Class NEET 2026 ALL Q #1911
COMPETENCY BASED
APPLY
4 Marks 2026 NTA-RE-NEET-2026 MCQ SINGLE
Two planets $P_{1}$ and $P_{2}$ with equal mass have radii $R_{1}$ and $R_{2}$, respectively, where $R_{2}=\frac{R_{1}}{2}$. The escape speeds of $P_{1}$ and $P_{2}$ are $v_{1}$ and $v_{2},$ respectively. Then $\frac{v_{2}}{v_{1}}$ is:
(A) 2
(B) $\frac{1}{\sqrt{2}}$
(C) 1
(D) $\sqrt{2}$
Correct Answer: D

AI Tutor Explanation

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Detailed Solution

Step 1: Recall the formula for escape velocity

The escape velocity $v$ of a planet with mass $M$ and radius $R$ is given by the formula: $$v = \sqrt{\frac{2GM}{R}}$$ where $G$ is the universal gravitational constant.

Step 2: Express escape velocities for both planets

Given that both planets have equal mass $M$, we can write: $$v_{1} = \sqrt{\frac{2GM}{R_{1}}}$$ $$v_{2} = \sqrt{\frac{2GM}{R_{2}}}$$

Step 3: Calculate the ratio

To find the ratio $\frac{v_{2}}{v_{1}}$, we divide the two expressions: $$\frac{v_{2}}{v_{1}} = \frac{\sqrt{\frac{2GM}{R_{2}}}}{\sqrt{\frac{2GM}{R_{1}}}} = \sqrt{\frac{R_{1}}{R_{2}}}$$

Step 4: Substitute the given relationship

We are given $R_{2} = \frac{R_{1}}{2}$, which implies $\frac{R_{1}}{R_{2}} = 2$. Substituting this into our ratio: $$\frac{v_{2}}{v_{1}} = \sqrt{2}$$

Final Answer: \sqrt{2}

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must utilize the standard formula for escape velocity and apply algebraic manipulation to compare two distinct physical systems.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific sequence of steps involving formula substitution and ratio calculation to arrive at the result.
Syllabus Audit: In the context of NEET, this is classified as COMPETENCY. It tests the student's ability to handle inverse relationships between physical variables, a core skill for competitive physics examinations.