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 $$\sqrt{\frac{d}{s}}$$. The value of $$b$$ should therefore be :

The given formula for the time period of oscillation is $$ t = \sqrt{r^{2b} s^c d^{a/2}} $$. We can simplify this expression by distributing the square root. Since $$ \sqrt{x^y} = x^{y/2} $$, we rewrite the formula as:

$$ t = \left( r^{2b} s^c d^{a/2} \right)^{1/2} = r^{2b \cdot \frac{1}{2}} \cdot s^{c \cdot \frac{1}{2}} \cdot d^{\frac{a}{2} \cdot \frac{1}{2}} = r^b \cdot s^{c/2} \cdot d^{a/4} $$

The problem states that the time period $$ t $$ is directly proportional to $$ \sqrt{\frac{d}{s}} $$. Expressing this proportionality mathematically:

$$ t \propto \sqrt{\frac{d}{s}} = \left( \frac{d}{s} \right)^{1/2} = d^{1/2} \cdot s^{-1/2} $$

This proportionality implies that when the radius $$ r $$ is held constant (as the observation does not involve varying $$ r $$), the exponents of $$ d $$ and $$ s $$ in the expression for $$ t $$ must match those in the proportionality expression. Therefore, comparing the exponents in $$ t = r^b \cdot s^{c/2} \cdot d^{a/4} $$ and $$ d^{1/2} \cdot s^{-1/2} $$, we set up the following equations:

For the exponent of $$ d $$:

$$ \frac{a}{4} = \frac{1}{2} $$

Solving for $$ a $$:

$$ a = 4 \times \frac{1}{2} = 2 $$

For the exponent of $$ s $$:

$$ \frac{c}{2} = -\frac{1}{2} $$

Solving for $$ c $$:

$$ c = 2 \times \left(-\frac{1}{2}\right) = -1 $$

Substituting $$ a = 2 $$ and $$ c = -1 $$ back into the expression for $$ t $$:

$$ t = r^b \cdot s^{-1/2} \cdot d^{2/4} = r^b \cdot s^{-1/2} \cdot d^{1/2} $$

Now, we must find $$ b $$. Since $$ t $$ has the dimension of time ($$ T $$), we use dimensional analysis. The dimensions of the variables are:

• $$ [t] = T $$
• $$ [r] = L $$ (length)
• $$ [s] = $$ surface tension = force per unit length. Force has dimensions $$ M L T^{-2} $$, so $$ [s] = \frac{M L T^{-2}}{L} = M T^{-2} $$
• $$ [d] = $$ density = mass per unit volume = $$ M L^{-3} $$

The dimensions of the right-hand side are:

$$ [r^b] = L^b $$

$$ [s^{-1/2}] = \left( M T^{-2} \right)^{-1/2} = M^{-1/2} T^{1} $$

$$ [d^{1/2}] = \left( M L^{-3} \right)^{1/2} = M^{1/2} L^{-3/2} $$

Multiplying these together:

$$ [r^b] \cdot [s^{-1/2}] \cdot [d^{1/2}] = L^b \cdot M^{-1/2} T^{1} \cdot M^{1/2} L^{-3/2} = L^{b - \frac{3}{2}} M^{-1/2 + 1/2} T^{1} = L^{b - \frac{3}{2}} M^{0} T^{1} $$

This must equal the dimension of $$ t $$, which is $$ T $$. Therefore, the exponent of $$ L $$ must be zero (since there is no length dimension in time), and the exponent of $$ T $$ is already 1, which matches. Setting the exponent of $$ L $$ to zero:

$$ b - \frac{3}{2} = 0 $$

Solving for $$ b $$:

$$ b = \frac{3}{2} $$

Thus, the value of $$ b $$ is $$ \frac{3}{2} $$. Comparing with the options:

A. $$ \frac{3}{4} $$

B. $$ \sqrt{3} $$

C. $$ \frac{3}{2} $$

D. $$ \frac{2}{3} $$

Hence, the correct answer is Option C.