Q145 / 180 · COMEDK 2026 (9 May Shift 1) · Physics
Current in a conductor is expressed as $I = 8t^3 + 3t^2 + 1$, where current I is measured in amperes and time t is measured in seconds. What is the charge that flows through a cross-section of the conductor between time t = 1s to t = 2s?
- A$29\ \text{C}$
- Correct answerB$38\ \text{C}$
- C$18\ \text{C}$
- D$49\ \text{C}$
Solution
The current in the conductor is given by $I = \dfrac{dq}{dt} = 8t^3 + 3t^2 + 1$
The charge $q$ that flows through a cross-section is the integral of current with respect to time.
$dq = I dt = (8t^3 + 3t^2 + 1) dt$
Integrating both sides from $t = 1$ s to $t = 2$ s:
$q = \int_{1}^{2} (8t^3 + 3t^2 + 1) dt$
$q = \left[ \dfrac{8t^4}{4} + \dfrac{3t^3}{3} + t \right]_{1}^{2}$
$q = \left[ 2t^4 + t^3 + t \right]_{1}^{2}$
$q = (2(2)^4 + (2)^3 + 2) - (2(1)^4 + (1)^3 + 1)$
$q = (2(16) + 8 + 2) - (2 + 1 + 1)$
$q = (32 + 8 + 2) - 4$
$q = 42 - 4 = 38 \text{ C}$
The charge $q$ that flows through a cross-section is the integral of current with respect to time.
$dq = I dt = (8t^3 + 3t^2 + 1) dt$
Integrating both sides from $t = 1$ s to $t = 2$ s:
$q = \int_{1}^{2} (8t^3 + 3t^2 + 1) dt$
$q = \left[ \dfrac{8t^4}{4} + \dfrac{3t^3}{3} + t \right]_{1}^{2}$
$q = \left[ 2t^4 + t^3 + t \right]_{1}^{2}$
$q = (2(2)^4 + (2)^3 + 2) - (2(1)^4 + (1)^3 + 1)$
$q = (2(16) + 8 + 2) - (2 + 1 + 1)$
$q = (32 + 8 + 2) - 4$
$q = 42 - 4 = 38 \text{ C}$
Hence, the correct answer is option: 'B'.