JEE Main 2025 — Current Electricity Question with Solution

To find the total charge $q$ that flows through the wire from $t=1\text{s}$ to $t=2\text{s}$, we can integrate the current function $I(t)=0.02t+0.01\text{A}$ over this interval.

The formula for the charge $q$ flowing through the wire is given by the integral of the current with respect to time:

$q=\int I(t)dt$

We integrate the function from $t=1$ to $t=2$:

$q={\int }_1^2(0.02t+0.01)dt$

To solve this, first find the antiderivative:

$q={\left[0.02\frac{t^2}{2}+0.01t\right]}_1^2$

Calculate the expression at the upper and lower bounds:

$={\left[0.01t^2+0.01t\right]}_1^2$

Plug in the values:

For $t=2$:

$=0.01(2{)}^2+0.01(2)=0.04+0.02=0.06$

For $t=1$:

$=0.01(1{)}^2+0.01(1)=0.01+0.01=0.02$

Subtract the result at $t=1$ from the result at $t=2$:

$q=0.06-0.02=0.04\text{C}$

Therefore, the charge that flows through the wire from $t=1\text{s}$ to $t=2\text{s}$ is $0.04\text{C}$.