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MHT CET 2024 (3 May, Shift 1) · Terminal Velocity

Terminal Velocity question from MHT CET 2024 (3 May, Shift 1)

Q31MHT CET 2024 · 3 MAY · SHIFT 1Difficult

Terminal Velocity

A metal ball of radius 9×10−4 m9 \times 10^{-4}\text{ m} and density 104 kg/m310^{4}\text{ kg/m}^{3} falls freely under gravity through a distance ' h ' and enters a tank of water. Considering that the metal ball has constant velocity, the value of hh is [coefficient of viscosity of water =8.1×10−4 Pa-s, g=10 m/s2= 8.1 \times 10^{-4}\text{ Pa-s},\ \mathrm{g} = 10\text{ m/s}^{2} density of water =103 kg/m3]= 10^{3}\text{ kg/m}^{3}]

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Answer: (a)

  1. Ball moves at constant velocity in water, so 2gh=vt=2r2(ρ−σ)g9η\sqrt{2gh} = v_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta}.
  2. vt=2×(9×10−4)2×(104−103)×109×8.1×10−4=2×81×10−8×9×1047.29×10−3=20 m/sv_t = \dfrac{2\times(9\times10^{-4})^2\times(10^4 - 10^3)\times10}{9\times8.1\times10^{-4}} = \dfrac{2\times81\times10^{-8}\times9\times10^{4}}{7.29\times10^{-3}} = 20\text{ m/s}
  3. h=vt22g=40020=20 mh = \dfrac{v_t^2}{2g} = \dfrac{400}{20} = 20\text{ m}
  4. So the answer is (a) 20 m.
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