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A spherical ball of density \mathrm{\rho} and radius \mathrm{0.003 \mathrm{~m}} is dropped into a tube containing a viscous fluid filled up to the \mathrm{0 \mathrm{~cm}} mark as shown in the figure. Viscosity of the fluid \mathrm{=1.260 \mathrm{~N} \cdot \mathrm{m}^{-2}} and its density \mathrm{\rho_L=\rho / 2=1260 \mathrm{~kg} \cdot \mathrm{m}^{-3}}. Assume the ball reaches a terminal speed by the \mathrm{10 \mathrm{~cm}} mark. The time taken by the ball to traverse the distance between the \mathrm{10 \mathrm{~cm} \, \, and \, 20 \mathrm{cm}} mark is.

                                                 

Option: 1

\mathrm{500 \mu \mathrm{s}}


Option: 2

\mathrm{500 \ \mathrm{ms}}


Option: 3

\mathrm{0.5 \ \mathrm{s}}


Option: 4

\mathrm{5 \ \mathrm{s}}


Answers (1)

best_answer

\mathrm{\begin{aligned} & \mathrm{V}_{\mathrm{T}}=\frac{2}{9} \frac{\mathrm{r}^2 \mathrm{~g}}{\eta}(\rho-\sigma) \\ & =\frac{2}{9} \frac{(0.003)^2 \times 10}{1.260}(1260) \\ & \mathrm{V}_{\mathrm{T}}=0.02 \mathrm{~m} / \mathrm{sec} . \\ & \therefore \text { Time }=\frac{0.1}{0.02}=5 \mathrm{sec} . \end{aligned}}

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