The time period T of an oscillating liquid drop is assumed to...

Let $T = k S^\alpha \rho^\beta R^\gamma$, where $k$ is a dimensionless constant.

Writing the dimensions of each quantity:
$[T] = [M^0 L^0 T^1]$
$[S] = [M^1 L^0 T^{-2}]$
$[\rho] = [M^1 L^{-3} T^0]$
$[R] = [M^0 L^1 T^0]$

Substituting these into the equation:
$[M^0 L^0 T^1] = [M T^{-2}]^\alpha [M L^{-3}]^\beta [L]^\gamma$
$[M^0 L^0 T^1] = [M^{\alpha+\beta} L^{-3\beta+\gamma} T^{-2\alpha}]$

Equating the powers of $M, L$, and $T$ on both sides:
For $T$: $-2\alpha = 1 \Rightarrow \alpha = -\dfrac{1}{2}$
For $M$: $\alpha + \beta = 0 \Rightarrow \beta = -\alpha = \dfrac{1}{2}$
For $L$: $-3\beta + \gamma = 0 \Rightarrow \gamma = 3\beta = \dfrac{3}{2}$

Therefore, the relation is $T \propto S^{-1/2} \rho^{1/2} R^{3/2}$.

The maximum percentage error in $T$ is given by:
$\dfrac{\Delta T}{T} \times 100 = \left|-\dfrac{1}{2}\right| \left(\dfrac{\Delta S}{S} \times 100\right) + \left|\dfrac{1}{2}\right| \left(\dfrac{\Delta \rho}{\rho} \times 100\right) + \left|\dfrac{3}{2}\right| \left(\dfrac{\Delta R}{R} \times 100\right)$

Substituting the given percentage errors:
$\% \text{ error in } T = \dfrac{1}{2}(3\%) + \dfrac{1}{2}(1\%) + \dfrac{3}{2}(2\%)$
$= 1.5\% + 0.5\% + 3\% = 5\%$.

Answer: $5\%$