JEE Advanced 2023 Paper 2, Physics Section 3 Q3: Stationary Waves in Air Column

For a string fixed at both ends with linear mass density $\mu$, the $p$-th harmonic has frequency

$f_p=\frac{p}{2L}\sqrt{\frac{T}{\mu }}$.

Two successive harmonics, $p$ and $p+1$, satisfy

$\frac{f_{p+1}}{f_p}=\frac{p+1}{p}=\frac{1000}{750}=\frac{4}{3}\Rightarrow p=3$.

Therefore from $f_3=750$ Hz with $L=1$ m,

$750=\frac{3}{2}\sqrt{\frac{T}{\mu }}\Rightarrow \sqrt{\frac{T}{\mu }}=500$.

With $\mu =2\times 10^{-5}$ kg m${}^{-1}$,

$T=500^2\cdot 2\times 10^{-5}=25\times 10^4\cdot 2\times 10^{-5}=5\text{N}$.