JEE Physics 2025 Question with Solution

Given: $I(t)=0.02t+0.01\,\text{A}$, and the time interval is from $t=1\,\text{s}$ to $t=2\,\text{s}$.

Find: The charge flowing through the wire in this interval.

Using the relation between charge and current:

$$q = \int I(t)\,dt$$

So,

$$q = \int_{1}^{2} (0.02t + 0.01)\,dt$$

Integrating,

$$q = \left[ 0.02 \frac{t^2}{2} + 0.01t \right]_{1}^{2}$$

Evaluating at the limits,

$$q = \left[ 0.01(4) + 0.01(2) \right] - \left[ 0.01(1) + 0.01(1) \right]$$ $$q = 0.04\,\text{C}$$

Therefore, the charge that flows through the wire is $0.04\,\text{C}$. The correct option is D.

Given: $I(t) = 0.02t + 0.01 \, \text{A}$.

Find: Total charge between $t = 1 \, \text{s}$ and $t = 2 \, \text{s}$.

The total charge passing through a conductor over a time interval is obtained by integrating current:

$$Q = \int_{t_1}^{t_2} I(t) \, dt$$

Substitute the given current function:

$$Q = \int_{1}^{2} (0.02t + 0.01) \, dt$$

Integrate each term separately:

$$Q = 0.02 \int_{1}^{2} t \, dt + 0.01 \int_{1}^{2} dt$$

Now use

$$\int t \, dt = \frac{t^2}{2}, \quad \int dt = t$$

Therefore,

$$Q = 0.02 \left[\frac{t^2}{2}\right]_{1}^{2} + 0.01 [t]_{1}^{2}$$

Evaluate the limits:

$$Q = 0.02 \left(\frac{4 - 1}{2}\right) + 0.01 (2 - 1)$$ $$Q = 0.02 \times 1.5 + 0.01 \times 1$$ $$Q = 0.03 + 0.01 = 0.04 \, \text{C}$$

Therefore, the total charge that flows through the wire is $0.04 \, \text{C}$. The correct option is D.