A table tennis ball has radius $$(3/2) \times 10^{-2}$$ m and mass $$(22/7) \times 10^{-3}$$ 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: $$\pi = 22/7$$, $$g = 10$$ ms$$^{-2}$$, density of water $$= 1 \times 10^3$$ kg m$$^{-3}$$, viscosity of water $$= 1 \times 10^{-3}$$ Pa·s.]

Radius of the ball, $$r = \frac{3}{2}\times 10^{-2}\;{\rm m}=0.015\;{\rm m}$$
Mass of the ball, $$m = \frac{22}{7}\times 10^{-3}\;{\rm kg}= \pi\times 10^{-3}\;{\rm kg}=3.1429\times 10^{-3}\;{\rm kg}$$

Volume of the ball
$$V = \frac{4}{3}\pi r^{3}$$
$$r^{3}=0.015^{3}=3.375\times 10^{-6}\,{\rm m^{3}}$$
$$V=\frac{4}{3}\pi(3.375\times 10^{-6})=4.18879\times3.375\times10^{-6}=1.414\times10^{-5}\;{\rm m^{3}}$$

Buoyant force in water
$$F_{b} = \rho_{\text w}gV = (1000)(10)(1.414\times10^{-5}) = 0.1414\;{\rm N}$$

Weight of the ball
$$W = mg = (3.1429\times10^{-3})(10)=0.03143\;{\rm N}$$

Net upward force (viscosity neglected)
$$F_{\text{net}} = F_{b}-W = 0.1414-0.03143 = 0.1100\;{\rm N}\approx 0.11\;{\rm N}$$

Option A  - Work done in pushing the ball down by $$d=0.7\;{\rm m}$$ (quasi-static)
To move slowly we must apply a downward force equal to the constant upward imbalance $$F_{b}-W$$. $$W_{\text{push}} = (F_{b}-W)\,d = (0.11)(0.7)=0.077\;{\rm J}$$
Hence Option A is correct.

Option B  - Speed $$v$$ at the water surface (viscosity neglected)
Acceleration inside water: $$a=\dfrac{F_{\text{net}}}{m}=\dfrac{0.11}{3.1429\times10^{-3}}\approx35\;{\rm m\,s^{-2}}$$
Using $$v^{2}=2as$$ with $$s=d=0.7\;{\rm m}$$:
$$v=\sqrt{2(35)(0.7)}=\sqrt{49}=7\;{\rm m\,s^{-1}}$$
Thus Option B is correct.

Option C  - Height $$H$$ above the surface
After emerging, only gravity acts. $$H=\dfrac{v^{2}}{2g}=\dfrac{49}{20}=2.45\;{\rm m}$$, not $$1.4\;{\rm m}$$. Therefore Option C is incorrect.

Option D  - Ratio of (net force) to (maximum viscous force)
Maximum possible viscous (Stokes) drag inside water occurs at speed $$v=7\;{\rm m\,s^{-1}}$$:
$$F_{\eta,\max}=6\pi\eta rv = 6\left(\frac{22}{7}\right)(1\times10^{-3})(0.015)(7)$$
$$(\frac{22}{7})\times7 = 22$$, so $$F_{\eta,\max}=6(22)(0.015)\times10^{-3}=1.98\times10^{-3}\;{\rm N}$$

Hence $$\dfrac{F_{\text{net}}}{F_{\eta,\max}}=\dfrac{0.11}{1.98\times10^{-3}}\approx55.56=\frac{500}{9}$$
Option D is therefore correct.

Final result: Option A, Option B and Option D are correct.