Question

If \(Q\), \(L\) and \(T\) represent the electric charge, inductance and time respectively, then the physical quantity having the dimensions of \( \dfrac{QL}{T^2} \) is:

• A. Magnetic energy
• B. Impedance
• C. Electric potential
• D. Electric field

Hint:

Remember: \[ Q = IT \] and \[ V = \frac{W}{Q} \] These relations are extremely useful in dimensional analysis problems involving electrical quantities.

Answer 1

The Correct Option is C

Step 1: Understand what we need.
We are given a mix of charge $Q$, inductance $L$ and time $T$. We must find which real physical quantity has the same dimensions as $\dfrac{QL}{T^2}$. The best trick here is to build everything from base quantities and then match.

Step 2: Write charge in base form.
Current is charge per unit time, so charge is current times time. This gives us \[ [Q] = A\,T \] Here $A$ is the dimension of electric current.

Step 3: Write inductance in base form.
Energy stored in an inductor is $\tfrac12 L I^2$. So inductance equals energy divided by current squared. Using energy dimension $ML^2T^{-2}$ we get \[ [L] = \frac{ML^2T^{-2}}{A^2} \]
Step 4: Put the pieces into the given expression.
Now plug both into $\dfrac{QL}{T^2}$. \[ \left[\frac{QL}{T^2}\right] = \frac{(AT)\left(\frac{ML^2T^{-2}}{A^2}\right)}{T^2} \]
Step 5: Simplify carefully.
Multiply the top first, then divide by $T^2$. The $A$ in charge cancels one $A$ in inductance, leaving one $A$ below. \[ = \frac{ML^2T^{-3}}{A} = ML^2T^{-3}A^{-1} \]
Step 6: Match with a known quantity.
Electric potential is work done per unit charge. So its dimension is energy divided by charge. \[ [V] = \frac{ML^2T^{-2}}{AT} = ML^2T^{-3}A^{-1} \] This is exactly the same as what we found. So the answer is electric potential. \[ \boxed{\text{Electric potential}} \]