$\text { Match the LIST-I with LIST-II }$
| List - I |
List - II |
||
|---|---|---|---|
| A. | $ \text { Gravitational constant } $ |
I. | $ \left[\mathrm{LT}^{-2}\right] $ |
| B. | $ \text { Gravitational potential energy } $ |
II. | $ \left[\mathrm{L}^2 \mathrm{~T}^{-2}\right] $ |
| C. | $ \text { Gravitational potential } $ |
III. | $ \left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right] $ |
| D. | $ \text { Acceleration due to gravity } $ |
IV. | $ \left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right] $ |
Solution
<p>Here’s how each quantity lines up with its dimensional formula:</p>
<p><p>Gravitational constant, $G$ </p>
<p>• From Newton’s law $F = G\frac{m_1m_2}{r^2}$ </p>
<p>• $[G] = \frac{[F]\,[r]^2}{[m]^2} = \frac{(MLT^{-2})\;L^2}{M^2} = M^{-1}L^3T^{-2}$ </p>
<p>⇒ IV</p></p>
<p><p>Gravitational potential energy, $U$ </p>
<p>• As work or $U = mgh$ </p>
<p>• $[U] = [F]\,[h] = (MLT^{-2})\;L = ML^2T^{-2}$ </p>
<p>⇒ III</p></p>
<p><p>Gravitational potential, $\phi$ </p>
<p>• Energy per unit mass: $\phi = \frac{U}{m}$ </p>
<p>• $[\phi] = \frac{ML^2T^{-2}}{M} = L^2T^{-2}$ </p>
<p>⇒ II</p></p>
<p><p>Acceleration due to gravity, $g$ </p>
<p>• Just an acceleration </p>
<p>• $[g] = LT^{-2}$ </p>
<p>⇒ I</p></p>
<p>Matching them gives </p>
<p>A–IV, B–III, C–II, D–I </p>
<p>That corresponds to <strong>Option A</strong>.</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
This question is part of PrepWiser's free JEE Main question bank. 129 more solved questions on Gravitation are available — start with the harder ones if your accuracy is >70%.