A spherical ball of radius 1 times 10^{-4} ,m | vedclass

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(C) The terminal velocity $V_T$ of a spherical ball in a viscous liquid is given by Stokes' Law: $V_T = \frac{2}{9} \frac{R^2 g (\rho_B - \rho_L)}{\et

Vedclass · Accepted Answer

$2518$

Vedclass · Answer

(C) The terminal velocity $V_T$ of a spherical ball in a viscous liquid is given by Stokes' Law: $V_T = \frac{2}{9} \frac{R^2 g (ho_B - ho_L)}{\eta}$.Given: $R = 10^{-4} \,m$, $ho_B = 10^5 \,kg/m^3$, $ho_L = 10^3 \,kg/m^3$, $\eta = 9.8 	imes 10^{-6} \,N s/m^2$, and $g = 9.8 \,m/s^2$.Substituting the values: $V_T = \frac{2}{9} 	imes \frac{(10^{-4})^2 	imes 9.8 	imes (10^5 - 10^3)}{9.8 	imes 10^{-6}}$.$V_T = \frac{2}{9} 	imes \frac{10^{-8} 	imes 9.8 	imes 99000}{9.8 	imes 10^{-6}} = \frac{2}{9} 	imes 10^{-2} 	imes 99000 = 220 \,m/s$.Since the velocity remains constant upon entering the water, the velocity after falling height $h$ must be equal to the terminal velocity: $V = \sqrt{2gh} = V_T$.$h = \frac{V_T^2}{2g} = \frac{(220)^2}{2 	imes 9.8} = \frac{48400}{19.6} \approx 2469 \,m$. Given the options, the closest value is $2518 \,m$.

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