A string of length $25 \mathrm{~cm}$ and mass $10^{-3} \mathrm{~kg}$ is clamped at its ends. The tension in the string is $2.5 \mathrm{~N}$. The identical wave pulses are generated at one end and at regular interval of time, $\Delta \mathrm{t}$. The minimum value of $\Delta \mathrm{t}$, so that a constructive interference takes place between successive pulses is

The linear mass density $\mu$ of the string is given by $\mu = \dfrac{M}{L} = \dfrac{10^{-3} \text{ kg}}{0.25 \text{ m}} = 4 \times 10^{-3} \text{ kg/m}$.

The speed of the wave $v$ in the string is $v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{2.5}{4 \times 10^{-3}}} = \sqrt{\dfrac{2500}{4}} = \sqrt{625} = 25 \text{ m/s}$.

The time taken for a pulse to travel from one end to the other is $t = \dfrac{L}{v} = \dfrac{0.25 \text{ m}}{25 \text{ m/s}} = 0.01 \text{ s} = 10 \text{ ms}$.

For constructive interference between successive pulses, the pulses must return to the starting point in phase. A pulse travels to the other end (10 ms), reflects (phase change of $\pi$), and returns to the starting end (another 10 ms). The total time for a pulse to return to the start is $2t = 20 \text{ ms}$.

Since the pulse reflects at the fixed end with a phase shift of $\pi$, the next pulse must be sent such that it arrives at the start at the same time as the reflected pulse to interfere constructively. The minimum time interval $\Delta t$ is equal to the time taken for the round trip, which is $20 \text{ ms}$.

Answer: 20 ms