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A spherical oil drop falls at a steady rate of \( 2 cm / s \) in still air. Find diameter of the drop. [Take \( g =980 cm / s ^{2} \). The coefficient of viscosity of air \( 4.8 \times 10^{-4} \) poise, Density of oil \( =0.8 glcm ^{3} \), Density of air \( =1 g / cm ^{3} J \) |
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Answer» V = 2 cm/s v = 0.02m/s \(\eta\) = 4.8 x 10-4 Density of oil = 0.8gl cm3 Density of air = 18/cm3 J We know that r2 = \(\frac{9\eta v}{29(p-\sigma)}\) r2 = \(\frac{9\times4.8\times10^{-5}\times0.02}{2\times9.8(1000-800)}\) r2 = \(\frac{0.864\times10^{-5}}{3920}\) r2 = 0.00022 x 10-5 r2 = 2.2 x 10-9 r = \(\sqrt{2.2\times10^{-9}}\) m hence diameter of spherical drop d = 2 x \(\sqrt{2.2\times10^{-9}}\) m |
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