JEE Main 2022 · 28 Jun · Shift 2 · Physics Q37

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PhysicsSingle correct+4 / -1Question 37 of 90

A water drop of radius 1 $\mu$ m falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is $1.8\times10^{-5}$ Nsm $^{-2}$ and its density is negligible as compared to that of water $10^6$ gm $^{-3}$ . Terminal velocity of the water drop is :

(Take acceleration due to gravity $=10$ ms $^{-2}$ )

145.4\times10^{-6}

^{-1}

118.0\times10^{-6}

^{-1}

132.6\times10^{-6}

^{-1}

123.4\times10^{-6}

^{-1}

Answer and worked solution

Official answer: D

Stokes. $v=\dfrac{2r^2\rho g}{9\eta}=\dfrac{2\times10^{-12}\times10^3\times10}{9\times1.8\times10^{-5}}\approx123.4\times10^{-6}$ m s $^{-1}$ (density of water $10^3$ kg m $^{-3}$ ).

(D)

Topics:Terminal velocity Viscosity and Stokes' law

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JEE Main 2022 · 28 Jun · Shift 2, Physics question 37