Question

In the equation \[ X=\frac12 E_rYZ^2 \] \(Z\) has the dimensions of \[ \frac12 LI^2 \] and \(X\) has the dimensions of energy. \(L\) stands for coefficient of self-induction and \(I\) for electric current. What are the dimensions of \(Y\)?

• A. \(M^{-1}L^{-1}T^{2}\)
• B. \(M^{-1}L^{-2}T^{2}\)
• C. \(M^{-1}L^{-2}T\)
• D. \(ML^{-2}T^{-2}\)

Hint:

In dimensional analysis, \[ [X]=[Y][Z]^n \] implies \[ [Y]=\frac{[X]}{[Z]^n}. \] Always substitute dimensions before simplifying powers.

Answer 1

The Correct Option is B

Concept: Dimensions of energy: \[ [X]=ML^2T^{-2} \] Energy stored in an inductor: \[ \frac12 LI^2 \] has dimensions of energy.

Step 1: Find dimensions of \(Z\). Given \(Z\) has dimensions of \[ \frac12 LI^2 \] which is energy. Hence, \[ [Z]=ML^2T^{-2} \]

Step 2: Use dimensional homogeneity. \[ [X]=[Y][Z]^2 \] Therefore, \[ ML^2T^{-2} = [Y] (ML^2T^{-2})^2 \] \[ ML^2T^{-2} = [Y] (M^2L^4T^{-4}) \] \[ [Y] = M^{-1}L^{-2}T^{2} \] \[\begin{aligned} \boxed{ M^{-1}L^{-2}T^{2} } \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.