1. Effect of the lift moving with uniform velocity

Since the lift is moving upward with uniform velocity $v$, its acceleration is zero. So the lift frame is effectively an inertial frame. Hence, the motion of the block on the incline is exactly the same as if the incline were at rest.

2. Time of sliding on a frictionless incline

For a frictionless incline of angle $\theta$, acceleration of the block along the plane is

$a = g\sin\theta$

If the block starts from rest and slides a distance $l$ along the incline, then

$l = \frac{1}{2} a t^2 = \frac{1}{2} g\sin\theta\, t^2$

So,

$t = \sqrt{\frac{2l}{g\sin\theta}}$

Thus,

$t \propto \frac{1}{\sqrt{\sin\theta}}$

3. Use the given case

For $\theta_1 = 30^\circ$, time is $t_1 = 2\,\text{s}$.

For $\theta_2 = 45^\circ$, let time be $t_2$. Then

$\frac{t_2}{t_1} = \sqrt{\frac{\sin 30^\circ}{\sin 45^\circ}}$

Substitute values:

$\sin 30^\circ = \frac{1}{2}, \qquad \sin 45^\circ = \frac{1}{\sqrt{2}}$

So,

$t_2 = 2\sqrt{\frac{1/2}{1/\sqrt{2}}}$

$t_2 = 2\sqrt{\frac{\sqrt{2}}{2}}$

$t_2 = 2\sqrt{\frac{1}{\sqrt{2}}}$

Numerically,

$\frac{1}{\sqrt{2}} \approx 0.707$

$\sqrt{0.707} \approx 0.84$

Therefore,

$t_2 \approx 2 \times 0.84 = 1.68\,\text{s}$

4. Check options

• A: $2.66\,\text{s}$
• B: $0.83\,\text{s}$
• C: $1.68\,\text{s}$
• D: $0.70\,\text{s}$

Hence, the correct option is:

$\boxed{\text{C }\; 1.68\,\text{s}}$