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  • A sphere is dropped under gravity through a fluid of viscosity . If the average acceleration i

A sphere is dropped under gravity through a fluid of viscosity \mathrm{\eta}. If the average acceleration is half of the initial acceleration, the time to attain the terminal velocity is \mathrm{( \rho= density \, \, of\, \, sphere, r= radius)}

Option: 1

\mathrm{\frac{4 \rho r^2}{9 \eta}}


Option: 2

\mathrm{\frac{9 \rho r^2}{4 \eta}}


Option: 3

\mathrm{\frac{4 \rho r}{9 \eta}}


Option: 4

\mathrm{\frac{9 \rho r}{4 \eta}}


Answers (1)

best_answer

\mathrm{\begin{aligned} & a_{\text {avg }} \quad=\frac{v_T-0}{T} \\\\ & a_{\text {air }}=\frac{m g-f_B}{m}=g-\frac{F_B}{m} \\\\ & =g-\frac{v \rho_L g}{v \rho_S g}=g-\frac{\rho_L}{\rho_S} g \\\\ & a_{\text {air }}=\frac{g\left(\rho_s-\rho_L\right)}{\rho_s}=\frac{2 v_T}{T} \\\\ & a_{\text {air }}=\frac{g\left(\rho_s-\rho_L\right)}{\rho_s} \\\\ & =2 \cdot \frac{2}{9} \frac{r^2}{\eta} \cdot \frac{9}{T}\left(\rho_s-\rho_r\right) \\\\ & \mathrm{T}=\frac{4}{9} \frac{\rho r^2}{\eta} \\\\ & \end{aligned}}

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