Physics · Current Electricity

JEE Main 2024 — 1 February, Shift 1 — Question 55

The current in a conductor is expressed as $I=3 t^{2}+4 t^{3}$ , where $I$ is in Ampere and $t$ is in second. The amount of electric charge that flows through a section of the conductor during $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$ is $\_\_\_\_\_\_\_$ C.

Answer: 22

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{q}=\int_{1}^{2} \mathrm{i} d t=\int_{1}^{2}\left(3 \mathrm{t}^{2}+4 \mathrm{t}^{3}\right) \mathrm{dt}$

\begin{array}{l}q = \left. {\left( {{t^3} + {t^4}} \right)} \right|_1^2\\q = 22C\end{array}

Answer key and solution verified before publishing.

Practise Current Electricity

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Physics
Chapter: Current Electricity
Topic: Current in Conductors, Drift Speed, Resistivity and Relaxation Time

Answer: 22

Step-by-step solution

Practise Current Electricity

More Current Electricity questions from this paper