JEE PYQ: Units & Measurements - Question ID 52b0fd2c755b (JEE Main 2019)

In dimensional analysis, any physical quantity can be expressed as a product of powers of the fundamental units. Here, we are asked to express the dimensional formula for linear momentum ($p$) in terms of three new fundamental units:

• Surface Tension ($S$): Force per unit length, with dimensions $[S] = [M T^{-2}]$.
• Moment of Inertia ($I$): Rotational inertia, with dimensions $[I] = [M L^2]$.
• Planck’s Constant ($h$): Quantum of action, with dimensions $[h] = [M L^2 T^{-1}]$.

Linear momentum has the standard dimensional formula: $[p] = [M L T^{-1}].$ Our goal is to express $[M]$, $[L]$, and $[T]$ in terms of $[S]$, $[I]$, and $[h]$, and then substitute them into $[p]$.

Step 1: Express $[M]$, $[L]$, and $[T]$ in terms of $[S]$, $[I]$, and $[h]$.

Assume the following dimensional relationships: $[M] = S^a I^b h^c, \quad [L] = S^d I^e h^f, \quad [T] = S^g I^h h^k.$ We will solve for the exponents $a, b, c, \dots, k$ by equating dimensions.

Step 2: Solve for $[M]$.

From $[S] = [M T^{-2}]$ and $[I] = [M L^2]$, we substitute into $[M] = S^a I^b h^c$: $[M] = (M T^{-2})^a (M L^2)^b (M L^2 T^{-1})^c.$ Equate exponents of $M$, $L$, and $T$:

• $M$: $1 = a + b + c$,
• $L$: $0 = 2b + 2c$,
• $T$: $0 = -2a - c$.

• From $L$: $b + c = 0 \implies b = -c$.
• From $T$: $2a + c = 0 \implies a = -\frac{c}{2}$.
• Substitute into $M$: $1 = -\frac{c}{2} - c + c = -\frac{c}{2} \implies c = -2$.
• Thus, $a = 1$, $b = 2$.

Step 3: Solve for $[L]$.

From $[I] = [M L^2]$, substitute $[M]$: $[L^2] = \frac{[I]}{[M]} = \frac{I}{S^1 I^2 h^{-2}} = S^{-1} I^{-1} h^2.$ Thus, $[L] = S^{-1/2} I^{-1/2} h^1.$

Step 4: Solve for $[T]$.

From $[S] = [M T^{-2}]$, substitute $[M]$: $[T^{-2}] = \frac{[S]}{[M]} = \frac{S}{S^1 I^2 h^{-2}} = S^0 I^{-2} h^2.$ Thus, $[T] = S^0 I^1 h^{-1}.$

Step 5: Substitute $[M]$, $[L]$, and $[T]$ into $[p] = [M L T^{-1}]$.

Substitute the expressions: $[p] = (S^1 I^2 h^{-2}) \cdot (S^{-1/2} I^{-1/2} h^1) \cdot (S^0 I^1 h^{-1})^{-1}.$ Simplify the exponents:

• $S$: $1 - \frac{1}{2} + 0 = \frac{1}{2}$.
• $I$: $2 - \frac{1}{2} - 1 = \frac{1}{2}$.
• $h$: $-2 + 1 + 1 = 0$.

Students often make the following mistakes:

• Incorrectly solving for $[M]$, $[L]$, or $[T]$: Forgetting to equate exponents for all three dimensions ($M$, $L$, $T$) leads to wrong exponents. Always verify each step by substituting back.
• Miscounting exponents in $[p]$: When multiplying terms like $S^a I^b h^c$, students may add exponents incorrectly. Double-check arithmetic.
• Ignoring Planck’s constant ($h$): Some students assume $h$ is dimensionless or irrelevant, leading to incorrect options like $h^0$ being missed. Always include all given fundamental units in the derivation.

Exam Tip: When faced with such problems, systematically express each standard dimension ($M$, $L$, $T$) in terms of the new fundamental units before substituting into the target quantity. This structured approach minimizes errors.