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(r^(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. `sqrt(3)`
C. `3//2`
D. `2//3`

Answer

To solve the problem, we start with the given formula for the time period \( t \) of the oscillation of a drop of liquid:

\[ t = \sqrt{r^{2b} s^c d^{\frac{a}{2}}} \]

We know that the time period \( t \) is directly proportional to \( \sqrt{\frac{d}{s}} \). This means we can express this relationship as:

\[ t \propto \sqrt{\frac{d}{s}} \]

This implies that:

\[ t = k \sqrt{\frac{d}{s}} \quad \text{for some constant } k \]

### Step 1: Equate the two expressions for \( t \)

From the given formula, we can rewrite it as:

\[ t = \sqrt{r^{2b}} \cdot \sqrt{s^c} \cdot \sqrt{d^{\frac{a}{2}}} \]

This can be expressed as:

\[ t = r^b s^{\frac{c}{2}} d^{\frac{a}{4}} \]

### Step 2: Compare dimensions

We know that the dimensions of \( t \) are \( [T] \) (time). Therefore, we need to equate the dimensions from both expressions.

1. From \( t = r^b s^{\frac{c}{2}} d^{\frac{a}{4}} \):
- Dimensions of \( r \) (radius) = \( [L] \)
- Dimensions of \( s \) (surface tension) = \( [M][T]^{-2} \)
- Dimensions of \( d \) (density) = \( [M][L]^{-3} \)

Thus, we can write:

\[ [T] = [L]^b \cdot [M]^{\frac{c}{2}} \cdot [T]^{-c} \cdot [M]^{\frac{a}{4}} \cdot [L]^{-\frac{3a}{4}} \]

### Step 3: Combine dimensions

Combining the dimensions gives:

\[ [T] = [L]^b \cdot [M]^{\frac{c}{2} + \frac{a}{4}} \cdot [T]^{-c - 2} \]

### Step 4: Set up equations for dimensions

Now we can set up the equations based on the dimensions:

1. For \( [T] \):
\[ -c - 2 = 1 \quad \Rightarrow \quad c = -3 \]

2. For \( [L] \):
\[ b - \frac{3a}{4} = 0 \quad \Rightarrow \quad b = \frac{3a}{4} \]

3. For \( [M] \):
\[ \frac{c}{2} + \frac{a}{4} = 0 \quad \Rightarrow \quad \frac{-3}{2} + \frac{a}{4} = 0 \quad \Rightarrow \quad a = 6 \]

### Step 5: Substitute \( a \) back to find \( b \)

Now substituting \( a = 6 \) into the equation for \( b \):

\[ b = \frac{3 \cdot 6}{4} = \frac{18}{4} = \frac{9}{2} \]

### Conclusion

Thus, the value of \( b \) is \( \frac{9}{2} \).