r a wave pulse to travel from one end of the string to the other and back. This time is essentially the period for which constructive interference will occur.

First, we find the wave speed, $$v$$, on the string. The wave speed is given by the formula:

$$ v = \sqrt{\frac{T}{\mu}} $$

where:

$$T$$ is the tension in the string, which is given as $$2.5 \mathrm{~N}$$.
$$\mu$$ is the linear mass density of the string.

The linear mass density $$\mu$$ can be calculated using the formula:

$$ \mu = \frac{m}{L} $$

Where:

$$m$$ is the mass of the string, given as $$10^{-3} \mathrm{~kg}$$.
$$L$$ is the length of the string, given as $$25 \mathrm{~cm}$$ (which we convert to meters as $$0.25 \mathrm{~m}$$).

Thus,

$$ \mu = \frac{10^{-3} \mathrm{~kg}}{0.25 \mathrm{~m}} = 4 \times 10^{-3} \mathrm{~kg/m} $$

Now, substituting $$\mu$$ and $$T$$ back into the wave speed formula:

$$ v = \sqrt{\frac{2.5 \mathrm{~N}}{4 \times 10^{-3} \mathrm{~kg/m}}} = \sqrt{625 \mathrm{~m^2/s^2}} = 25 \mathrm{~m/s} $$

Now that we have the wave speed, we can calculate the time it takes for the wave pulse to travel the length of the string and back (i.e., a round trip). The total distance for a round trip is $$2L$$, so the time interval $$\Delta t$$ is given by:

$$ \Delta t = \frac{2L}{v} $$

Substituting the given values:

$$ \Delta t = \frac{2 \times 0.25 \mathrm{~m}}{25 \mathrm{~m/s}} = \frac{0.5 \mathrm{~m}}{25 \mathrm{~m/s}} = 0.02 \mathrm{~s} = 20 \mathrm{~ms} $$

Thus, the minimum value of $$\Delta \mathrm{t}$$ for constructive interference between successive pulses is:

Option D: 20 ms

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