If $u_1$ and $u_2$ are the units selected in two systems of measurement and $n_1$ and $n_2$ are their numerical values, then:

The velocity $v$ of a particle at time $t$ is given by ${v}={at}+\frac{{b}}{{t}+{c}}.$ The dimensions of ${a}$, ${b}$, and ${c}$ are respectively:
1. ${\left[{LT}^{-2}\right],[{L}],[{T}]}$
2. ${\left[{L}^2\right],[{T}] \text { and }\left[{LT}^2\right]}$
3. ${\left[{LT}^2\right],[{LT}] \text { and }[{L}]}$
4. ${[{L}],[{LT}], \text { and }\left[{T}^2\right]}$

Dimensions of electric current are:
1. $\left[M^0L^0T^{-1}Q\right]$
2. $\left[M^1L^2T^{-1}Q\right]$
3. $\left[M^2L^1T^{-1}Q\right]$
4. $\left[M^2L^2T^{-1}Q\right]$

The position of a body with acceleration $a$ is given by $x= Ka^{m}t^{n}$ (assume $t$ to be time). The values of $m$ and $n$ will be:
1. $m=1,~n=1$
2. $m=1,~n=2$
3. $m=2,~n=1$
4. $m=2,~n=2$

In the relation, $y=a \cos (\omega t-k x)$, the dimensional formula for $k$ will be:
1. ${\left[M^0 L^{-1} T^{-1}\right]}$
2. ${\left[M^0 L T^{-1}\right]}$
3. ${\left[M^0 L^{-1} T^0\right]}$
4. ${\left[M^0 L T\right]}$

When units of mass, length, and time are taken as $10~\text{kg}, 60~\text{m}~\text{and}~60~\text{s}$ respectively, the new unit of energy becomes $x$ times the initial SI unit of energy. The value of $x$ will be:
1. $10$
2. $20$
3. $60$
4. $120$