JEE Main 2020 — Units and Dimensions Question with Solution

(a) Dimensions of quantity are given bellow

$$\begin{gathered} {\mu }_0=\left[\mathrm{T} \mathrm{m} {\mathrm{A}}^{-1}\right]=\left[\mathrm{Wb} {\mathrm{m}}^{-1} {\mathrm{A}}^{-1}\right]=\left[\mathrm{V} \mathrm{s} {\mathrm{m}}^{-1} {\mathrm{C}}^{-1} \mathrm{s}\right]=\left[\mathrm{N} {\mathrm{C}}^{-2} {\mathrm{s}}^2\right] \\ {\epsilon }_0=\left[{\mathrm{C}}^2 {\mathrm{N}}^{-1}{\mathrm{m}}^{-2}\right] \end{gathered}$$

Now the dimension of $x$ is given by

$$\begin{gathered} \left[x\right]=\frac{1}{\left[\sqrt{{\mu }_0{\in }_0}\right]}=\sqrt{\frac{1}{\left[\mathrm{N} {\mathrm{C}}^{-2} {\mathrm{s}}^2\right]\left[{\mathrm{C}}^2 {\mathrm{N}}^{-1}{\mathrm{m}}^{-2}\right]}} \\ \left[x\right]=\mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$

(b) The dimension of electric and magnetic field is given by,

$$\begin{gathered} B=\left[\mathrm{T}\right]=\left[\mathrm{Wb} {\mathrm{m}}^{-2}\right]=\left[\mathrm{V} \mathrm{s} {\mathrm{m}}^{-2}\right] \\ E=\left[\mathrm{V} {\mathrm{m}}^{-1}\right] \end{gathered}$$

Now the dimension of $y$ is given by,

 $\left[y\right]=\frac{\left[E\right]}{\left[B\right]}=\frac{\left[\mathrm{V} {\mathrm{m}}^{-1}\right]}{\left[\mathrm{V} \mathrm{s} {\mathrm{m}}^{-2}\right]}=\mathrm{m} {\mathrm{s}}^{-1}$

(c) The dimension of capacitance and resistance is given by, 

$$$$\begin{gathered} C=\left[\mathrm{C} {\mathrm{V}}^{-1}\right] \\ R=\left[\mathrm{V} {\mathrm{C}}^{-1} \mathrm{s}\right] \end{gathered}$$

Now the dimension of $z$ is given by

 $\left[z\right]=\frac{\left[l\right]}{\left[CR\right]}=\frac{\left[\mathrm{m}\right]}{\left[\mathrm{C} {\mathrm{V}}^{-1}\right]\left[\mathrm{V} {\mathrm{C}}^{-1} \mathrm{s}\right]}=\mathrm{m} {\mathrm{s}}^{-1}$

All quantities are having the dimensions of velocity.