MHTCET 18 Apr 2026 Shift 1 Paper

Concept:The time period of a spring-mass system is $T = 2\pi\sqrt{\frac{m}{K_{\text{eq}}}}$, where $K_{\text{eq}}$ depends on how the springs are combined.Explanation:For figure (a), only one spring is attached, so $K_a = K$.Thus, $T_a = 2\pi\sqrt{\frac{m}{K}}$.For figure (b), the two identical springs are connected in series.The equivalent spring constant is $K_b = \frac{K \cdot K}{K + K} = \frac{K}{2}$.So, $T_b = 2\pi\sqrt{\frac{m}{K/2}} = 2\pi\sqrt{\frac{2m}{K}} = \sqrt{2}\,T_a$.For figure (c), the two identical springs are connected in parallel.The equivalent spring constant is $K_c = K + K = 2K$.So, $T_c = 2\pi\sqrt{\frac{m}{2K}} = \frac{T_a}{\sqrt{2}}$.From these results, $T_a = \sqrt{2}\,T_c$ and $T_b = \sqrt{2}\,T_a$.Substituting $T_a$ into the relation for $T_b$:$T_b = \sqrt{2}(\sqrt{2}\,T_c) = 2T_c$.Answer:Option D: $T_b = 2T_c$