In the formula \(X=3 Y Z^{2}, X\) and \(Z\) have the dimensions of capacitance and magnetic induction,…
- \(M^{-3} L^{-2} T^{-2} A^{-4}\)
- \(M L^{-2} \vec{A}\)
- \(M^{-3} L^{-2} T^{8} A^{4}\)
- \(M^{-3} L^{-2} T^{4} A^{4}\)
Solution
\(\therefore X=\frac{Q^{2}}{W} \cdot\) Now, \(Z=B=\frac{F}{I L}\)
\(\therefore \quad Y=\frac{X}{3 Z^{2}}=\frac{1}{3} \frac{Q^{2}}{W} \times \frac{I^{2} L^{2}}{F^{2}}\)
\(=\frac{A^{2} T^{2} A^{2} L^{2}}{\left[M L^{2} T^{-2}\right]\left[M^{2} L^{2} T^{-4}\right]}=\left[M^{-3} L^{-2} T^{8} A^{4}\right]\) ^
Asked in: JEE Mains - Units and Dimensions - Chapter Test