JEE PYQ: Units & Measurements - Question ID 4c5472dca7e2 (JEE Main 2026)

Core Formula & Concept:

The question asks for the dimensional formula of the universal gravitational constant $G$ in terms of Planck’s constant $h$, distance $L$, mass $M$, and time $T$. To solve this, we rely on two fundamental concepts:

1. Dimensional Homogeneity: Every physical equation must be dimensionally consistent. This means the dimensions on both sides of an equation must match.
2. Newton’s Law of Universal Gravitation: The gravitational force $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by: $F = G \frac{m_1 m_2}{r^2}$ where $G$ is the universal gravitational constant.
3. Planck’s Constant $h$: Planck’s constant relates the energy $E$ of a photon to its frequency $\nu$ via $E = h \nu$. Its dimensions are: $[h] = [E] \cdot [T] = [M L^2 T^{-2}] \cdot [T] = [M L^2 T^{-1}]$

Our goal is to express $[G]$ in terms of $[h]$, $[L]$, $[M]$, and $[T]$.

Step-by-Step Derivation:

Step 1: Express $[G]$ from Newton’s Law

From Newton’s law: $F = G \frac{m_1 m_2}{r^2}$ Rearranging for $G$: $G = \frac{F r^2}{m_1 m_2}$ Taking dimensions: $[G] = \frac{[F] [L]^2}{[M]^2}$ We know that force $F$ has dimensions $[M L T^{-2}]$. Substituting: $[G] = \frac{[M L T^{-2}] [L]^2}{[M]^2} = \frac{[M L^3 T^{-2}]}{[M]^2} = [M^{-1} L^3 T^{-2}]$ So, the dimensions of $G$ are $[M^{-1} L^3 T^{-2}]$.

Step 2: Express $[G]$ in terms of $[h]$, $[L]$, $[M]$, and $[T]$

We know: $[h] = [M L^2 T^{-1}]$ We need to express $[M^{-1} L^3 T^{-2}]$ using $[h]$, $[L]$, $[M]$, and $[T]$. Let’s assume: $[G] = [h]^a [L]^b [M]^c [T]^d$ Substitute $[h]$: $[M^{-1} L^3 T^{-2}] = [M L^2 T^{-1}]^a [L]^b [M]^c [T]^d$ Expanding the right side: $= [M^a L^{2a} T^{-a}] [L]^b [M]^c [T]^d = [M^{a+c} L^{2a+b} T^{-a+d}]$ Now, equate the exponents of $M$, $L$, and $T$ on both sides:

• For mass $M$: $a + c = -1$
• For length $L$: $2a + b = 3$
• For time $T$: $-a + d = -2$

Step 3: Solve the system of equations

We have three equations:

1. $a + c = -1$
2. $2a + b = 3$
3. $-a + d = -2$

We have four variables ($a, b, c, d$) but only three equations. However, we can express $b$, $c$, and $d$ in terms of $a$:

• From (1): $c = -1 - a$
• From (2): $b = 3 - 2a$
• From (3): $d = -2 + a$

We need to choose $a$ such that the expression for $[G]$ matches one of the given options. Let’s try $a = 1$:

• $c = -1 - 1 = -2$
• $b = 3 - 2(1) = 1$
• $d = -2 + 1 = -1$

Substituting back: $[G] = [h]^1 [L]^1 [M]^{-2} [T]^{-1} = [h L M^{-2} T^{-1}]$ This matches option B: $[h T^{-1} L M^{-2}]$.

Step 4: Verification

Let’s verify by substituting $[h] = [M L^2 T^{-1}]$ into option B: $[h T^{-1} L M^{-2}] = [M L^2 T^{-1}] \cdot [T^{-1}] \cdot [L] \cdot [M^{-2}] = [M^{1-2} L^{2+1} T^{-1-1}] = [M^{-1} L^3 T^{-2}]$ This matches the known dimensions of $G$. Hence, option B is correct.

Common Traps & Exam Tip:

Students often make the following mistakes:

1. Incorrectly expressing $[h]$: Some confuse $h$ with reduced Planck’s constant $\hbar = h / 2\pi$, leading to wrong dimensions. Always use $[h] = [M L^2 T^{-1}]$.
2. Miscounting exponents: When substituting $[h]$ into the expression, students may miscount the exponents of $M$, $L$, or $T$. Double-check each step.
3. Assuming $[G]$ directly from options: Some try to reverse-engineer the answer by testing options without deriving $[G]$ first. This can lead to errors if the options are close. Always derive the dimensions systematically.
4. Ignoring negative exponents: Students may overlook the negative exponents in $[G] = [M^{-1} L^3 T^{-2}]$, leading to incorrect substitutions.

Exam Tip: When expressing one physical constant in terms of others, always:

1. Start with the known dimensions of the target constant (here, $[G] = [M^{-1} L^3 T^{-2}]$).
2. Express the given constants (here, $[h]$) in terms of $M$, $L$, and $T$.
3. Set up an equation and solve for the exponents systematically.
4. Verify by substituting back into the expression.