Dimensionsal methods provide three major advantages in verification , deviation , and changing the system of units . Any empirical formula that is derived based on this method has to be verified and propportionality constants found by experimental means . The presence or absence of certain factors - non dimensional constants or variables - cannot be identified by this method . So every dimensionally correct relation cannot be taken as perfectly correct.
The time period of oscillation of a drop depends on surface tension `sigma` , density of the liquid `rho` , and radius ` r ` . The relation is

Options

A. a. ` sqrt (( rho r^(2))/( sigma))`
B. b. ` sqrt (( r^(2))/( rho sigma))`
C. c. ` sqrt (( r^(3) rho)/( sigma))`
D. d. ` sqrt (( rho sigma)/( r^(3)))`

Explanation

To find the relation for the time period of oscillation of a drop based on surface tension (σ), density (ρ), and radius (r), we can use dimensional analysis. Here’s a step-by-step solution:

### Step 1: Identify the dimensions of each variable
- **Time period (T)**: The dimension is [T] or \( T^1 \).
- **Surface tension (σ)**: The dimension is \( [σ] = \frac{F}{L} = \frac{M L T^{-2}}{L} = M T^{-2} \) or \( [σ] = M^1 L^0 T^{-2} \).
- **Density (ρ)**: The dimension is \( [ρ] = \frac{M}{L^3} = M^1 L^{-3} \).
- **Radius (r)**: The dimension is \( [r] = L^1 \).

### Step 2: Set up the dimensional equation
Assume the time period \( T \) is proportional to the variables \( σ \), \( ρ \), and \( r \) as follows:
\[ T \propto σ^a ρ^b r^c \]
This implies:
\[ [T] = [σ]^a [ρ]^b [r]^c \]

### Step 3: Substitute the dimensions into the equation
Substituting the dimensions we identified:
\[ T^1 = (M^1 T^{-2})^a (M^1 L^{-3})^b (L^1)^c \]

This expands to:
\[ T^1 = M^{a+b} L^{-3b+c} T^{-2a} \]

### Step 4: Equate the dimensions
Now, we equate the powers of M, L, and T from both sides:
1. For mass (M): \( 0 = a + b \) (Equation 1)
2. For length (L): \( 0 = -3b + c \) (Equation 2)
3. For time (T): \( 1 = -2a \) (Equation 3)

### Step 5: Solve the equations
From Equation 3:
\[ a = -\frac{1}{2} \]

Substituting \( a \) into Equation 1:
\[ 0 = -\frac{1}{2} + b \]
\[ b = \frac{1}{2} \]

Now substituting \( b \) into Equation 2:
\[ 0 = -3\left(\frac{1}{2}\right) + c \]
\[ c = \frac{3}{2} \]

### Step 6: Write the final relation
Now that we have the values of \( a \), \( b \), and \( c \):
- \( a = -\frac{1}{2} \)
- \( b = \frac{1}{2} \)
- \( c = \frac{3}{2} \)

Thus, the relation becomes:
\[ T = k \cdot σ^{-\frac{1}{2}} \cdot ρ^{\frac{1}{2}} \cdot r^{\frac{3}{2}} \]

Where \( k \) is a dimensionless constant.

### Final Expression
The final expression for the time period \( T \) is:
\[ T = k \sqrt{\frac{r^3 ρ}{σ}} \]