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 regarding the time period \( T \) of a liquid drop with 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: Establish the relationship
We start by assuming that the square of the time period \( T^2 \) is proportional to the powers of the density \( \rho \), radius \( r \), and surface tension \( \sigma \):
\[ T^2 \propto \rho^a r^b \sigma^c \]
where \( a \), \( b \), and \( c \) are the powers we need to determine.

### Step 2: Write the dimensional formulas
Next, we write the dimensional formulas for each quantity:
- Density \( \rho \): \([M L^{-3}]\)
- Radius \( r \): \([L]\)
- Surface tension \( \sigma \): \([M T^{-2}]\)

### Step 3: Substitute the dimensional formulas
Now, substituting these into our proportionality expression gives:
\[ [T^2] = k \cdot [\rho^a] \cdot [r^b] \cdot [\sigma^c] \]
This translates to:
\[ [T^2] = k \cdot (M L^{-3})^a \cdot (L)^b \cdot (M T^{-2})^c \]
Simplifying this, we have:
\[ [T^2] = k \cdot M^{a+c} \cdot L^{-3a + b} \cdot T^{-2c} \]

### Step 4: Equate dimensions
Since \( T^2 \) has the dimension \([T^2]\), we equate the dimensions:
1. For mass \( M \): \( a + c = 0 \)
2. For length \( L \): \( -3a + b = 0 \)
3. For time \( T \): \( -2c = 2 \)

### Step 5: Solve the equations
From the third equation, we find:
\[ c = -1 \]
Substituting \( c = -1 \) into the first equation:
\[ a - 1 = 0 \implies a = 1 \]
Now substituting \( a = 1 \) into the second equation:
\[ -3(1) + b = 0 \implies b = 3 \]

### Step 6: Write the final expression
Now that we have \( a = 1 \), \( b = 3 \), and \( c = -1 \), we can substitute these back into our original proportionality:
\[ T^2 = k \cdot \rho^1 \cdot r^3 \cdot \sigma^{-1} \]
Assuming \( k = 1 \) (a dimensionless constant), we get:
\[ T^2 = \frac{\rho r^3}{\sigma} \]

### Step 7: Identify the correct option
Thus, the correct expression for \( T^2 \) is:
\[ T^2 = \frac{\rho r^3}{\sigma} \]

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

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