The current flowing through a wire as a function fo time is given as: I = 3t² + 2t + 5 The charge flowing through any cross-section of the wire in first 2 s is:

The correct option is 22 C.

To find the total charge flowing through any cross-section of the wire, we use the relationship between electric current and electric charge. Electric current is defined as the rate of flow of charge with respect to time:

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

This equation can be rearranged to express the small amount of charge, $dq$, that flows in an infinitesimal time interval, $dt$:

$$dq=I\cdot dt$$

To determine the total charge $q$ that flows through the wire in the first 2 seconds (from $t=0$ to $t=2\text{s}$), we integrate both sides of the equation over this time range:

$$q={\int }_0^{2}Idt$$

We are given the current as a function of time: $I=3t^2+2t+5$. Substituting this expression into our integral, we get:

$$q={\int }_0^2(3t^2+2t+5)dt$$

We now integrate each term individually with respect to $t$:

$$q={[3\left(\frac{t^3}{3}\right)+2\left(\frac{t^2}{2}\right)+5t]}_0^2$$

Simplifying the coefficients gives:

$$q={[t^3+t^2+5t]}_0^2$$

Now, evaluate the antiderivative at the upper limit of integration (2) and subtract the value at the lower limit (0):

$$q=\left(2^3+2^2+5\left(2\right)\right)-\left(0^3+0^2+5\left(0\right)\right)$$

$$q=(8+4+10)-0$$

$$q=22\text{C}$$

Therefore, the charge flowing through any cross-section of the wire in the first 2 seconds is 22 C.