JEE PYQ: Units & Measurements - Question ID 11102bfa0a1e (JEE Main 2020)

In dimensional analysis, we express physical quantities in terms of fundamental units. Young’s modulus ($Y$) is defined as the ratio of stress to strain:

However, since strain ($\Delta L / L$) is dimensionless, Young’s modulus has the same dimensions as stress, i.e., force per unit area:

But in this problem, we are asked to express $[Y]$ in terms of the new fundamental units: speed ($V$), area ($A$), and force ($F$). To do this, we must express all other dimensions (like length, mass, and time) in terms of $F$, $A$, and $V$.

Step 1: Express fundamental dimensions in terms of $F$, $A$, and $V$

We know the standard dimensions of force, area, and speed:

• $[F] = MLT^{-2}$
• $[A] = L^2$
• $[V] = LT^{-1}$

We need to express $M$, $L$, and $T$ in terms of $F$, $A$, and $V$.

Step 2: Solve for $L$ in terms of $A$

From $[A] = L^2$, we get:

Step 3: Solve for $T$ in terms of $V$ and $A$

From $[V] = LT^{-1}$ and $L = A^{1/2}$, we substitute:

Step 4: Solve for $M$ in terms of $F$, $A$, and $V$

From $[F] = MLT^{-2}$ and substituting $L = A^{1/2}$ and $T = A^{1/2} V^{-1}$:

Step 5: Express Young’s modulus in terms of $F$, $A$, and $V$

Young’s modulus has dimensions $[Y] = \frac{[F]}{[A]} = F A^{-1}$ in standard units. However, we must verify if any additional dimensions of $V$ appear when expressing $Y$ in the new system.

Since $Y$ is force per unit area, and we have already expressed all fundamental dimensions in terms of $F$, $A$, and $V$, we substitute:

Notice that $V$ does not appear in the expression because $Y$ does not inherently depend on speed. Thus, the dimension of $V$ in the expression for $Y$ is $V^0$ (i.e., dimensionless).

Final Dimensional Formula:

This matches Option A.

Trap 1: Overcomplicating the derivation. Many students try to express Young’s modulus using its full definition ($Y = \frac{F L}{A \Delta L}$) and unnecessarily involve strain. Since strain is dimensionless, it is sufficient to consider $Y$ as force per unit area.

Trap 2: Incorrectly assuming $V$ must appear in the final expression. Since $Y$ does not depend on speed, $V$ should have an exponent of $0$. Students often force $V$ into the expression by incorrectly manipulating dimensions.

Trap 3: Miscounting exponents. When substituting $L = A^{1/2}$ and $T = A^{1/2} V^{-1}$, students may make algebraic errors in solving for $M$. Double-check each step to avoid sign or power mistakes.

Exam Tip: Always verify that the derived dimensions reduce to the standard form when $F$, $A$, and $V$ are expressed in terms of $M$, $L$, and $T$. For example:

This matches the standard dimensions of Young’s modulus, confirming the correctness of the answer.