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.]
$(A)$ 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 $500 / 9$.
IIT 2024, Advanced
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$\text { (A) } w_{\text {all }}=k_f-k_{ i }=0$
$w_g+w_B+w_v+w_{\text {ext }}=0$
$mgd -\rho_{ w } \cdot v \cdot g d-6 \pi \eta rvd + w _{\text {ext }}=0$
$\quad \text { (slowly } v =0 \text { ) }$
$w _{\text {ext }}=\rho_{ w } vgd - mgd =\left(1000 \times \frac{4}{3} \times \frac{22}{7} \times\left(\frac{3}{2} \times 10^{-2}\right)^3-\frac{22}{7} \times 10^{-3}\right) gd$
$w _{\text {ext }}=\frac{22}{7} \times 10^{-3}\left[\frac{9}{2}-1\right] \times 10 \times 0.7=\frac{22}{7} \times 10^{-3} \times \frac{7}{2} \times 7$
$w _{\text {ext }}=77 \times 10^{-3} J =0.077 J$
$\text { (B) } wg _{ g }+ w _{ B }= k _{ f }- k _{ i } \quad\left( k _{ i }=0\right)$
$\frac{1}{2} \times \frac{22}{7} \times 10^{-3} v ^2=77 \times 10^{-3}$
$v ^2=\frac{77 \times 7 \times 2}{22}$
$v =7 m / s$
$(C)$ $H =\frac{ v ^2}{2 g }=\frac{49}{20}=2.45 m$
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