1. Motion when the disc slips down the incline

If the disc slides without rolling, friction does not contribute to rotation, so it behaves like a particle.

The acceleration down the incline is $a_1 = g\sin\theta$

Given that it starts from rest and moves a distance $L$ along the incline, $L = \frac{1}{2} a_1 t_1^2$ so, $L = \frac{1}{2}(g\sin\theta)t_1^2$

Thus, $t_1 = \sqrt{\frac{2L}{g\sin\theta}}$

2. Motion when the disc rolls down the incline

For rolling without slipping, $a_2 = \frac{g\sin\theta}{1+\frac{I}{mR^2}}$

For a solid circular disc, $I = \frac{1}{2}mR^2$

Hence, $a_2 = \frac{g\sin\theta}{1+\frac{1}{2}} = \frac{g\sin\theta}{\frac{3}{2}} = \frac{2}{3}g\sin\theta$

Now using $L = \frac{1}{2} a_2 t_2^2$ we get $L = \frac{1}{2}\left(\frac{2}{3}g\sin\theta\right)t_2^2$

So, $t_2 = \sqrt{\frac{2L}{a_2}} = \sqrt{\frac{2L}{\frac{2}{3}g\sin\theta}}$

3. Find the ratio $\dfrac{t_2}{t_1}$

$\frac{t_2}{t_1} = \sqrt{\frac{a_1}{a_2}}$

Substituting, $\frac{t_2}{t_1} = \sqrt{\frac{g\sin\theta}{\frac{2}{3}g\sin\theta}} = \sqrt{\frac{3}{2}}$

Given, $\frac{t_2}{t_1} = \sqrt{\frac{3}{x}}$

Therefore, $\sqrt{\frac{3}{x}} = \sqrt{\frac{3}{2}}$

Squaring both sides, $\frac{3}{x} = \frac{3}{2}$

Hence, $x=2$

4. Comparison with stored answer

Derived answer: $2$

Stored correct answer: $2$

They match.