Correct Answer - Option 4 :

\(\frac{1}{2}\sqrt {\frac{T}{{ml}}} \)
__Concept:__

The frequency of oscillation of a string stretched under tension T is given by

\(f =\frac{1}{2l}\sqrt {\frac{T}{μ}}{}\)

Where f = frequency of oscillation, n = number of loops, l = Length of string, T =Tension, and μ= linear density

**Calculation:**

Given, mass of string m

length of string l

linear mass density

\(\mu = \frac{m}{l}\)

\(\implies f =\frac{1}{2l}\sqrt{\frac{T}{(m/l)}}{}\)

\(\implies f =\frac{1}{2l}\sqrt{\frac{Tl}{(m)}}{}\)

\(\implies f =\frac{1}{2}\sqrt{\frac{Tl}{(ml^2)}}{}\)

Writing l = √ (l)^{2}

\(\implies f =\frac{1}{2}\sqrt{\frac{T}{(ml)}}{}\)

So, the correct option is \(\frac{1}{2}\sqrt {\frac{T}{{ml}}} \)