JEE PYQ: Units & Measurements - Question ID 9224071c3980 (JEE Main 2021)

In the chapter Units & Measurements, we learn how to express any physical quantity in terms of the fundamental dimensions: mass ($M$), length ($L$), time ($T$), electric current ($I$), thermodynamic temperature ($\Theta$), amount of substance ($N$), and luminous intensity ($J$). Here we are concerned only with $M$, $L$, $T$, and $I$.

The key formulas we need are:

• Capacitance $C$ is defined by $Q = C\,V$, where $Q$ is charge and $V$ is potential difference.
• Charge $Q$ has dimensions $[I\,T]$.
• Potential difference $V$ is work per unit charge, so $[V] = \frac{[M\,L^2\,T^{-3}\,I^{-1}]}{[I\,T]} = [M\,L^2\,T^{-3}\,I^{-1}]$.

1. Dimensions of Capacitance $C$:
   From $Q = C\,V$ we get $[C] = \frac{[Q]}{[V]} = \frac{[I\,T]}{[M\,L^2\,T^{-3}\,I^{-1}]}.$ Simplify the exponents: $[C] = M^{-1}\,L^{-2}\,T^{4}\,I^{2}.$
2. Dimensions of Voltage $V$:
   As already noted, $[V] = M\,L^2\,T^{-3}\,I^{-1}.$
3. Dimensions of $\lambda = C/V$:
   Divide the dimensions of $C$ by those of $V$: $[\lambda] = \frac{[C]}{[V]} = \frac{M^{-1}\,L^{-2}\,T^{4}\,I^{2}}{M^{1}\,L^{2}\,T^{-3}\,I^{-1}}.$ Combine the exponents for each base dimension:
   • Mass: $-1 - 1 = -2$
   • Length: $-2 - 2 = -4$
   • Time: $4 - (-3) = 7$
   • Current: $2 - (-1) = 3$
   Hence $[\lambda] = M^{-2}\,L^{-4}\,I^{3}\,T^{7}.$
4. Matching with the given options:
   The derived dimension $M^{-2}\,L^{-4}\,I^{3}\,T^{7}$ exactly matches option C.

Many students confuse the dimensions of capacitance and voltage, especially the signs of the exponents for current and time. A frequent mistake is to forget that dividing by $V$ reverses the sign of each exponent in $[V]$. Always write out the division explicitly to avoid sign errors.

Exam Tip: When in doubt, re-derive $[C]$ and $[V]$ from first principles rather than memorizing them. This ensures you never mix up the signs.