Given data

• Length of rectangular loop: $l = 2.5\,\text{m}$
• Width of rectangular loop: $b = 2\,\text{m}$
• Magnetic field: $B = 4\,\text{T}$
• Angle with magnetic field: $60^\circ$
• Time taken to remove loop: $\Delta t = 10\,\text{s}$

Area of the loop

$A = l \times b = 2.5 \times 2 = 5\,\text{m}^2$

Magnetic flux initially linked with the loop

Magnetic flux is

$\Phi = BA\cos\theta$

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

The loop is said to be placed at $60^\circ$ to the magnetic field, which in such problems is commonly taken as the angle between the plane of the loop and $\vec B$.

Therefore, angle between $\vec B$ and the normal is

$\theta = 30^\circ$

So,

$\Phi_i = BA\cos 30^\circ = 4 \times 5 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\,\text{Wb}$

Final flux

When the loop is completely removed from the magnetic field,

$\Phi_f = 0$

Average induced emf

By Faraday's law,

$\mathcal{E}_{\text{avg}} = -\frac{\Delta \Phi}{\Delta t}$

$\Delta \Phi = \Phi_f - \Phi_i = 0 - 10\sqrt{3} = -10\sqrt{3}$

Hence,

$\mathcal{E}_{\text{avg}} = -\frac{-10\sqrt{3}}{10} = \sqrt{3}\,\text{V} \approx 1.73\,\text{V}$

Choosing the nearest option

The induced emf is approximately

$1.73\,\text{V} \approx 2\,\text{V}$

Since the sign in such MCQs usually depends on chosen orientation, the physically meaningful choice is the magnitude. Thus the nearest option is:

Option B: $+2\,\text{V}$

Check against stored answer

Stored correct answer is C: $+1\,\text{V}$.

My derived result is approximately $+1.73\,\text{V}$, which matches neither exactly, but is much closer to $+2\,\text{V}$ than to $+1\,\text{V}$.

So I disagree with the stored answer. The likely intended answer is B.