The frequency ($$\nu$$) of an oscillating liquid drop may depend upon radius ($$r$$) of the drop, density ($$\rho$$) of liquid and the surface tension ($$s$$) of the liquid as: $$\nu = r^a \rho^b s^c$$. The values of $$a$$, $$b$$ and $$c$$ respectively are

Using dimensional analysis, the frequency $$\nu = r^a \rho^b s^c$$.

Dimensions of each quantity:

$$[\nu] = T^{-1}$$

$$[r] = L$$

$$[\rho] = ML^{-3}$$

$$[s] = MT^{-2}$$ (surface tension = force per unit length)

Writing the dimensional equation:

$$T^{-1} = L^a \cdot (ML^{-3})^b \cdot (MT^{-2})^c$$

$$T^{-1} = M^{b+c} \cdot L^{a-3b} \cdot T^{-2c}$$

Comparing powers of each dimension:

For $$M$$: $$0 = b + c$$ ... (i)

For $$L$$: $$0 = a - 3b$$ ... (ii)

For $$T$$: $$-1 = -2c$$ ... (iii)

From (iii): $$c = \frac{1}{2}$$

From (i): $$b = -c = -\frac{1}{2}$$

From (ii): $$a = 3b = 3 \times \left(-\frac{1}{2}\right) = -\frac{3}{2}$$

Therefore, $$(a, b, c) = \left(-\frac{3}{2}, -\frac{1}{2}, \frac{1}{2}\right)$$

The correct answer is Option 1.