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Question

(A) Given: Radius of copper ball $r = 3.0 \,mm = 3 	imes 10^{-3} \,m$, viscosity of oil $\eta = 1 \,kg / ms$, density of oil $ho = 1.5 	imes 10^3

Vedclass · Accepted Answer

$15 	imes 10^{-2} \,m / s$

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(A) Given: Radius of copper ball $r = 3.0 \,mm = 3 	imes 10^{-3} \,m$, viscosity of oil $\eta = 1 \,kg / ms$, density of oil $ho = 1.5 	imes 10^3 \,kg / m^3$, density of copper $\sigma = 9 	imes 10^3 \,kg / m^3$, and acceleration due to gravity $g = 10 \,m / s^2$.The formula for terminal velocity $v_T$ is given by Stokes' Law:$v_T = \frac{2}{9} \frac{r^2(\sigma - ho)g}{\eta}$Substituting the values:$v_T = \frac{2}{9} 	imes \frac{(3 	imes 10^{-3})^2 	imes (9 	imes 10^3 - 1.5 	imes 10^3) 	imes 10}{1}$$v_T = \frac{2}{9} 	imes (9 	imes 10^{-6}) 	imes (7.5 	imes 10^3) 	imes 10$$v_T = 2 	imes 10^{-6} 	imes 7.5 	imes 10^4$$v_T = 15 	imes 10^{-2} \,m / s$Thus, the terminal velocity is $15 	imes 10^{-2} \,m / s$.

$A$ copper ball of radius $3.0 \,mm$ falls in an oil tank of viscosity $1 \,kg / ms$. Then, the terminal velocity of the copper ball will be (Density of oil $= 1.5 \times 10^3 \,kg / m^3$, Density of copper $= 9 \times 10^3 \,kg / m^3$ and $g = 10 \,m / s^2$.)

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