CBSE Class 11-science Answered

L=0.25m, m = 0.0025kg

The first overtone in the string has a frequency of n_string = 2.(1/2L square root of T/m). Note that this is twice the fundamental frequency.

The fundamental frequency of the pipe closed at one end n_pipe = v/4L

Here v = 330m/s and L = 0.4m which is the length of the closed pipe.

Find n_pipe.

The frequency of the string will be n_pipe +8 because the question says that on decreasing the tension the beat frequency decreases. If the frequency of the string was n_pipe - 8 then on decreasing the tension the beat frequency would increase.

Once you know n_pipe, L of the string, mass per unit length = 0.0025kg/0.25m then the value of T can be found by subsituting the known values.