Q179 / 180 · COMEDK 2025 (10 May Shift 1) · Physics
If $R$ and $L$ denote resistance and inductance of a material, then the dimension of $L R$ will be:
- Correct answerA$M^2 L^4 T^{-5} A^{-4}$
- B$M^{-1} L^4 T A^{-3}$
- C$M L T A^{-1}$
- D$M^0 L^0 T^0 A^0$
Solution
The dimension of resistance $R$ is given by $V = IR$, so $[R] = [V][I]^{-1}$. Since $V = W/q$, $[V] = [M L^2 T^{-2}][A T]^{-1} = [M L^2 T^{-3} A^{-1}]$. Thus, $[R] = [M L^2 T^{-3} A^{-2}]$.
The dimension of inductance $L$ is given by $V = L(dI/dt)$, so $[L] = [V][T][I]^{-1} = [M L^2 T^{-3} A^{-1}][T][A]^{-1} = [M L^2 T^{-2} A^{-2}]$.
The dimension of the product $LR$ is $[L][R] = [M L^2 T^{-2} A^{-2}] \times [M L^2 T^{-3} A^{-2}] = [M^2 L^4 T^{-5} A^{-4}]$.
Answer: $M^2 L^4 T^{-5} A^{-4}$
The dimension of inductance $L$ is given by $V = L(dI/dt)$, so $[L] = [V][T][I]^{-1} = [M L^2 T^{-3} A^{-1}][T][A]^{-1} = [M L^2 T^{-2} A^{-2}]$.
The dimension of the product $LR$ is $[L][R] = [M L^2 T^{-2} A^{-2}] \times [M L^2 T^{-3} A^{-2}] = [M^2 L^4 T^{-5} A^{-4}]$.
Answer: $M^2 L^4 T^{-5} A^{-4}$
Hence, the correct answer is option: 'A'.