1. Magnetic flux through the rotating loop

If the loop rotates in a uniform magnetic field with angular speed $\omega$, the magnetic flux is

$\Phi = BA \cos\theta$

where $\theta$ is the angle between the normal to the plane of the loop and the magnetic field.

At $t=0$, the plane of the loop is perpendicular to the magnetic field. This means the normal to the plane is parallel to $\vec B$, so

$\theta=0 \quad \Rightarrow \quad \Phi = BA \; (\text{maximum})$

2. Induced emf

By Faraday's law,

$e = \left| -\frac{d\Phi}{dt} \right|$

Let

$\theta = \omega t$

Then

$\Phi = BA\cos(\omega t)$

So,

$e = \left| \frac{d}{dt}(BA\cos \omega t) \right| = BA\omega |\sin \omega t|$

Thus, the magnitude of induced emf is

$e = E_0 |\sin \omega t|$

where $E_0=BA\omega$.

3. Time period of rotation

Given period of rotation,

$T = 10\,\text{s}$

Hence,

$\omega = \frac{2\pi}{T} = \frac{2\pi}{10} = \frac{\pi}{5}\,\text{rad/s}$

4. Condition for maximum emf

Maximum value occurs when

$|\sin \omega t| = 1$

So,

$\omega t = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$

First time:

$t = \frac{\pi/2}{\pi/5} = \frac{5}{2} = 2.5\,\text{s}$

5. Condition for minimum emf

Minimum value occurs when

$|\sin \omega t| = 0$

So,

$\omega t = 0, \pi, 2\pi, \dots$

Ignoring the initial instant $t=0$, the next minimum occurs at

$t = \frac{\pi}{\pi/5} = 5.0\,\text{s}$

6. Match with options

Maximum emf at $2.5\,\text{s}$ and minimum emf at $5.0\,\text{s}$.

Therefore, the correct option is:

$\boxed{\text{D}}$