1. Geometry and magnetic field

   • The square loop is initially in the $XZ$ plane, so its area vector is along $\pm \hat{j}$.
   • The hinge is along the $X$-axis.
   • The loop rotates about the $X$-axis with angular speed $\omega$.
   • The magnetic field exists only in the region $y\ge 0$ and is $\vec B(t)=B_0\cos(\omega t)\,\hat k.$

2. Position of the loop during rotation

   Let the side of the square be $a$.

   At time $t$, the loop has rotated by angle $\theta=\omega t.$

   A point of the loop that was initially at height $z$ above the hinge goes to $y=-z\sin\theta, \qquad z'=z\cos\theta$ for clockwise rotation as seen from $+X$.

   Since initially $z\ge 0$ over the loop, the condition for a point to lie in the magnetic-field region $y\ge 0$ becomes $-z\sin\theta \ge 0.$

   Thus:

   • if $\sin\theta<0$, the entire loop is in $y\ge 0$,
   • if $\sin\theta>0$, only the hinge line is in $y\ge 0$, so effectively no area is inside the field.

   Therefore, over one cycle:

   • for $0<\omega t<\pi$, enclosed area in field $=0$,
   • for $\pi<\omega t<2\pi$, enclosed area in field $=a^2$.

3. Flux through the loop

   The area vector rotates with the loop. Initially we may take the area vector as $\hat n=\hat j$; after clockwise rotation by angle $\theta$ about $x$-axis, $\hat n = \cos\theta\,\hat j - \sin\theta\,\hat k.$

   Hence,

   = B_0\cos\omega t(-\sin\omega t).$$ So when the whole loop is inside the field region, $$\Phi(t)=a^2 B_0\cos\omega t(-\sin\omega t) =-a^2B_0\sin\omega t\cos\omega t.$$ When the loop is outside the field region, $\Phi=0$. Therefore, $$\Phi(t)= \begin{cases} 0, & 0<\omega t<\pi,\\[4pt] -a^2B_0\sin\omega t\cos\omega t, & \pi<\omega t<2\pi. \end{cases}$$

4. Induced emf

   Using Faraday's law, $V=-\frac{d\Phi}{dt}.$

   For $0<\omega t<\pi$, $V=0.$

   For $\pi<\omega t<2\pi$, $V=-\frac{d}{dt}\left(-a^2B_0\sin\omega t\cos\omega t\right).$

   Now, $\frac{d}{dt}(\sin\omega t\cos\omega t)=\omega\cos 2\omega t.$

   Hence, $V=a^2B_0\omega\cos 2\omega t, \qquad (\pi<\omega t<2\pi).$

5. Nature of the graph

   So the emf is:

   • zero during the first half-rotation,
   • a cosine-like variation during the next half-rotation,
   • then repeats periodically.

   Also note at $t=\pi/\omega$ and $2\pi/\omega$, the flux changes abruptly because the loop enters/leaves the magnetic-field region, so the graph corresponds to the option showing:

   • zero for half a period,
   • then oscillatory cosine segment for the next half.

6. Matching with options

   This matches Option A.

Final Answer

The correct plot is A.