In the formula $X=3 Y Z^{2}, X$ and $Z$ have the dimensions of capacitance and magnetic induction,…
- $M^{-3} L^{-2} T^{-2} Q^{-4}$
- $M L^{-2} Q$
- $M^{-3} L^{-2} T^{2} Q^{4}$
- $M^{-3} L^{-2} T^{4} Q^{4}$
Solution
$\mathrm{X}=3 \mathrm{Y} \mathrm{Z}^{2}$
$\mathrm{Y}=\frac{1}{3} \frac{\mathrm{X}}{\mathrm{Z}^{2}}$
since $\mathrm{X}$ is the capacitance. So, its dimensions are: $\mathrm{X}=\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{2} \mathrm{Q}^{2}$
and $\mathrm{Z}$ is the magnetic induction and its dimensions are $\mathrm{Z}=\mathrm{MT}^{-1} \mathrm{Q}^{-1}$
Therefore, $\mathrm{Y}$ can be expressed as:
$\mathrm{Y}=\frac{\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{2} \mathrm{Q}^{2}}{\left[\mathrm{MT}^{-1} \mathrm{Q}^{-1}\right]^{2}}=\mathrm{M}^{-3} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{Q}^{4}$ .
Asked in: JEE Mains - Units and Dimensions - Test 3