A lead shot of 1, mm diameter falls through a | vedclass

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(A) When a lead shot falls through a viscous liquid like glycerine,it experiences three forces: gravitational force $(mg)$ acting downwards,buoyant fo

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(A) When a lead shot falls through a viscous liquid like glycerine,it experiences three forces: gravitational force $(mg)$ acting downwards,buoyant force $(F_B)$ acting upwards,and viscous drag force $(F_v)$ acting upwards.The net force on the lead shot is $F_{net} = mg - F_B - F_v$.Initially,the velocity $v$ is zero,so the viscous force $F_v = 6\pi\eta rv$ is zero. The net force is maximum,causing the lead shot to accelerate.As the velocity $v$ increases,the viscous force $F_v$ also increases. Consequently,the net force $F_{net}$ decreases,which means the acceleration decreases.Eventually,the viscous force becomes large enough such that the net force becomes zero $(mg - F_B - F_v = 0)$. At this point,the lead shot attains a constant terminal velocity $(v_t)$.Therefore,the velocity increases initially and then levels off to a constant value as the distance covered increases. This behavior is correctly represented by the graph in option $A$.

$A$ lead shot of $1\, mm$ diameter falls through a long column of glycerine. The variation of its velocity $v$ with distance covered is represented by

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