Q120 / 150 · MHT CET 2026 (16 April Shift 1) · Physics
An alternating e.m.f. is given by $e = e_0 \sin\omega t$. In what time the e.m.f. will have half its maximum value, if 'e' Starts from zero?
- A$\dfrac{\text{T}}{4}$
- B$\dfrac{\text{T}}{8}$
- Correct answerC$\dfrac{\text{T}}{12}$
- D$\dfrac{\text{T}}{16}$
Solution
Given alternating e.m.f. is $e = e_0 \sin\omega t$.
The maximum value of e.m.f. is $e_0$.
We need to find the time $t$ when $e = \dfrac{e_0}{2}$.
Substituting $e = \dfrac{e_0}{2}$ in the given equation:
$\dfrac{e_0}{2} = e_0 \sin\omega t$
$\sin\omega t = \dfrac{1}{2}$
The smallest positive value for $\omega t$ is $\dfrac{\pi}{6}$.
$\omega t = \dfrac{\pi}{6}$
Substituting $\omega = \dfrac{2\pi}{T}$:
$\dfrac{2\pi}{T} t = \dfrac{\pi}{6}$
$t = \dfrac{T}{12}$
Answer: $\dfrac{\text{T}}{12}$
The maximum value of e.m.f. is $e_0$.
We need to find the time $t$ when $e = \dfrac{e_0}{2}$.
Substituting $e = \dfrac{e_0}{2}$ in the given equation:
$\dfrac{e_0}{2} = e_0 \sin\omega t$
$\sin\omega t = \dfrac{1}{2}$
The smallest positive value for $\omega t$ is $\dfrac{\pi}{6}$.
$\omega t = \dfrac{\pi}{6}$
Substituting $\omega = \dfrac{2\pi}{T}$:
$\dfrac{2\pi}{T} t = \dfrac{\pi}{6}$
$t = \dfrac{T}{12}$
Answer: $\dfrac{\text{T}}{12}$
Hence, the correct answer is option: 'C'.