1. Given data

• Mass of sphere: $m=2\,\text{kg}$
• Radius: $R=0.5\,\text{m}$
• Initial speed: $u=1\,\text{m s}^{-1}$
• Inclination angle: $\theta=30^\circ$
• Rolling without slipping

We need the total time taken by the sphere to go up the incline, stop, and return to the starting point.

2. Acceleration of a rolling sphere on an incline

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

For a solid sphere, $I=\frac{2}{5}mR^2$

So, $a=\frac{g\sin\theta}{1+\frac{2}{5}}=\frac{g\sin\theta}{\frac{7}{5}}=\frac{5}{7}g\sin\theta$

Since the sphere is moving upward, this acts as a retardation.

Now, $\sin 30^\circ=\frac12$

Hence, $a=\frac{5}{7}\cdot g\cdot \frac12=\frac{5g}{14}$

Taking $g=9.8\,\text{m s}^{-2}$, $a=\frac{5\times 9.8}{14}=3.5\,\text{m s}^{-2}$

3. Time to go up

At the highest point, final speed becomes zero. Using $v=u-at$

With $v=0$, $u=1$ and $a=3.5$, $0=1-3.5t$ $t_{\text{up}}=\frac{1}{3.5}=0.2857\,\text{s}$

4. Time to come down

The acceleration while coming down has the same magnitude, $3.5\,\text{m s}^{-2}$, and the sphere starts from rest at the top.

The motion is symmetric because the rolling condition remains valid throughout. Thus, $t_{\text{down}}=t_{\text{up}}=0.2857\,\text{s}$

5. Total time

$T=t_{\text{up}}+t_{\text{down}}=2\times 0.2857=0.5714\,\text{s}$

So, $T\approx 0.57\,\text{s}$

6. Checking options

• A: $0.60\,\text{s}$ ❌
• B: $0.52\,\text{s}$ ❌
• C: $0.80\,\text{s}$ ❌
• D: $0.57\,\text{s}$ ✅

Therefore, the correct option is D.