A metal block of base area 0.2; m^{2} is conn | vedclass

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(B) Given: Area $A = 0.2\; m^{2}$,mass $m = 0.02\; kg$,thickness $l = 0.6\; mm = 0.6 	imes 10^{-3}\; m$,velocity $v = 0.17\; m/s$,$g = 10\; m/s^{2}$.

Vedclass · Accepted Answer

$3.45 	imes 10^{-3} \; Pa \cdot s$

Vedclass · Answer

(B) Given: Area $A = 0.2\; m^{2}$,mass $m = 0.02\; kg$,thickness $l = 0.6\; mm = 0.6 	imes 10^{-3}\; m$,velocity $v = 0.17\; m/s$,$g = 10\; m/s^{2}$.The block moves with a constant speed,so the net force on it is zero. The tension $T$ in the string is equal to the weight of the hanging mass:$T = m \cdot g = 0.02\; kg 	imes 10\; m/s^{2} = 0.2\; N$.This tension is balanced by the viscous force $F$ acting on the block:$F = \eta A \frac{v}{l} \implies T = \eta A \frac{v}{l}$.Rearranging for the coefficient of viscosity $\eta$:$\eta = \frac{T \cdot l}{A \cdot v} = \frac{0.2 	imes 0.6 	imes 10^{-3}}{0.2 	imes 0.17}$.$\eta = \frac{0.6 	imes 10^{-3}}{0.17} \approx 3.53 	imes 10^{-3} \; Pa \cdot s$.Comparing with the given options,the closest value is $3.45 	imes 10^{-3} \; Pa \cdot s$.

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