A block $$A$$ takes 2 s to slide down a frictionless incline of $$30^\circ$$ and length $$l$$, kept inside a lift going up with uniform velocity $$v$$. If the incline is changed to $$45^\circ$$, the time taken by the block, to slide down the incline, will be approximately:

A block takes 2 s to slide down a frictionless incline of $$30^\circ$$ and length $$l$$, inside a lift going up with uniform velocity. We need to find the time when the incline angle changes to $$45^\circ$$. Since the lift moves with uniform velocity (zero acceleration), there is no pseudo force, and the effective gravitational acceleration remains $$g$$. For a frictionless incline at angle $$\theta$$, the component of gravitational acceleration along the incline is $$a = g\sin\theta$$. The block starts from rest and slides a distance $$l$$, giving $$l = \tfrac{1}{2}(g\sin 30^\circ)\,t_1^2 = \tfrac{1}{2}\cdot g\cdot \tfrac{1}{2}\cdot (2)^2 = g\;. $$ Therefore: $$l = g\quad\text{(i)}$$

When the incline angle becomes $$45^\circ$$, the block slides the same length $$l$$, so $$l = \tfrac{1}{2}(g\sin 45^\circ)\,t_2^2 = \tfrac{1}{2}\cdot g\cdot \tfrac{1}{\sqrt{2}}\cdot t_2^2\;. $$ Using equation (i), we set $$g = \tfrac{g}{2\sqrt{2}}\;t_2^2\,, $$ which leads to $$1 = \tfrac{t_2^2}{2\sqrt{2}}\quad\Longrightarrow\quad t_2^2 = 2\sqrt{2} = 2\times1.414 = 2.828\;. $$ Hence $$t_2 = \sqrt{2.828}\approx1.68\text{ s}\;. $$

We can also express this as $$t_2 = 2^{3/4} = (8)^{1/4} \approx 1.68\text{ s}\;. $$

Verification using the ratio method: From the two cases: $$\frac{t_2^2}{t_1^2} = \frac{\sin 30^\circ}{\sin 45^\circ} = \frac{1/2}{1/\sqrt{2}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\;, $$ so $$t_2^2 = \frac{t_1^2}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\quad\Longrightarrow\quad t_2 = \sqrt{2\sqrt{2}}\approx1.68\text{ s}\;. $$

Answer: Option C: 1.68 s