A metal sphere of diameter \(7 cm\) falls through aliquid of coefficient of viscosity \(0.8 Pa.s\). When itsvelocity is \(20 cm.s^{-1}\) the viscous force acting on itis nearly ----\(N.\)

(A) \(0.11\)

(B) \(0.22\)

(C) \(0.3\)

(D) \(0.01\)

(A) \(0.11\)

(B) \(0.22\)

(C) \(0.3\)

(D) \(0.01\)

Answer: A

Viscous force on the metal sphere = \(6 \pi \eta r V\)

\(= 6 \times \times 0.8 \times \dfrac{3.5}{100} \times \dfrac{4}{3} \pi \left ( \dfrac{3.5}{100} \right )^{3}\times \dfrac{20}{100}N\)

\(= 0.1055 \approx 0.11 N\)

Viscous force on the metal sphere = \(6 \pi \eta r V\)

\(= 6 \times \times 0.8 \times \dfrac{3.5}{100} \times \dfrac{4}{3} \pi \left ( \dfrac{3.5}{100} \right )^{3}\times \dfrac{20}{100}N\)

\(= 0.1055 \approx 0.11 N\)

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