MHTCET 11 Apr 2026 Shift 1 Paper

Concept:A pulse reflected twice from fixed ends returns with the same phase, so constructive interference needs one full round trip of the string.Explanation:The linear mass density of the string is$$\mu = \frac{m}{L}$$Substitute $m = 10^{-3}\ \mathrm{kg}$ and $L = 0.5\ \mathrm{m}$:$$\mu = \frac{10^{-3}}{0.5} = 2 \times 10^{-3}\ \mathrm{kg\ m^{-1}}$$The speed of a wave pulse on the string is$$v = \sqrt{\frac{T}{\mu}}$$With $T = 0.8\ \mathrm{N}$:$$v = \sqrt{\frac{0.8}{2 \times 10^{-3}}} = \sqrt{400} = 20\ \mathrm{m\ s^{-1}}$$At a fixed end, the pulse inverts after reflection. To regain the same phase, the pulse must undergo two reflections, covering a distance $2L$.Thus, the minimum time interval between successive pulses is$$\Delta t_{\min} = \frac{2L}{v}$$Substitute $L = 0.5\ \mathrm{m}$ and $v = 20\ \mathrm{m\ s^{-1}}$:$$\Delta t_{\min} = \frac{2(0.5)}{20} = \frac{1}{20} = 0.05\ \mathrm{s}$$Answer:The minimum value of $\Delta t$ is $0.05\ \mathrm{s}$, which corresponds to option D.