A liquid drop of density `rho` , radius `r`, and surface tension `sigma` oscillates with time period `T` . Which of the following expressions for `T^(2)` is correct?

Options

A. `( rho r^(3))/( sigma)`
B. `( rho sigma)/( r^(3))`
C. `( r^(3) sigma)/( rho)`
D. None of these

Explanation

To solve the problem of determining the correct expression for \( T^2 \) in terms of the given quantities (density \( \rho \), radius \( r \), and surface tension \( \sigma \)), we will use dimensional analysis. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the quantities and their dimensions
We have three quantities:
- Density \( \rho \) has dimensions \( [\rho] = M L^{-3} \)
- Radius \( r \) has dimensions \( [r] = L \)
- Surface tension \( \sigma \) has dimensions \( [\sigma] = M T^{-2} \)

### Step 2: Set up the relationship for the time period \( T \)
We assume that the time period \( T \) is related to these quantities by the equation:
\[ T = k \cdot \rho^a \cdot r^b \cdot \sigma^c \]
where \( k \) is a dimensionless constant and \( a \), \( b \), and \( c \) are the powers we need to determine.

### Step 3: Write the dimensions of \( T \)
The dimension of time \( T \) is:
\[ [T] = T \]

### Step 4: Substitute the dimensions into the equation
Substituting the dimensions of \( \rho \), \( r \), and \( \sigma \) into the equation gives:
\[ [T] = (M L^{-3})^a \cdot (L)^b \cdot (M T^{-2})^c \]
This can be expanded to:
\[ [T] = M^{a+c} \cdot L^{-3a+b} \cdot T^{-2c} \]

### Step 5: Equate the dimensions
Now we equate the dimensions on both sides:
- For mass \( M \): \( a + c = 0 \)
- For length \( L \): \( -3a + b = 0 \)
- For time \( T \): \( -2c = 1 \)

### Step 6: Solve the equations
From the third equation \( -2c = 1 \):
\[ c = -\frac{1}{2} \]

Substituting \( c \) into the first equation:
\[ a - \frac{1}{2} = 0 \implies a = \frac{1}{2} \]

Substituting \( a \) into the second equation:
\[ -3 \left(\frac{1}{2}\right) + b = 0 \implies b = \frac{3}{2} \]

### Step 7: Write the expression for \( T^2 \)
Now we have:
- \( a = \frac{1}{2} \)
- \( b = \frac{3}{2} \)
- \( c = -\frac{1}{2} \)

Substituting these values back into the equation for \( T \):
\[ T = k \cdot \rho^{1/2} \cdot r^{3/2} \cdot \sigma^{-1/2} \]

To find \( T^2 \):
\[ T^2 = k^2 \cdot \rho \cdot r^3 \cdot \sigma^{-1} \]

### Final Expression
Thus, the expression for \( T^2 \) is:
\[ T^2 = k^2 \cdot \frac{\rho \cdot r^3}{\sigma} \]

### Conclusion
The correct expression for \( T^2 \) is proportional to \( \frac{\rho r^3}{\sigma} \).