I have tried the part a of this problem nearly 6 times. In my fifth and 6th attempt I got the same answer and I was confident about my solution but I am getting the wrong answer nonetheless.

Here's my **attempt**:

Call the vertical diagonal p and the horizontal diagonal q.

Let the upper and lower vertices move towards the centre with velocity

$v$ just as the sidewards vertices move away with $u$.

From constraint relation, $2v\cos \alpha \cos\theta= u \implies v =

\dfrac{u}{ \sin 2\theta}$

Area of the figure at any instant is given by$A= \dfrac 1 2 pq$

$\dfrac{dA}{dt}= \frac 12 \left(p

\dfrac{dq}{dt}+q\dfrac{dp}{dt}\right)$Now note the following relations:

$\dfrac{dq}{dt}= 2u$

$\dfrac{dp}{dt}= -2v$

$p = 2a \sin \theta$

$q = 2a \cos \theta$

$\theta = 30^\circ$

Substituting, we get: $\dfrac{dA}{dt} = -au$

$\implies \epsilon = -\dfrac{d\phi}{dt}= -B\dfrac{dA}{dt}= Bau $

But answer is : $2Bau$

Please let me know my error.