WBJEE 2026 Physics and Chemistry Paper

Concept:A sphere falling in a viscous liquid experiences increasing viscous drag, so its acceleration gradually reduces and it reaches a constant terminal velocity.Explanation:The forces acting are weight $mg$ and viscous drag $6\pi\eta r v$.The equation of motion is:$$m\frac{dv}{dt} = mg - 6\pi\eta r v$$Here, $v$ is the instantaneous velocity and the drag force increases as $v$ increases.Solving this differential equation gives:$$v = v_t \left(1 - e^{-t/\tau}\right)$$where $v_t$ is the terminal velocity and $\tau$ is a time constant.At $t=0$, $v=0$.As time increases, $v$ initially rises quickly, then its slope decreases and it approaches $v_t$ asymptotically.Thus, the $v-t$ graph starts from the origin, rises with a decreasing slope, and finally becomes horizontal at the terminal velocity.Answer:The correct graph is the one that starts at the origin, rises with a decreasing slope, and becomes horizontal as it reaches the terminal velocity.