JEE Main 2026 — Units and Dimensions Question with Solution

The dimensional formula of the universal gravitational constant $G$ is obtained from Newton's law of gravitation $F=\frac{G{m}_1{m}_2}{r^2}$.

$[G]=\frac{[F][r^2]}{[m_1{m}_2]}=\frac{[ML{T}^{-2}][L^2]}{[M^2]}=[M^{-1}{L}^3{T}^{-2}]$

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

$[h]=\frac{[E]}{[\nu ]}=\frac{[M{L}^2{T}^{-2}]}{[T^{-1}]}=[M{L}^2{T}^{-1}]$

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

Substituting the dimensions, we get:

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

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

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 $[h{T}^{-1}L{M}^{-2}]$.

Answer: $[h{T}^{-1}L{M}^{-2}]$