KCET 2026 Physics Solved Paper

Concept:The passenger's weight equals the net gravitational pull of Earth and Moon, which changes with position as the spaceship travels at constant velocity.Explanation:Let the Earth–Moon distance be $d$ and the spaceship's distance from Earth's centre be $r$, so its distance from the Moon's centre is $d - r$.Since the spaceship moves at constant velocity, $r$ increases uniformly with time, so the graph mirrors the force variation with $r$.The net gravitational force on the passenger is:$$F_{net} = \frac{GM_e M}{r^2} - \frac{GM_m M}{(d-r)^2}$$At Earth's surface ($t = 0$), the Moon's pull is negligible, so $F_{net} = Mg$.As the ship moves away from Earth, Earth's pull weakens while the Moon's pull strengthens, so the net force decreases.At the neutral point the two pulls cancel, giving $F_{net} = 0$.Since $M_e \approx 81M_m$, the neutral point lies much closer to the Moon ($r \approx 0.9d$), so the zero occurs near $t = t_0$, not early in the journey.After the neutral point, the Moon's pull dominates and the force magnitude rises; at the Moon's surface, $|F_{net}| = Mg'$.Also, $g' < g$, so the final weight $Mg'$ is lower than the starting weight $Mg$.Thus the correct curve starts at $Mg$, falls to zero near the Moon, and then rises to $Mg'$.This behaviour matches curve C.Answer:C