MHTCET 20 Apr 2026 Shift 2 Paper

Concept:When a metal sphere falls through a viscous liquid, its velocity increases exponentially and approaches a constant terminal velocity, so the $v$–$t$ graph starts steep and gradually flattens.Explanation:Three forces act on the sphere: weight $mg$ downward, buoyant force $F_B = \frac{4}{3}\pi r^3 \sigma g$ upward, and viscous drag $F_v = 6\pi \eta r v$ upward.The equation of motion is:$$m\frac{dv}{dt} = mg - \frac{4}{3}\pi r^3 \sigma g - 6\pi \eta r v$$At $t = 0$, the velocity $v = 0$, so the viscous drag is zero. Hence, the net downward force is maximum and the initial acceleration is maximum.As the sphere speeds up, the viscous drag $6\pi \eta r v$ increases linearly with $v$, reducing the net downward force. Therefore, the acceleration decreases continuously, and the slope of the $v$–$t$ graph keeps decreasing.When the drag force equals the effective weight, the net force becomes zero and the sphere attains a constant terminal velocity $v_T$.The velocity variation is given by:$$v = v_T\left(1 - e^{-kt/m}\right), \quad k = 6\pi \eta r$$This curve starts from the origin, rises rapidly, and then gradually levels off toward $v_T$. Among the given graphs, graph (d) shows exactly this behaviour.Answer:Option C: graph (d).