A spherical liquid drop is placed on a horizontal plane . A small distrubance cause the volume of the drop to oscillate . The time period oscillation (T) of the liquid drop depends on radius (r) of the drop , density `(rho)` and surface tension tension (S) of the liquid. Which amount the following will be be a possible expression for T (where k is a dimensionless constant)?

Options

A. `ksqrt((rhor)/S)`
B. `ksqrt((rho^2r)/S)`
C. `ksqrt((rhor^3)/S)`
D. `ksqrt((rhor^3)/S^2)`

Answer

To solve the problem of finding the time period of oscillation (T) of a spherical liquid drop based on its radius (r), density (ρ), and surface tension (S), we can use dimensional analysis. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the dimensions of the variables
1. **Time Period (T)**: The dimension of time is given as:
\[ [T] = [M^0 L^0 T^1] \]

2. **Surface Tension (S)**: Surface tension is defined as force per unit length. The dimension of force is \( [M L T^{-2}] \), so:
\[ [S] = \frac{[M L T^{-2}]}{[L]} = [M^1 L^0 T^{-2}] \]

3. **Radius (r)**: The radius is a length, so:
\[ [r] = [M^0 L^1 T^0] \]

4. **Density (ρ)**: Density is mass per unit volume. The dimension of volume is \( [L^3] \), hence:
\[ [\rho] = \frac{[M]}{[L^3]} = [M^1 L^{-3} T^0] \]

### Step 2: Formulate the relationship
We assume that the time period \( T \) is a function of \( S \), \( r \), and \( \rho \):
\[ T \propto S^x r^y \rho^z \]
where \( x \), \( y \), and \( z \) are the powers we need to determine.

### Step 3: Write the dimensions of the right-hand side
Substituting the dimensions into the equation:
\[ [T] = [M^1 L^0 T^{-2}]^x [M^0 L^1 T^0]^y [M^1 L^{-3} T^0]^z \]
This expands to:
\[ [T] = [M^{x + z} L^{y - 3z} T^{-2x}] \]

### Step 4: Equate the dimensions
Now we equate the dimensions on both sides:
1. For mass (M):
\[ 0 = x + z \quad \text{(1)} \]

2. For length (L):
\[ 0 = y - 3z \quad \text{(2)} \]

3. For time (T):
\[ 1 = -2x \quad \text{(3)} \]

### Step 5: Solve the equations
From equation (3):
\[ x = -\frac{1}{2} \]

Substituting \( x \) into equation (1):
\[ 0 = -\frac{1}{2} + z \implies z = \frac{1}{2} \]

Substituting \( z \) into equation (2):
\[ 0 = y - 3 \left(\frac{1}{2}\right) \implies y = \frac{3}{2} \]

### Step 6: Write the expression for T
Now substituting the values of \( x \), \( y \), and \( z \) back into the equation for \( T \):
\[ T = k S^{-\frac{1}{2}} r^{\frac{3}{2}} \rho^{\frac{1}{2}} \]

### Step 7: Rearranging the expression
This can be rewritten as:
\[ T = k \sqrt{\frac{r^3 \rho}{S}} \]

### Conclusion
The possible expression for the time period \( T \) of the liquid drop is:
\[ T = k \sqrt{\frac{r^3 \rho}{S}} \]