
- A$1$
- B$1.5$
- C$4.5$
- ✓$0.3$

Area under graph and $V$ axis $=$ work done
$=\frac{1}{2} \times(30+10) \times 10^3 \times(25-10) \times 10^{-6}$
$=0.3 \,J$
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($1$) 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 ($2$)
$\sqrt{6}$ times faster.
$\sqrt{6}$ times slower.

