JEE Advanced 2024 — Paper 2 — PHY-II solution

JEE Advanced2024Paper 2PHY-II

A table tennis ball has radius $(3/2) \times 1 0^{- 2}$ m and mass $(22/7) \times 1 0^{- 3}$ 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 $= 1 \times 1 0^{3} kg m^{- 3}$ , viscosity of water $= 1 \times 1 0^{- 3} Pa-s$ .]

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 $500/9$ .

Solution

Volume of the ball: $V = \frac{4}{3} π R^{3} = \frac{4}{3} \cdot \frac{22}{7} \cdot (\frac{3}{2} \times 1 0^{- 2})^{3} = \frac{99}{7} \times 1 0^{- 6} m^{3} .$

The buoyant force on the ball is $F_{B} = ρ_{w} V g$ and its weight is $W = m g$ . Work done in pushing the ball down a distance $d$ :

$W_{push} = (F_{B} - W) d = (ρ_{w} V g - m g) d .$

Numerically, $ρ_{w} V g = 1 0^{3} \cdot \frac{99}{7} \times 1 0^{- 6} \cdot 10 \approx 0.1414 N$ and $m g = \frac{22}{7} \times 1 0^{- 3} \cdot 10 \approx 0.0314 N$ . So $W_{push} \approx 0.11 \times 0.7 = 0.077$ J. (A) is correct.

By energy conservation, the same $0.077$ J converts to kinetic energy as the ball just emerges from the water:

$\frac{1}{2} m v^{2} = 0.077 \Rightarrow v = \frac{2 \times 0.077}{( 22/7 ) \times 1 0 ^{- 3}} = 7 m/s .$ (B) is correct.

Height above the water surface: $H = \frac{v ^{2}}{2 g} = \frac{49}{20} = 2.45 m$ , not $1.4$ m. (C) is incorrect.

Net non-viscous force in water: $F_{net} = F_{B} - W = 0.11 N .$

Maximum Stokes' viscous drag occurs at $v = 7 m/s$ :

$F_{v}^{m a x} = 6 π η r v = 6 \cdot \frac{22}{7} \cdot 1 0^{- 3} \cdot \frac{3}{2} \times 1 0^{- 2} \cdot 7 = 1.98 \times 1 0^{- 3} N .$

Ratio: $\frac{F _{net}}{F _{v}^{m a x}} = \frac{0.11}{1.98 \times 1 0 ^{- 3}} \approx \frac{500}{9} .$ (D) is correct.

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