Subjected to a constant torque of 0.10 N-m , a uniform square plate...

The moment of inertia of a uniform square plate of mass $m$ and side length $a$ about its diagonal is given by:

$I = \dfrac{ma^2}{12}$

Substituting the given values, $m = 2.0 \text{ kg}$ and $a = 10 \text{ cm} = 0.1 \text{ m}$:

$I = \dfrac{2.0 \times (0.1)^2}{12} = \dfrac{0.02}{12} = \dfrac{1}{600} \text{ kg-m}^2$

The angular momentum $L$ after time $t = 5 \text{ s}$ under a constant torque $\tau = 0.10 \text{ N-m}$ starting from rest is:

$L = \tau t = 0.10 \times 5 = 0.5 \text{ kg-m}^2/\text{s}$

The kinetic energy $K$ of the plate is given by:

$K = \dfrac{L^2}{2I}$

Substituting the values of $L$ and $I$:

$K = \dfrac{(0.5)^2}{2 \times \left(\dfrac{1}{600}\right)} = \dfrac{0.25}{\dfrac{1}{300}} = 0.25 \times 300 = 75 \text{ J}$