In figure, a crate slides down an inclined right angled trough. Acceleration of the crate is

The correct option is C $2.1{\text{m/s}}^2$
$\text{FBD}$ of crate:


Here,
$N=$ Normal reaction on the crate applied by trough,
$f=$ Friction force on the crate opposite to the sliding motion.

Balancing forces for front view of the crate, we have,

$mg\cos {45}^{\circ }=\sqrt{N^2+N^2}$

$\Rightarrow \sqrt{2}N=\frac{mg}{\sqrt{2}}$

$\Rightarrow N=\frac{mg}{2}$

Now, balancing forces for side view, we have,

$mg\sin {45}^{\circ }-2f=ma......\left(1\right)$

where, $a=$ acceleration of the crate parallel to the inclined plane in the downwards direction.
$f=\mu N=0.5N=0.5\frac{mg}{2}$

Substituting the values in $\left(1\right)$, we have,

$mg\sin {45}^{\circ }-2\mu N=ma$

$\Rightarrow ma=mg\times \frac{1}{\sqrt{2}}-2\times 0.5\times \frac{mg}{2}$

$\Rightarrow a=\frac{g}{\sqrt{2}}-0.5g$

$\Rightarrow a=0.71-0.5g$

$\Rightarrow a=0.21g$

$\Rightarrow a=0.21\times 10$

$\therefore a=2.1{\text{m/s}}^2$

Hence, option $\left(a\right)$ is correct answer.