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NEET PHYSICSEasy

The terminal velocity of a copper ball of radius 5 mm5 \text{ mm} falling through a tank of oil at room temperature is 10 cm s−110 \text{ cm s}^{-1}. If the viscosity of oil at room temperature is 0.9 kg m−1 s−10.9 \text{ kg m}^{-1} \text{ s}^{-1}, the viscous drag force is:

A

8.48×10−3 N8.48 \times 10^{-3} \text{ N}

B

8.48×10−5 N8.48 \times 10^{-5} \text{ N}

C

4.23×10−3 N4.23 \times 10^{-3} \text{ N}

D

4.23×10−6 N4.23 \times 10^{-6} \text{ N}

Step-by-Step Solution

According to Stokes' Law, the viscous drag force (FF) acting on a spherical body of radius rr moving with terminal velocity vv in a fluid of viscosity η\eta is given by the formula: F=6πηrvF = 6\pi \eta r v

Given values: Viscosity, η=0.9 kg m−1 s−1\eta = 0.9 \text{ kg m}^{-1} \text{ s}^{-1} Radius, r=5 mm=5×10−3 mr = 5 \text{ mm} = 5 \times 10^{-3} \text{ m}

  • Velocity, v=10 cm s−1=0.1 m s−1v = 10 \text{ cm s}^{-1} = 0.1 \text{ m s}^{-1}

Calculation: Substitute the values into the formula: F=6×3.14159×0.9×(5×10−3)×0.1F = 6 \times 3.14159 \times 0.9 \times (5 \times 10^{-3}) \times 0.1 F=6×0.9×0.5×10−3×3.14159F = 6 \times 0.9 \times 0.5 \times 10^{-3} \times 3.14159 F=5.4×5×10−4×3.14159F = 5.4 \times 5 \times 10^{-4} \times 3.14159 F=2.7×10−3×3.14159F = 2.7 \times 10^{-3} \times 3.14159 F≈8.48×10−3 NF \approx 8.48 \times 10^{-3} \text{ N}

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