The dimension of 1/2 ε₀E² is M^aL^bT^c, then the value of a−2b+c is:

The correct option is 1.

To find the value of $a-2b+c$, we first need to determine the dimensions of the expression $\frac{1}{2}{\epsilon }_0{E}^2$.

The expression $\frac{1}{2}{\epsilon }_0{E}^2$ represents the energy density (energy per unit volume) of an electric field in a vacuum.
Let us write the dimensional formula for energy density:
$\text{Energy Density}=\frac{\text{Energy}}{\text{Volume}}$

The dimensional formula for energy (which has the same dimensions as work, $\text{Force}\times \text{Distance}$) is:
$\left[\text{Energy}\right]=\left[{\text{ML}}^2{\text{T}}^{-2}\right]$

The dimensional formula for volume is:
$\left[\text{Volume}\right]=\left[{\text{L}}^3\right]$

Therefore, the dimensional formula for energy density is:
$\left[\frac{1}{2}{\epsilon }_0{E}^2\right]=\frac{\left[{\text{ML}}^2{\text{T}}^{-2}\right]}{\left[{\text{L}}^3\right]}$
$\left[\frac{1}{2}{\epsilon }_0{E}^2\right]=\left[{\text{ML}}^{-1}{\text{T}}^{-2}\right]$

Comparing this with the given dimensional form ${\text{M}}^a{\text{L}}^b{\text{T}}^c$, we get:
$a=1$
$b=-1$
$c=-2$

Now, we calculate the value of the expression $a-2b+c$:
$a-2b+c=1-2(-1)+(-2)$
$a-2b+c=1+2-2$
$a-2b+c=1$