JEE Main 2026 — Physics Mechanics

2026mcqmedium

Dimensions of universal gravitational constant $(G)$ in terms of Planck's constant $(h)$ , distance $(L)$ , mass $(M)$ and time $(T)$ are _______.

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

A
$[hTLM^{-2}]$
B
$[hT^{-1}LM^{-2}]$
C
$[hTL^2 M^{-2}]$
D
$[h^{-1}T^{-1}LM^{-2}]$

Did you get this right?

Solution

The dimensional formula of the universal gravitational constant $G$ is obtained from Newton's law of gravitation $F = \dfrac{G m_1 m_2}{r^2}$ .

$[G] = \dfrac{[F][r^2]}{[m_1 m_2]} = \dfrac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$

The dimensional formula of Planck's constant $h$ is obtained from the energy of a photon $E = h\nu$ .

$[h] = \dfrac{[E]}{[\nu]} = \dfrac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$

Let the dimensions of $G$ be expressed as $[G] = [h]^a [L]^b [M]^c [T]^d$ .

Substituting the dimensions, we get:

$[M^{-1}L^3T^{-2}] = [ML^2T^{-1}]^a [L]^b [M]^c [T]^d$

$[M^{-1}L^3T^{-2}] = [M^{a+c} L^{2a+b} T^{-a+d}]$

Equating the powers of $M$ , $L$ , and $T$ on both sides:

$a + c = -1$

$2a + b = 3$

$-a + d = -2$

Assuming $a = 1$ (as per the given options), we get:

$c = -1 - 1 = -2$

$b = 3 - 2(1) = 1$

$d = -2 + 1 = -1$

Therefore, the dimensions of $G$ are $[h^1 L^1 M^{-2} T^{-1}]$ , which can be written as $[hT^{-1}LM^{-2}]$ .

Answer: $[hT^{-1}LM^{-2}]$

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JEE Main 2026 — Physics Mechanics

2026mcqmedium

Dimensions of universal gravitational constant $(G)$ in terms of Planck's constant $(h)$ , distance $(L)$ , mass $(M)$ and time $(T)$ are _______.

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

A
$[hTLM^{-2}]$
B
$[hT^{-1}LM^{-2}]$
C
$[hTL^2 M^{-2}]$
D
$[h^{-1}T^{-1}LM^{-2}]$

Did you get this right?

Solution

The dimensional formula of the universal gravitational constant $G$ is obtained from Newton's law of gravitation $F = \dfrac{G m_1 m_2}{r^2}$ .

$[G] = \dfrac{[F][r^2]}{[m_1 m_2]} = \dfrac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$

The dimensional formula of Planck's constant $h$ is obtained from the energy of a photon $E = h\nu$ .

$[h] = \dfrac{[E]}{[\nu]} = \dfrac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$

Let the dimensions of $G$ be expressed as $[G] = [h]^a [L]^b [M]^c [T]^d$ .

Substituting the dimensions, we get:

$[M^{-1}L^3T^{-2}] = [ML^2T^{-1}]^a [L]^b [M]^c [T]^d$

$[M^{-1}L^3T^{-2}] = [M^{a+c} L^{2a+b} T^{-a+d}]$

Equating the powers of $M$ , $L$ , and $T$ on both sides:

$a + c = -1$

$2a + b = 3$

$-a + d = -2$

Assuming $a = 1$ (as per the given options), we get:

$c = -1 - 1 = -2$

$b = 3 - 2(1) = 1$

$d = -2 + 1 = -1$

Therefore, the dimensions of $G$ are $[h^1 L^1 M^{-2} T^{-1}]$ , which can be written as $[hT^{-1}LM^{-2}]$ .

Answer: $[hT^{-1}LM^{-2}]$

Your note

JEE Main Mechanics in other years

Mechanics 2025 Mechanics 2024 Mechanics 2023 Mechanics 2022 Mechanics 2021 Mechanics 2020 Mechanics 2019 Mechanics 2018 Mechanics 2017 Mechanics 2016 Mechanics 2015 Mechanics 2014

Keep practicing Mechanics

Work through every JEE Main Mechanics PYQ, year by year.

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JEE Main 2026 — Physics Mechanics

Options

Did you get this right?

Solution

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JEE Main Mechanics in other years

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Electromagnetism

Modern Physics

Optics

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Waves & Oscillations

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JEE Main 2026 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics