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JEE Advanced 2024 · Mechanical Properties of Fluids

A table tennis ball has radius (32)×10−2 m\left(\frac{3}{2}\right)\times 10^{-2}\,\text{m} and mass (227)×10−3 kg\left(\frac{22}{7}\right)\times 10^{-3}\,\text{kg}. It is slowly pushed down into a swimming pool to a depth of d=0.7 md = 0.7\,\text{m} below the water surface and then released from rest. It emerges from the water surface at speed vv, without getting wet, and rises up to a height HH. Which of the following option(s) is(are) correct? [Given: π=227\pi = \frac{22}{7}, g=10 ms−2g = 10\,\text{ms}^{-2}, density of water =1×103 kg m−3= 1 \times 10^{3}\,\text{kg m}^{-3}, viscosity of water =1×10−3 Pa-s= 1 \times 10^{-3}\,\text{Pa-s}.]

Answers: A. The work done in pushing the ball to the depth dd is 0.077 J0.077\,\text{J}. · B. If we neglect the viscous force in water, then the speed v=7 m/sv = 7\,\text{m/s}. · D. The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 5009\frac{500}{9}.

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Written solution

Step 1: Calculate the volume, buoyant force, and weight of the ball.

The volume of the spherical ball is:

V=43πr3V = \frac{4}{3}\pi r^3

Substitute the given values:

V=43×227×(1.5×10−2)3=1.414×10−5 m3V = \frac{4}{3} \times \frac{22}{7} \times \left(1.5 \times 10^{-2}\right)^3 = 1.414 \times 10^{-5} \text{ m}^3

The buoyant force acting on the ball is:

FB=ρwVgF_B = \rho_w V g FB=103×1.414×10−5×10=0.1414 NF_B = 10^3 \times 1.414 \times 10^{-5} \times 10 = 0.1414 \text{ N}

The weight of the ball is:

W=mgW = mg W=(227×10−3)×10=0.03143 NW = \left(\frac{22}{7} \times 10^{-3}\right) \times 10 = 0.03143 \text{ N}

Step 2: Calculate the work done to push the ball to depth dd.

The net upward force (excluding viscosity) is:

Fnet=FB−WF_{\text{net}} = F_B - W Fnet=0.1414−0.0314=0.11 NF_{\text{net}} = 0.1414 - 0.0314 = 0.11 \text{ N}

The work done to slowly push the ball to depth d=0.7 md = 0.7 \text{ m} is:

Wpush=Fnet×dW_{\text{push}} = F_{\text{net}} \times d Wpush=0.11×0.7=0.077 JW_{\text{push}} = 0.11 \times 0.7 = 0.077 \text{ J}

This matches option (1).

Step 3: Determine the speed vv of the ball as it emerges from the water surface.

Neglecting viscous forces, the work done on the ball is converted into kinetic energy:

Fnetd=12mv2F_{\text{net}} d = \frac{1}{2}mv^2 v=2Fnetdmv = \sqrt{\frac{2 F_{\text{net}} d}{m}} v=2×0.077227×10−3=49=7 m s−1v = \sqrt{\frac{2 \times 0.077}{\frac{22}{7} \times 10^{-3}}} = \sqrt{49} = 7 \text{ m s}^{-1}

This matches option (2).

Step 4: Determine the maximum height HH reached by the ball.

After emerging from the water, the ball rises under gravity:

H=v22gH = \frac{v^2}{2g} H=722×10=2.45 mH = \frac{7^2}{2 \times 10} = 2.45 \text{ m}

This does not match option (3).

Step 5: Calculate the ratio of the net force to the maximum viscous force.

The viscous force is given by Stokes' law:

Fvisc=6πηrvF_{\text{visc}} = 6\pi\eta r v

The viscous force is maximum when the speed is maximum, which occurs at the surface where v=7 m s−1v = 7 \text{ m s}^{-1}:

Fvisc,max=6×227×10−3×1.5×10−2×7=1.98×10−3 NF_{\text{visc,max}} = 6 \times \frac{22}{7} \times 10^{-3} \times 1.5 \times 10^{-2} \times 7 = 1.98 \times 10^{-3} \text{ N}

The ratio of the net force (excluding viscosity) to the maximum viscous force is:

FnetFvisc,max=0.111.98×10−3=5009\frac{F_{\text{net}}}{F_{\text{visc,max}}} = \frac{0.11}{1.98 \times 10^{-3}} = \frac{500}{9}

This matches option (4).

Answer: Correct options are (1), (2) and (4).

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