The instantaneous current from a a.c source is $I=10 \sin 314t$. The frequency of the source would be-

50 Hz

The correct answer is Option (2) - 50 Hz

The equation of an AC is given as -

$I=I_0\sin (ωt)$

where,

$I_0$​ = peak current

$ω$ = angular frequency

$t$ = time

$I=10\sin(314t)$ [given]

$∴ω=314\,rad/s$

$⇒f=\frac{ω}{2\pi}=\frac{314}{\pi}$

$f=\frac{314}{2×3.14}=50Hz$