝑣 = 7 m/s. (C) If we neglect the viscous force in water, then the height 𝐻 = 1.4 m. (D) The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500/9.

 To solve this problem, we need to analyze the motion of the table tennis ball as it is pushed into the water, released, and rises back to the surface. We will consider the forces acting on the ball, including buoyancy, gravity, and viscous drag, and calculate the work done, speed, height, and force ratios.

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Step 1: Work Done in Pushing the Ball to Depth $d$

The work done in pushing the ball to a depth $d$ is equal to the change in potential energy of the ball-water system. The ball displaces a volume of water equal to its own volume, and the buoyant force acts upward. The work done is:

$$W=\text{Buoyant Force}\times d=({\rho }_{w}Vg)\cdot d$$

where:

• ${\rho }_w=1\times 10^3{\text{kg/m}}^3$ is the density of water,

• $V=\frac{4}{3}\pi r^3$ is the volume of the ball,

• $r=\frac{3}{2}\times 10^{-2}\text{m}$ is the radius of the ball,

• $g=10{\text{m/s}}^2$ is the acceleration due to gravity,

• $d=0.7\text{m}$ is the depth.

Substituting the values:

$$V=\frac{4}{3}\pi {(\frac{3}{2}\times 10^{-2})}^3=\frac{4}{3}\cdot \frac{22}{7}\cdot (\frac{27}{8}\times 10^{-6})=\frac{4\cdot 22\cdot 27}{3\cdot 7\cdot 8}\times 10^{-6}=\frac{2376}{168}\times 10^{-6}=14.14\times 10^{-6}{\text{m}}^3$$

The buoyant force is:

$$F_b={\rho }_{w}Vg=(1\times 10^3)\cdot (14.14\times 10^{-6})\cdot 10=0.1414\text{N}$$

The work done is:

$$W=F_b\cdot d=0.1414\cdot 0.7=0.099\text{J}$$

This does not match option (A), so (A) is incorrect.

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Step 2: Speed $v$ and Height $H$ (Neglecting Viscous Force)

If we neglect viscous forces, the ball's motion is governed by buoyancy and gravity. The net upward force on the ball is:

$$F_{\text{net}}=F_b-mg$$

where $m=\frac{22}{7}\times 10^{-3}\text{kg}$ is the mass of the ball. Substituting the values:

$$F_{\text{net}}=0.1414-(\frac{22}{7}\times 10^{-3}\cdot 10)=0.1414-0.0314=0.11\text{N}$$

The acceleration of the ball is:

$$a=\frac{F_{\text{net}}}{m}=\frac{0.11}{\frac{22}{7}\times 10^{-3}}=\frac{0.11\cdot 7}{22\times 10^{-3}}=\frac{0.77}{22\times 10^{-3}}=35{\text{m/s}}^2$$

Using the kinematic equation $v^2=2ad$, the speed $v$ at the surface is:

$$v=\sqrt{2\cdot 35\cdot 0.7}=\sqrt{49}=7\text{m/s}$$

This matches option (B), so (B) is correct.

The maximum height $H$ reached by the ball is given by:

$$H=\frac{v^2}{2g}=\frac{7^2}{2\cdot 10}=\frac{49}{20}=2.45\text{m}$$

This does not match option (C), so (C) is incorrect.

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Step 3: Ratio of Forces

The net force excluding viscous force is $F_{\text{net}}=0.11\text{N}$. The maximum viscous force is given by Stokes' law:

$$F_{\text{viscous}}=6\pi \eta rv$$

where:

• $\eta =1\times 10^{-3}\text{Pa-s}$ is the viscosity of water,

• $r=\frac{3}{2}\times 10^{-2}\text{m}$ is the radius of the ball,

• $v=7\text{m/s}$ is the speed of the ball.

Substituting the values:

$$F_{\text{viscous}}=6\pi \cdot (1\times 10^{-3})\cdot (\frac{3}{2}\times 10^{-2})\cdot 7=6\cdot \frac{22}{7}\cdot 10^{-3}\cdot \frac{3}{2}\cdot 10^{-2}\cdot 7=6\cdot 22\cdot 10^{-3}\cdot \frac{3}{2}\cdot 10^{-2}=6\cdot 22\cdot 1.5\cdot 10^{-5}=198\cdot 10^{-5}=0.00198\text{N}$$

The ratio of the net force to the maximum viscous force is:

$$\frac{F_{\text{net}}}{F_{\text{viscous}}}=\frac{0.11}{0.00198}=\frac{110}{1.98}=\frac{500}{9}$$

This matches option (D), so (D) is correct.

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Final Answer:

The correct options are:

$$\boxed{{\displaystyle \text{(B) and (D)}}}$$