A string of length 1 m and mass $2\times {10}^{-5}\text{ kg}$ is under tension $T$ . When the string vibrates, two successive harmonics are found to occur at frequencies 750 Hz and 1000 Hz. The value of tension T ______ is Newton.

The correct answer is 5.

Step 1: Understand the given values
Length of the string, $L=1\text{ m}$
Mass of the string, $m=2\times {10}^{-5}\text{ kg}$
Frequencies of two successive harmonics:
$f_n=750\text{ Hz}$
$f_{n+1}=1000\text{ Hz}$

Step 2: Find the fundamental frequency
The difference between the frequencies of two consecutive harmonics on a string fixed at both ends is equal to the fundamental frequency ($f_1$):

$f_1=f_{n+1}-f_n$

$f_1=1000-750=250\text{ Hz}$

Step 3: Calculate the linear mass density ($\mu$)
Linear mass density is mass per unit length:

$\mu =\frac{m}{L}=\frac{2\times {10}^{-5}\text{ kg}}{1\text{ m}}$

Step 4: Use the formula for fundamental frequency to calculate Tension ($T$)
The fundamental frequency of a vibrating string is given by:

$f_1=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

Substitute the known values into the equation:

$250=\frac{1}{2\times 1}\sqrt{\frac{T}{2\times {10}^{-5}}}$

$250=\frac{1}{2}\sqrt{\frac{T}{2\times {10}^{-5}}}$

$500=\sqrt{\frac{T}{2\times {10}^{-5}}}$

Squaring both sides:

${\left(500\right)}^2=\frac{T}{2\times {10}^{-5}}$

$250000=\frac{T}{2\times {10}^{-5}}$

$T=250000\times 2\times {10}^{-5}$

$T=5\text{ N}$

Thus, the value of tension T is 5 Newton.