Assertion (A): $\sqrt\dfrac{1}{4}+\dfrac{1}{\sqrt{0.01}}- \sqrt[3]{27^2} = \dfrac{1}{2}$

Reason (R): $\sqrt[n]{a^m}=a^\dfrac{m}{n}$, where a ≠ 0.

1. A is true, R is false.
2. A is false, R is true.
3. Both A and R are true.
4. Both A and R are false.

A is false, R is true.

Explanation

Given,

$\sqrt\dfrac{1}{4}+\dfrac{1}{\sqrt{0.01}}- \sqrt[3]{27^2} = \dfrac{1}{2}$

$\sqrt\dfrac{1}{4}+\dfrac{1}{\sqrt{0.01}}- \sqrt[3]{27^2}\\[1em] = \dfrac{1}{2} +\dfrac{1}{0.1} - \sqrt[3]{729}\\[1em] = \dfrac{1}{2} + 10 - 9\\[1em] = \dfrac{1}{2} + 1\\[1em] = \dfrac{1 + 2}{2} \\[1em] = \dfrac{3}{2}$

So, $\sqrt\dfrac{1}{4}+\dfrac{1}{\sqrt{0.01}}- \sqrt[3]{27^2} = \dfrac{1}{2}$ ≠ $\dfrac{3}{2}$

∴ Assertion (A) is false.

Given, $\sqrt[n]{a^m}=a^\dfrac{m}{n}$

According to laws of exponents, any non-zero number raised to the power of m and square root of n can be expressed as number to the power $\dfrac{m}{n}$.

$\sqrt[n]{a^m}=a^\dfrac{m}{n}$

∴ Reason (R) is true.

Hence, Assertion (A) is false, Reason (R) is true.