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JEE Advanced 2024 · Paper 2, Physics question 6

A table tennis ball has radius (3/2)×10−2 m(3/2)\times10^{-2}\,\mathrm{m} and mass (22/7)×10−3 kg(22/7)\times10^{-3}\,\mathrm{kg}. It is slowly pushed down into a swimming pool to a depth of d=0.7 md = 0.7\,\mathrm{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=22/7\pi = 22/7, g=10 m s−2g = 10\,\mathrm{m\,s^{-2}}, density of water =1×103 kg m−3= 1\times10^3\,\mathrm{kg\,m^{-3}}, viscosity of water =1×10−3= 1\times10^{-3} Pa-s.]

Choose one or more options
Answer and worked solution

Official answer: A, B, D

Data.

  • V=43πr3=43⋅227⋅(1.5×10−2)3=997×10−6 m3V=\frac43\pi r^3=\frac43\cdot\frac{22}{7}\cdot(1.5\times10^{-2})^3=\frac{99}{7}\times10^{-6}\ \mathrm{m^3}
  • Buoyancy: FB=ρVg=0.997 N≈0.1414 NF_B=\rho Vg=\frac{0.99}{7}\ \mathrm N\approx0.1414\ \mathrm N
  • Weight: mg=0.227 N≈0.0314 Nmg=\frac{0.22}{7}\ \mathrm N\approx0.0314\ \mathrm N
  • Net upward force: F=0.777=0.11 NF=\frac{0.77}{7}=0.11\ \mathrm N

Options.

  • (A) Pushing slowly to d=0.7d=0.7 m takes W=Fd=0.11×0.7=0.077W=Fd=0.11\times0.7=0.077 J. ✓
  • (B) Without viscosity, the same 0.0770.077 J becomes kinetic energy: 12mv2=0.077\frac12mv^2=0.077 gives v2=0.154(22/7)×10−3=49v^2=\frac{0.154}{(22/7)\times10^{-3}}=49, so v=7v=7 m/s. ✓
  • (C) H=v22g=2.45H=\frac{v^2}{2g}=2.45 m, not 1.41.4 m. ✗
  • (D) The maximum viscous force is at the maximum speed (77 m/s): Fv=6πηrv=6⋅227⋅10−3⋅1.5×10−2⋅7=1.98×10−3F_v=6\pi\eta rv=6\cdot\frac{22}{7}\cdot10^{-3}\cdot1.5\times10^{-2}\cdot7=1.98\times10^{-3} N. The ratio is 0.111.98×10−3=5009\frac{0.11}{1.98\times10^{-3}}=\frac{500}{9}. ✓
Topics:Buoyancy and Archimedes' principleViscosity and Stokes' lawWork-energy theorem
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