If the time period (T) of vibration of a liquid drop depends on surface tension (S), radius (r) of the drop and density (ρ) of the liquid, then the expression of T is

To find the correct relation for the time period of vibration of a liquid drop using dimensional analysis, we can express the time period as a product of the given physical quantities raised to unknown powers.

Let the relationship be:
$T=K\cdot S^a\cdot r^b\cdot {\rho }^c$
where:

1. $T$ is the time period of vibration.
2. $S$ is the surface tension of the liquid.
3. $r$ is the radius of the drop.
4. $\rho$ is the density of the liquid.
5. $K$ is a dimensionless constant.
6. $a$, $b$, and $c$ are exponents to be determined.

Let us write the dimensional formulas for each of these physical quantities:
- Time period ($T$): $\left[T\right]=\left[M^0{L}^0{T}^1\right]$
- Surface tension ($S$, force per unit length): $\left[S\right]=\frac{\left[\text{Force}\right]}{\left[\text{Length}\right]}=\frac{\left[ML{T}^{-2}\right]}{\left[L\right]}=\left[M^1{L}^0{T}^{-2}\right]$
- Radius ($r$): $\left[r\right]=\left[M^0{L}^1{T}^0\right]$
- Density ($\rho$, mass per unit volume): $\left[\rho \right]=\left[M^1{L}^{-3}{T}^0\right]$

Now, substituting the dimensional formulas into our initial equation:
$\left[M^0{L}^0{T}^1\right]={\left[M^1{L}^0{T}^{-2}\right]}^a\cdot {\left[M^0{L}^1{T}^0\right]}^b\cdot {\left[M^1{L}^{-3}{T}^0\right]}^c$
Combining the exponents on the right-hand side:
$\left[M^0{L}^0{T}^1\right]=\left[M^{a+c}{L}^{b-3c}{T}^{-2a}\right]$

By comparing the powers of $M$, $L$, and $T$ on both sides, we get a system of linear equations:

1. For $T$:
$-2a=1\Rightarrow a=-\frac{1}{2}$

2. For $M$:
$a+c=0\Rightarrow c=-a=\frac{1}{2}$

3. For $L$:
$b-3c=0\Rightarrow b=3c=3\cdot \left(\frac{1}{2}\right)=\frac{3}{2}$

Substituting the values of $a$, $b$, and $c$ back into our initial equation:
$T=K\cdot S^{-1/2}\cdot r^{3/2}\cdot {\rho }^{1/2}$
Simplifying the fractional powers under a square root:
$T=K\sqrt{\frac{\rho r^3}{S}}$

Thus, the correct expression is indeed $T=K\sqrt{\frac{\rho r^3}{S}}$.