JEE PYQ: Units & Measurements - Question ID f05785abe0b2 (JEE Main 2020)

Core Formula & Concept:

In dimensional analysis, every physical quantity can be expressed in terms of the fundamental dimensions: mass ($M$), length ($L$), time ($T$), electric current ($A$), thermodynamic temperature ($K$), amount of substance ($mol$), and luminous intensity ($cd$). The given problem involves the dimensional formula of a derived quantity $x$ defined as: $x = \frac{IFv^2}{WL^4}$ where:

• $I$ = Moment of inertia (dimensions: $ML^2$)
• $F$ = Force (dimensions: $MLT^{-2}$)
• $v$ = Velocity (dimensions: $LT^{-1}$)
• $W$ = Work (dimensions: $ML^2T^{-2}$)
• $L$ = Length (dimensions: $L$)

Our goal is to find the dimensional formula of $x$ and match it with one of the given options.

Step-by-Step Derivation:

Step 1: Write the dimensional formula for each quantity in $x$
We express each term in the expression for $x$ in terms of $M$, $L$, and $T$:

• $[I] = ML^2$
• $[F] = MLT^{-2}$
• $[v] = LT^{-1} \Rightarrow [v^2] = L^2T^{-2}$
• $[W] = ML^2T^{-2}$
• $[L] = L \Rightarrow [L^4] = L^4$

Step 2: Substitute the dimensions into the expression for $x$
The expression for $x$ is: $x = \frac{IFv^2}{WL^4}$ Substituting the dimensions: $[x] = \frac{[I][F][v^2]}{[W][L^4]} = \frac{(ML^2)(MLT^{-2})(L^2T^{-2})}{(ML^2T^{-2})(L^4)}$ Step 3: Simplify the dimensional expression
Multiply the dimensions in the numerator and denominator:

• Numerator: $(ML^2)(MLT^{-2})(L^2T^{-2}) = M^{1+1}L^{2+1+2}T^{-2-2} = M^2L^5T^{-4}$
• Denominator: $(ML^2T^{-2})(L^4) = ML^{2+4}T^{-2} = ML^6T^{-2}$

Now, divide the numerator by the denominator: $[x] = \frac{M^2L^5T^{-4}}{ML^6T^{-2}} = M^{2-1}L^{5-6}T^{-4+2} = ML^{-1}T^{-2}$ Step 4: Compare with the dimensional formulas of the given options
We now compare $[x] = ML^{-1}T^{-2}$ with the dimensional formulas of the options:

• A: Coefficient of viscosity
  The coefficient of viscosity ($\eta$) has dimensions $ML^{-1}T^{-1}$. This does not match $[x]$.
• B: Force constant
  The force constant ($k$) has dimensions $MT^{-2}$ (since $F = kx \Rightarrow [k] = \frac{[F]}{[x]} = \frac{MLT^{-2}}{L} = MT^{-2}$). This does not match $[x]$.
• C: Energy density
  Energy density is energy per unit volume. Energy has dimensions $ML^2T^{-2}$, and volume has dimensions $L^3$. Thus, energy density has dimensions: $\frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}$ This matches $[x]$.
• D: Planck's constant
  Planck's constant ($h$) has dimensions $ML^2T^{-1}$. This does not match $[x]$.

Conclusion: The dimensional formula of $x$ matches that of energy density, which corresponds to option C.

Common Traps & Exam Tip:

1. Incorrect simplification of dimensions: Students often make mistakes in adding or subtracting exponents when multiplying or dividing dimensions. For example, they might incorrectly simplify $L^2 \cdot L^4$ as $L^6$ (correct) but then forget to subtract exponents when dividing, leading to errors like $L^{2-4} = L^{-2}$ instead of $L^{5-6} = L^{-1}$ in this problem.
2. Confusing force constant with energy density: The force constant has dimensions $MT^{-2}$, while energy density has $ML^{-1}T^{-2}$. Students might overlook the $L^{-1}$ term and incorrectly select option B.
3. Miscounting exponents in $v^2$: Some students forget that $v^2$ has dimensions $L^2T^{-2}$ and mistakenly write $LT^{-2}$, leading to an incorrect final dimension for $x$.
4. Not verifying all options: Students might stop after finding a partial match (e.g., matching $MT^{-2}$ with force constant) without checking the full dimensional formula. Always verify all options before selecting the answer.

Exam Tip: When solving dimensional analysis problems, always:

• Write down the dimensions of each quantity clearly.
• Simplify step-by-step, keeping track of exponents.
• Double-check the final dimensions against all options, even if one seems to match early.