Q116 / 200 · NEET 2023 (Manipur) · Physics

The viscous drag acting on a metal sphere of diameter $1 \mathrm{~mm}$, falling through a fluid of viscosity $0.8 \mathrm{~Pa} s$ with a velocity of $2 \mathrm{~m} \mathrm{~s}^{-1}$ is equal to

  • A
    $1.5 \times 10^{-3} \mathrm{~N}$
  • B
    $20 \times 10^{-3} \mathrm{~N}$
  • Correct answer
    C
    $15 \times 10^{-3} \mathrm{~N}$
  • D
    $30 \times 10^{-3} \mathrm{~N}$

Solution

Viscous drag acting on sphere is given by $\begin{aligned} F_V & =6 \pi \eta r V \\ & =6 \times \frac{22}{7} \times 0.8 \times \frac{10^{-3}}{2} \times 2 \\ & =15.08 \times 10^{-3} \mathrm{~N} \\ & \simeq 15 \times 10^{-3} \mathrm{~N}\end{aligned}$

Hence, the correct answer is option: 'C'.

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