If the force is given by $F = at + bt^2$ where $t$ is time, the dimensions of $a$ and $b$ are:

The $SI$ unit of energy is $J = kg \, m^{2} \, s^{-2}$; that of speed $v$ is $m \, s^{-1}$ and of acceleration $a$ is $m \, s^{-2}$. Which of the formulae for kinetic energy $(K)$ given below can you rule out on the basis of dimensional arguments ($m$ stands for the mass of the body):
$(a)$ $K = m^{2} v^{3}$
$(b)$ $K = (1/2) m v^{2}$
$(c)$ $K = m a$
$(d)$ $K = (3/16) m v^{2}$
$(e)$ $K = (1/2) m v^{2} + m a$

In the relation $P = \frac{\alpha}{\beta} e^{-\frac{\alpha Z}{k\theta}}$,$P$ is pressure,$Z$ is distance,$k$ is Boltzmann constant,and $\theta$ is temperature. The dimensional formula of $\beta$ is:

$A$ force defined by $F = \alpha t^2 + \beta t$ acts on a particle at a given time $t$. The factor which is dimensionless,if $\alpha$ and $\beta$ are constants,is:

If $E$,$M$,$L$,and $G$ denote energy,mass,angular momentum,and the universal gravitational constant respectively,then the quantity $\left(\frac{EL^{2}}{G^{2} M^{5}}\right)$ has the dimensions of:

The dimensions of the product $\mu_{0} \varepsilon_{0}$ are related to those of velocity as