1. Use energy conservation for rolling without slipping

For a solid sphere rolling without slipping down an incline,

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

For a solid sphere,

$I = \frac{2}{5}mr^2, \qquad \omega = \frac{v}{r}$

So,

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2$

$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$

Hence,

$v^2 = \frac{10}{7}gh$

So the square of the speed is directly proportional to the vertical height fallen:

$v^2 \propto h$

2. Find the vertical heights in the two cases

The incline length is constant at $L$.

Vertical height:

$h = L\sin\theta$

Case 1: $\theta = 30^\circ$

$h_1 = L\sin 30^\circ = L\cdot \frac{1}{2} = \frac{L}{2}$

Thus,

$v_1^2 \propto h_1 = \frac{L}{2}$

Case 2: $\theta = 45^\circ$

$h_2 = L\sin 45^\circ = L\cdot \frac{1}{\sqrt{2}}$

Thus,

$v_2^2 \propto h_2 = \frac{L}{\sqrt{2}}$

3. Take the ratio

$v_1^2 : v_2^2 = h_1 : h_2 = \frac{L}{2} : \frac{L}{\sqrt{2}}$

Cancel $L$:

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

Multiply both terms by $2\sqrt{2}$:

$= \sqrt{2} : 2$

Now divide by $\sqrt{2}$:

$= 1 : \sqrt{2}$

4. Check options

• A: $1:3$ ❌
• B: $1:2$ ❌
• C: $1:\sqrt{2}$ ✅
• D: $1:\sqrt{3}$ ❌

Therefore, the correct answer is:

$\boxed{1:\sqrt{2}}$

So, Option C is correct.