JEE Advanced 2024 Paper 2 Q06 Physics Fluids & Surface Tension Viscosity Hard

JEE Advanced 2024 Paper 2 · Q06 · Viscosity

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: $\pi = 22/7$, $g = 10$ m s$^{-2}$, density of water $= 1 \times 10^3$ kg m$^{-3}$, viscosity of water $= 1 \times 10^{-3}$ Pa$\cdot$s.]

  1. A. The work done in pushing the ball to the depth $d$ is $0.077$ J.
  2. B. If we neglect the viscous force in water, then the speed $v = 7$ m/s.
  3. C. If we neglect the viscous force in water, then the height $H = 1.4$ m.
  4. D. The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500/9$.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, D

Solution

Volume of ball $V = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\cdot\dfrac{22}{7}\cdot(1.5\times 10^{-2})^3 = \dfrac{4}{3}\cdot\dfrac{22}{7}\cdot 3.375\times 10^{-6} = 1.4143\times 10^{-5}$ m$^3$.

Mass of water displaced $m_w = \rho_w V = 10^3 \times 1.4143\times 10^{-5} = 1.4143\times 10^{-2}$ kg. Buoyancy $F_B = \rho_w V g = 0.14143$ N. Weight $mg = (22/7)\times 10^{-3}\times 10 = 3.143\times 10^{-3}\times 10 \approx 3.143\times 10^{-2}$ N. Hmm — let me redo: $mg = (22/7)\times 10^{-3}\times 10 = 220/7\times 10^{-3} \approx 3.143\times 10^{-2}$ N.

(A) Work done pushing the ball down (against net upward force $F_B - mg$, plus initial work to submerge but since slowly): $W = (F_B - mg)\cdot d = (0.14143 - 0.03143)\cdot 0.7 = 0.11\cdot 0.7 = 0.077$ J. Correct.

(B) Neglecting viscosity: at release the net upward force is $F_{\text{net}} = F_B - mg = 0.11$ N. Acceleration $a = F_{\text{net}}/m = 0.11/(3.143\times 10^{-3}) = 35$ m/s$^2$ upward. From depth $d = 0.7$ m, $v^2 = 2 a d = 2\times 35\times 0.7 = 49$ ⇒ $v = 7$ m/s. Correct.

(C) After emerging, only gravity decelerates: $H = v^2/(2g) = 49/20 = 2.45$ m, not 1.4 m. So (C) is incorrect.

(D) Maximum viscous (Stokes) force o…[truncated]

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