If the voltage applied to a load is 100 sin 500t and current through the load is 10sin (500t + 60°), then the power consumed by the load is:

Correct option is C

​$$\text{Given, } V = 100 \sin 500t \\I = 10 \sin (500t + 60^\circ) \\P = \frac{100}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times \cos 60^\circ \\P = 250 \, \text{W}$$​​