Write dimensions of all quantities

• Time period: $[T]$
• Density: $[\rho] = M L^{-3}$
• Radius: $[r] = L$
• Surface tension: $[S] = \frac{\text{force}}{\text{length}} = \frac{M L T^{-2}}{L} = M T^{-2}$

Check the dimensional formula of the given expression in Assertion (A)

Given, $T = K\sqrt{\frac{\rho r^3}{S^{3/2}}}$ where $K$ is dimensionless.

First find dimensions inside the square root:

• $[\rho r^3] = (M L^{-3})(L^3) = M$
• $[S^{3/2}] = (M T^{-2})^{3/2} = M^{3/2} T^{-3}$

Therefore, $\left[\frac{\rho r^3}{S^{3/2}}\right] = \frac{M}{M^{3/2}T^{-3}} = M^{-1/2}T^3$

Now taking square root, $\left[\sqrt{\frac{\rho r^3}{S^{3/2}}}\right] = M^{-1/4} T^{3/2}$

This is not equal to $[T]$.

Hence the expression is not dimensionally correct.

So, Assertion (A) is false.

Check the Reason (R)

Reason says: using dimensional analysis, the RHS has different dimensions than that of time period.

From the above calculation, $\text{RHS dimension} = M^{-1/4}T^{3/2} \neq T$

So Reason (R) is true.

Conclusion

• Assertion (A) is false
• Reason (R) is true

Therefore, the correct option is: $\boxed{\text{D}}$