1. For rolling without slipping on an incline

For a rigid body rolling down an incline of angle $\theta$,

$a = \frac{g\sin\theta}{1 + \dfrac{I}{mr^2}}$

where:

• $a$ = linear acceleration of the center,
• $I$ = moment of inertia about the center.

2. Moment of inertia of a uniform solid cylinder

For a solid cylinder,

$I = \frac{1}{2}mr^2$

So,

$\frac{I}{mr^2} = \frac{1}{2}$

3. Substitute into the rolling acceleration formula

$a = \frac{g\sin\theta}{1 + \frac{1}{2}} = \frac{g\sin\theta}{\frac{3}{2}} = \frac{2}{3}g\sin\theta$

Given $\theta = 45^\circ$,

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

Hence,

$a = \frac{2}{3}g\cdot \frac{1}{\sqrt{2}} = \frac{2g}{3\sqrt{2}}$

Rationalizing,

$a = \frac{\sqrt{2}g}{3}$

4. Check options

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

Therefore, the correct answer is:

$\boxed{\frac{\sqrt{2}g}{3}}$