JEE PYQ: Units & Measurements - Question ID 44c6928cdd2b (JEE Main 2021)

In dimensional analysis, we express every physical quantity in terms of a chosen set of base quantities. The usual base quantities are mass ($M$), length ($L$), and time ($T$). However, the question asks us to treat velocity ($V$), time ($T$), and force ($F$) as the new base quantities. Our goal is to express the dimension of mass in terms of these new base dimensions.

The key formula we use is Newton’s second law: $F = m \, a$ where $F$ is force, $m$ is mass, and $a$ is acceleration. Acceleration is the rate of change of velocity, so $a = \frac{dV}{dt} \quad \Longrightarrow \quad [a] = \frac{[V]}{[T]} = V\,T^{-1}.$

Step 1: Express mass in terms of force and acceleration
From $F = m\,a$, we solve for mass: $m = \frac{F}{a}.$ Therefore, the dimension of mass is $[m] = \frac{[F]}{[a]}.$

Step 2: Express acceleration in terms of the new base quantities
Acceleration is the time derivative of velocity, so its dimension is $[a] = \frac{[V]}{[T]} = V\,T^{-1}.$

Step 3: Substitute back to find the dimension of mass
Substitute $[a] = V\,T^{-1}$ into $[m] = \frac{[F]}{[a]}$: $[m] = \frac{F}{V\,T^{-1}} = F\,V^{-1}\,T.$

Step 4: Match with the given options
The derived dimension is $[m] = F\,T\,V^{-1}$, which corresponds exactly to option B.

1. Confusing base and derived quantities: Students sometimes forget that $V$, $T$, and $F$ are now the base dimensions and try to express $F$ in terms of $M$, $L$, and $T$. This leads to circular reasoning. 2. Sign errors in exponents: A frequent mistake is writing $[a] = V^{-1}T$ instead of $V\,T^{-1}$. Always remember acceleration is velocity per unit time, not the other way around. 3. Misapplying Newton’s second law: Some students write $F = m\,v$ instead of $F = m\,a$, which gives an incorrect dimension for mass.

Exam Tip: When the base quantities change, always start from the definition of the quantity you need (here, mass via $F = m\,a$) and express everything in terms of the new base dimensions.