If the time period t of the oscillation of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula `t=sqrt(t^(2b)s^(c)d^(a//2))`. It is observed that the time period is directly proportional to `sqrt((d)/(s)).` The value of b should therefore be:

Options

A. `(3)/(4)`
B. `sqrt3`
C. `(3)/(2)`
D. `(2)/(3)`

Answer

To solve the problem, we need to analyze the given formula for the time period \( t \) of oscillation of a drop of liquid and determine the value of \( b \) based on the proportionality condition provided.

### Step-by-Step Solution:

1. **Understand the Given Formula**:
The time period \( t \) is given by the formula:
\[ t = \sqrt{r^{2b} s^{c} d^{\frac{a}{2}}} \]
where \( r \) is the radius, \( s \) is the surface tension, and \( d \) is the density of the liquid.

2. **Identify the Proportionality Condition**:
It is stated that the time period \( t \) is directly proportional to \( \sqrt{\frac{d}{s}} \):
\[ t \propto \sqrt{\frac{d}{s}} \]
This implies that:
\[ t = k \sqrt{\frac{d}{s}} \]
for some constant \( k \).

3. **Express the Proportionality in Terms of Dimensions**:
From the proportionality, we can express:
\[ t = k \sqrt{d} \cdot s^{-\frac{1}{2}} \]

4. **Equate the Two Expressions for Time Period**:
Now, we can equate the two expressions for \( t \):
\[ \sqrt{r^{2b} s^{c} d^{\frac{a}{2}}} = k \sqrt{d} \cdot s^{-\frac{1}{2}} \]

5. **Square Both Sides**:
Squaring both sides gives:
\[ r^{2b} s^{c} d^{\frac{a}{2}} = k^2 d \cdot s^{-1} \]

6. **Rearranging the Equation**:
Rearranging the equation, we have:
\[ r^{2b} s^{c + 1} d^{\frac{a}{2} - 1} = k^2 \]
This indicates that the left-hand side must be dimensionless.

7. **Analyzing Dimensions**:
We know the dimensions of each term:
- Density \( d \): \( [M L^{-3}] \)
- Surface tension \( s \): \( [M T^{-2} L^{-1}] \)
- Radius \( r \): \( [L] \)

We can express the dimensions of \( t \) as:
\[ [T] = [L^{b} M^{0} T^{0}] [M^{1} T^{-2} L^{-1}]^{c} [M^{1} L^{-3}]^{\frac{a}{2}} \]

8. **Setting Up the Dimension Equation**:
From the equation, we separate the dimensions of mass \( M \), length \( L \), and time \( T \):
- For \( M \): \( 0 + c + \frac{a}{2} = 0 \)
- For \( L \): \( b - c - \frac{3a}{2} = 0 \)
- For \( T \): \( -2c = 1 \)

9. **Solving for \( c \)**:
From \( -2c = 1 \), we find:
\[ c = -\frac{1}{2} \]

10. **Substituting \( c \) into the Mass Equation**:
Substituting \( c \) into the mass equation:
\[ 0 - \frac{1}{2} + \frac{a}{2} = 0 \implies a = 1 \]

11. **Substituting \( a \) into the Length Equation**:
Now substituting \( a = 1 \) into the length equation:
\[ b + \frac{1}{2} - \frac{3}{2} = 0 \implies b = 1 \]

### Final Answer:
The value of \( b \) is \( 1 \).