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Physics · Fluid Mechanics

JEE Advanced 2024 — Paper 2 — Question 17

A table tennis ball has radius (3/2)×10−2 m(3 / 2) \times 10^{-2} \mathrm{~m} and mass (22/7)×10−3 kg(22 / 7) \times 10^{-3} \mathrm{~kg}. It is slowly pushed down into a swimming pool to a depth of d=0.7 m\mathrm{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: m=22/7, g=10 ms−2\mathrm{m}=22 / 7, \mathrm{~g}=10 \mathrm{~ms}^{-2}, density of water =1×103kgm−3=1 \times 10^{3} \mathrm{kgm}^{-3}, viscosity of water =1×10−3 Pa=1 \times 10^{-3} \mathrm{~Pa}-s.]

  1. Option A:

    The work done in pushing the ball to the depth d is 0.077 J

    Correct
  2. Option B:

    If we neglect the viscous force in water, then the speed v=7 m/sv=7 \mathrm{~m} / \mathrm{s}.

    Correct
  3. Option C:

    If we neglect the viscous force in water, then the height H=1.4 m\mathrm{H}=1.4 \mathrm{~m}.

  4. Option D:

    The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500/9500 / 9.

    Correct

Answer: A, B, D

Step-by-step solution

(A) Wext +Wg+WB=ΔK(∵\mathrm{W}_{\text {ext }}+\mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{B}}=\Delta \mathrm{K}(\because WD by viscous force =0)=0)

Wext +mgh−ρvgh=0W_{\text {ext }}+m g h-\rho v g h=0

Wext =ρℓvgh−mgh=gh(ρgV−m)W_{\text {ext }}=\rho_{\ell} \mathrm{vgh}-\mathrm{mgh}=\mathrm{gh}\left(\rho_{\mathrm{g}} \mathrm{V}-\mathrm{m}\right)

=10×0.7[103×43π(1.5×10−2)3−π×10−3]=0.77 J=10 \times 0.7\left[10^{3} \times \frac{4}{3} \pi\left(1.5 \times 10^{-2}\right)^{3}-\pi \times 10^{-3}\right]=0.77 \mathrm{~J}

(B) Wg+WB=12mv2−0\mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{B}}=\frac{1}{2} \mathrm{mv}^{2}-0

0.77=12(227×10−3)v2⇒v=7 m/s0.77=\frac{1}{2}\left(\frac{22}{7} \times 10^{-3}\right) v^{2} \Rightarrow v=7 \mathrm{~m} / \mathrm{s}

(C) H=v22 g=(7)22×10=2.45 m\mathrm{H}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}}=\frac{(7)^{2}}{2 \times 10}=2.45 \mathrm{~m} (D) Ratio =∣ρℓVg−Mg6πηrV∣=5009 J=\left|\frac{\rho_{\ell} \mathrm{Vg}-\mathrm{Mg}}{6 \pi \eta \mathrm{rV}}\right|=\frac{500}{9} \mathrm{~J}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 2
Subject
Physics
Chapter
Fluid Mechanics
Topic
Buoyancy and Archimedes' Principle