JEE PYQ: Current Electricity Question 14
Question 14 - 2021 (24 Feb 2021 Shift 1)
A current through a wire depends on time as $i = \alpha_0 t + \beta t^2$ where $\alpha_0 = 20$ A/s and $\beta = 8$ As$^{-2}$. Find the charge crossed through a section of the wire in 15 s.
(1) 2100 C
(2) 260 C
(3) 2250 C
(4) 11250 C
Show Answer
Answer: (4)
Solution
$Q = \int_0^{15} (\alpha_0 t + \beta t^2),dt = \left[\frac{\alpha_0 t^2}{2} + \frac{\beta t^3}{3}\right]_0^{15} = \frac{20 \times 225}{2} + \frac{8 \times 3375}{3} = 2250 + 9000 = 11250$ C.
BETA
