JEE Advance - Physics (2024 - Paper 2 Online - No. 6)
A table tennis ball has radius $(3 / 2) \times 10^{-2} \mathrm{~m}$ and mass $(22 / 7) \times 10^{-3} \mathrm{~kg}$. It is slowly pushed down into a swimming pool to a depth of $d=0.7 \mathrm{~m}$ 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: $\pi=22 / 7, g=10 \mathrm{~m} \mathrm{~s}^{-2}$, density of water $=1 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$, viscosity of water $=1 \times 10^{-3} \mathrm{~Pa}$-s.]
Explanation
$$\begin{aligned} & \text { (A) } w_{\text {all }}=k_f-k_i=0 \\ & w_g+w_B+w_v+w_{\text {ext }}=0 \\ & \mathrm{mgd}-\rho_w \cdot v \cdot g d-6 \pi \eta r v d+w_{\text {ext }}=0 \\ & \quad(\text { slowly } v=0) \end{aligned}$$
$$\begin{aligned} & \mathrm{w}_{\text {ext }}=\rho_{\mathrm{w}} \operatorname{vgd}-\mathrm{mgd}=\left(1000 \times \frac{4}{3} \times \frac{22}{7} \times\left(\frac{3}{2} \times 10^{-2}\right)^3-\frac{22}{7} \times 10^{-3}\right) \mathrm{gd} \\ & \mathrm{w}_{\text {ext }}=\frac{22}{7} \times 10^{-3}\left[\frac{9}{2}-1\right] \times 10 \times 0.7=\frac{22}{7} \times 10^{-3} \times \frac{7}{2} \times 7 \\ & \mathrm{~W}_{\text {ext }}=77 \times 10^{-3} \mathrm{~J}=0.077 \mathrm{~J} \end{aligned}$$
$$\begin{aligned} & \text { (B) } \mathrm{w}_{\mathrm{g}}+\mathrm{w}_{\mathrm{B}}=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}} \quad\left(\mathrm{k}_{\mathrm{i}}=0\right) \\ & \frac{1}{2} \times \frac{22}{7} \times 10^{-3} \mathrm{v}^2=77 \times 10^{-3} \\ & \mathrm{v}^2=\frac{77 \times 7 \times 2}{22} \\ & \mathrm{v}=7 \mathrm{~m} / \mathrm{s} \end{aligned}$$
(C) $$\mathrm{H}=\frac{\mathrm{v}^2}{2 \mathrm{~g}}=\frac{49}{20}=2.45 \mathrm{~m}$$
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