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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) \(0.01\)

Answer

Answer: A
Viscous force on the metal sphere = \(6 \pi \eta r V\)
\(= 6 \times \times 0.8 \times \dfrac{3.5}{100} \times \dfrac{4}{3} \pi \left ( \dfrac{3.5}{100} \right )^{3}\times \dfrac{20}{100}N\)
\(= 0.1055 \approx 0.11 N\)
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