The current in a device varies with time 't' as I = 6t, where I is in mA and t is in s. The amount of charge that passes through the device during t = 0 s to t = 3 s is

27 mC

The correct answer is Option (3) → 27 mC

$I = \frac{dQ}{dt}$

$I = 6t \text{ mA} = 6 \times 10^{-3} t \text{ A}$

$Q = \int_0^3 I \, dt = \int_0^3 6 \times 10^{-3} t \, dt = 6 \times 10^{-3} \int_0^3 t \, dt$

$Q = 6 \times 10^{-3} \left[ \frac{t^2}{2} \right]_0^3 = 6 \times 10^{-3} \cdot \frac{9}{2} = 27 \times 10^{-3} \text{ C}$

Answer: $Q = 0.027 \text{ C}$