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(B) According to Stokes' Law, the viscous force $F$ acting on a spherical object of radius $r$ moving with a terminal velocity $v$ in a fluid of visco

Vedclass · Accepted Answer

$6.6 	imes 10^{-8} \ N$

Vedclass · Answer

(B) According to Stokes' Law, the viscous force $F$ acting on a spherical object of radius $r$ moving with a terminal velocity $v$ in a fluid of viscosity $\eta$ is given by $F = 6 \pi \eta r v$.Given:Diameter $d = 1 \ mm = 10^{-3} \ m$, so radius $r = 0.5 	imes 10^{-3} \ m$.Terminal velocity $v = 0.7 \ ms^{-1}$.Coefficient of viscosity $\eta = 2 	imes 10^{-5} \ Pa \cdot s$.Substituting these values into the formula:$F = 6 	imes 3.14 	imes (2 	imes 10^{-5}) 	imes (0.5 	imes 10^{-3}) 	imes 0.7$$F = 6 	imes 3.14 	imes 10^{-5} 	imes 0.5 	imes 10^{-3} 	imes 0.7$$F = 6 	imes 0.5 	imes 0.7 	imes 3.14 	imes 10^{-8}$$F = 2.1 	imes 3.14 	imes 10^{-8}$$F = 6.594 	imes 10^{-8} \ N \approx 6.6 	imes 10^{-8} \ N$.Thus, the correct option is $B$.

$A$ rain drop of diameter $1 \ mm$ falls with a terminal velocity of $0.7 \ ms^{-1}$ in air. If the coefficient of viscosity of air is $2 \times 10^{-5} \ Pa \cdot s$, the viscous force on the rain drop is

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