The question presents an assertion and a reason concerning the safety features of an aircraft in the context of lightning strikes. Let's analyze both statements:
Now, let's evaluate the statements:
Thus, both statements are correct and the reason provided in (R) is indeed the correct explanation for (A). The outer metal body protects the passengers from lightning by utilizing the property that the electric field inside the conductor (the aircraft body) is zero, thanks to charge redistribution on the conductor's surface.
The correct answer is: Both (A) and (R) are correct and (R) is the correct explanation of (A).
Step 1: Understanding Assertion (A)
The outer body of an aircraft is made of metal, which provides a conductive path for electric charge to flow. In case of lightning strikes, the metal body of the aircraft channels the electrical current around the passengers, preventing it from entering the interior. This is why passengers are safe inside a metal-bodied aircraft during lightning strikes.
Therefore, assertion (A) is correct.
Step 2: Understanding Reason (R)
The electric field inside a conductor in electrostatic equilibrium is zero. This is known as the principle of electrostatic shielding. When an external electric field is applied, the free charges in the conductor rearrange themselves to cancel the electric field inside the conductor.
Therefore, reason (R) is also correct.
Step 3: Connecting Assertion and Reason
The metal body of the aircraft behaves as a conductor. According to the principle of electrostatic shielding, the electric field inside the conducting body (the aircraft) is zero, which protects the passengers from the effects of lightning strikes.
Thus, reason (R) explains why assertion (A) is true.



Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
An organic compound (X) with molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ is not readily oxidised. On reduction it gives $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}(\mathrm{Y})\right.$ which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis give 2,3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are: