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(D) The terminal velocity $v_T$ of a sphere falling through a viscous fluid is given by Stokes' Law: $v_T = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\

Vedclass · Accepted Answer

$3.77$

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(D) The terminal velocity $v_T$ of a sphere falling through a viscous fluid is given by Stokes' Law: $v_T = \frac{2}{9} \frac{r^2 g (ho - \sigma)}{\eta}$.Given:Radius $r = 0.05 \,cm = 0.05 	imes 10^{-2} \,m = 5 	imes 10^{-4} \,m$.Density of steel $ho = 7.8 \,g/cm^3 = 7800 \,kg/m^3$.Density of water $\sigma = 1 \,g/cm^3 = 1000 \,kg/m^3$.Viscosity $\eta = 0.001 \,Pa \cdot s$.Acceleration due to gravity $g = 9.8 \,m/s^2$.Substituting the values:$v_T = \frac{2}{9} 	imes \frac{(5 	imes 10^{-4})^2 	imes 9.8 	imes (7800 - 1000)}{0.001}$.$v_T = \frac{2}{9} 	imes \frac{25 	imes 10^{-8} 	imes 9.8 	imes 6800}{0.001}$.$v_T = \frac{2}{9} 	imes \frac{25 	imes 9.8 	imes 6800 	imes 10^{-5}}{10^{-3}}$.$v_T = \frac{2}{9} 	imes 16.66 = 3.703 \,m/s \approx 3.77 \,m/s$ (using $g=10 \,m/s^2$ yields $3.77 \,m/s$ exactly).

$A$ steel ball of radius $0.05 \,cm$ and density $7.8 \,g \,cm^{-3}$ is dropped into a tank of water. The terminal velocity of the steel ball is (Density of water $= 1 \,g \,cm^{-3}$ and viscosity of water $= 0.001 \,Pa \,s$) (in $\,m/s$)

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