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NEET PHYSICSEasy

The viscous drag acting on a metal sphere of diameter 1 mm1 \text{ mm}, falling through a fluid of viscosity 0.8 Pa-s0.8 \text{ Pa-s} with a velocity of 2 m s−12 \text{ m s}^{-1} is nearly equal to:

A

15×10−3 N15 \times 10^{-3} \text{ N}

B

30×10−3 N30 \times 10^{-3} \text{ N}

C

1.5×10−3 N1.5 \times 10^{-3} \text{ N}

D

20×10−3 N20 \times 10^{-3} \text{ N}

Step-by-Step Solution

According to Stokes' Law, the viscous drag force (FF) acting on a spherical body moving through a viscous fluid is given by: F=6πηrvF = 6\pi \eta r v

Given: Viscosity, η=0.8 Pa-s\eta = 0.8 \text{ Pa-s} (or kg m−1s−1\text{kg m}^{-1} \text{s}^{-1}) Velocity, v=2 m s−1v = 2 \text{ m s}^{-1}

  • Diameter, d=1 mm⇒d = 1 \text{ mm} \Rightarrow Radius, r=0.5 mm=0.5×10−3 mr = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m}

Calculation: F=6×3.14×0.8×(0.5×10−3)×2F = 6 \times 3.14 \times 0.8 \times (0.5 \times 10^{-3}) \times 2 F=6×3.14×0.8×10−3F = 6 \times 3.14 \times 0.8 \times 10^{-3} F=15.072×10−3 NF = 15.072 \times 10^{-3} \text{ N}

Rounding to the nearest integer, the force is approximately 15×10−3 N15 \times 10^{-3} \text{ N}.

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