
- A$\sqrt {\frac{{10gh}}{9}} $
- ✓$\sqrt {\frac{{20gh}}{19}} $
- C$\sqrt {\frac{{15gh}}{139}} $
- D$\sqrt {\frac{{18gh}}{159}} $

$\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right) \mathrm{R}=\frac{1}{2} \mathrm{MR}^{2} \frac{\mathrm{a}}{\mathrm{R}}$
$\mathrm{Mg}-\mathrm{T}_{2}=\mathrm{Ma}$
$\mathrm{Mg}=\mathrm{Ma}\left(\frac{2}{5}+\frac{1}{2}+1\right)$
$g=a \times \frac{19}{10}$
$a=\frac{10 g}{19}$
$v=\sqrt{2 \mathrm{as}}$
$v=\sqrt{2 \times \frac{10 g h}{19}}$
$v=\sqrt{\frac{20 g h}{19}}$
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Assertion $A$ : When a body is projected at an angle $45^{\circ}$, it's range is maximum.
Reason $R$ : For maximum range, the value of $\sin 2 \theta$ should be equal to one.
In the light of the above statements, 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 ($2$)