Question

If $ C $ be the capacitance and $ V $ be the electric potential, then the dimensional formula of $ CV^2 $ is

• A. \( [MLT^{-2}A^0] \)
• B. \( [MLT^{-2}A^{-1}] \)
• C. \( [M^{1}L^{2}T^{-2}A^0] \)
• D. \( [ML^{-3}T^1A] \)

Hint:

For dimensional analysis, remember to use the dimensional formulas for basic quantities like charge, potential, and capacitance. Multiply or divide these to get the dimensional formula for more complex expressions.

Answer 1

The Correct Option is A

We know the dimensional formulas of capacitance \( C \) and electric potential \( V \):
1. Capacitance \( C \): The dimensional formula for capacitance is: \[ C = \frac{Q}{V} \] where \( Q \) is charge and \( V \) is electric potential. The dimensional formula for charge \( Q \) is \( [A T] \), and the dimensional formula for electric potential \( V \) is \( [ML^2 T^{-3} A^{-1}] \).
Thus, the dimensional formula for capacitance \( C \) is: \[ [C] = \frac{[A T]}{[ML^2 T^{-3} A^{-1}]} = [M^{-1} L^{-2} T^4 A^2] \]
2. Electric potential \( V \): The dimensional formula for electric potential is already provided as: \[ [V] = [ML^2 T^{-3} A^{-1}] \] Now, the expression \( CV^2 \) has dimensions: \[ [C V^2] = [M^{-1} L^{-2} T^4 A^2] \times [ML^2 T^{-3} A^{-1}]^2 \] Simplifying: \[ [C V^2] = [M^{-1} L^{-2} T^4 A^2] \times [M^2 L^4 T^{-6} A^{-2}] \] \[ [C V^2] = [M^{1} L^2 T^{-2} A^0] \]
Thus, the dimensional formula of \( CV^2 \) is: \[ \text{(A) } [MLT^{-2}A^0] \] Therefore, the correct answer is: \[ \text{(A) } [MLT^{-2}A^0] \]