Two solid spheres of radii 2 mm and 4 mm ar | vedclass

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(B) Let the radii of the spheres be $r_A = 2 \ mm = 2 	imes 10^{-3} \ m$ and $r_B = 4 \ mm = 4 	imes 10^{-3} \ m$. The density of the spheres is $

Vedclass · Accepted Answer

$4 \ cm \cdot s^{-1}$

Vedclass · Answer

(B) Let the radii of the spheres be $r_A = 2 \ mm = 2 	imes 10^{-3} \ m$ and $r_B = 4 \ mm = 4 	imes 10^{-3} \ m$. The density of the spheres is $ho_s = 2800 \ kg \cdot m^{-3}$ and the density of the liquid is $ho_f = 1.3 	imes 1000 = 1300 \ kg \cdot m^{-3}$. The coefficient of viscosity is $\eta = 1 \ Pa \cdot s$.At terminal velocity $v$,the net force on the system is zero. The forces acting on the system are the total weight acting downwards and the total buoyant force and total viscous drag acting upwards.Total weight $W = (m_A + m_B)g = \frac{4}{3} \pi (r_A^3 + r_B^3) ho_s g$.Total buoyant force $F_B = \frac{4}{3} \pi (r_A^3 + r_B^3) ho_f g$.Total viscous drag $F_v = 6 \pi \eta r_A v + 6 \pi \eta r_B v = 6 \pi \eta v (r_A + r_B)$.Equating forces: $W = F_B + F_v \Rightarrow \frac{4}{3} \pi (r_A^3 + r_B^3) g (ho_s - ho_f) = 6 \pi \eta v (r_A + r_B)$.Substituting values: $\frac{4}{3} \pi (8 + 64) 	imes 10^{-9} 	imes 10 	imes (2800 - 1300) = 6 \pi 	imes 1 	imes v 	imes (2 + 4) 	imes 10^{-3}$.$\frac{4}{3} 	imes 72 	imes 10^{-8} 	imes 1500 = 6 	imes 6 	imes 10^{-3} 	imes v$.$96 	imes 10^{-5} 	imes 1500 = 36 	imes 10^{-3} 	imes v \Rightarrow 1.44 = 0.036 	imes v$.$v = \frac{1.44}{0.036} = 40 	imes 10^{-3} \ m \cdot s^{-1} = 0.04 \ m \cdot s^{-1} = 4 \ cm \cdot s^{-1}$.

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