The emf induced in the second coil is given by:
\[ \text{EMF} = -M \frac{di}{dt} \]
Substitute \( i = i_0 \sin \omega t \):
\[ \frac{di}{dt} = i_0 \omega \cos \omega t \]
Thus, the maximum emf (when \( \cos \omega t = 1 \)) is:
\[ \text{EMF}_{\text{max}} = Mi_0 \omega = (0.002) \times (5) \times (50\pi) = \frac{\pi}{2} \, \text{V} \]
Therefore, \(\alpha = 2\).
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32