where $X = \frac{{{A^2}{B^{\frac{1}{2}}}}}{{{C^{\frac{1}{3}}}{D^3}}}$, will be
[Figure shows position of reference $'O'$ when jaws of screw gauge are closed]
Given pitch $=0.1 \,{cm}$.
| List - $I$ $(Number)$ | $List - II$ $(Significant figure)$ |
| $(A)$ $1001$ | $(I)$ $3$ |
| $(B)$ $010.1$ | $(II)$ $4$ |
| $(C)$ $100.100$ | $(III)$ $5$ |
| $(D)$ $0.0010010$ | $(IV)$ $6$ |
Choose the correct answer from the options given below :
($1$) The relation between $[E]$ and $[B]$ is
$(A)$ $[ E ]=[ B ][ L ][ T ]$ $(B)$ $[ E ]=[ B ][ L ]^{-1}[ T ]$ $(C)$ $[ E ]=[ B ][ L ][ T ]^{-1}$ $(D)$ $[ E ]=[ B ][ L ]^{-1}[ T ]^{-1}$
($2$) The relation between $\left[\varepsilon_0\right]$ and $\left[\mu_0\right]$ is
$(A)$ $\left[\mu_0\right]=\left[\varepsilon_0\right][ L ]^2[ T ]^{-2}$ $(B)$ $\left[\mu_0\right]=\left[\varepsilon_0\right][ L ]^{-2}[ T ]^2$ $(C)$ $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[ L ]^2[ T ]^{-2}$ $(D)$ $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[ L ]^{-2}[ T ]^2$
Give the answer or quetion ($1$) and ($2$)