1. Set up the geometry

Let the total length of the inclined plane be $L$.

• Upper half length $= \dfrac{L}{2}$ is smooth.
• Lower half length $= \dfrac{L}{2}$ is rough.
• Inclination of plane $= \phi$.

The body starts from rest at the top and comes to rest again at the bottom.

So, by work-energy theorem,

$\text{Net work done over the whole path} = 0$

because initial and final kinetic energies are both zero.

2. Work done in the smooth upper half

On the smooth part, there is no friction.

Component of weight along the incline is

$mg\sin\phi$

Hence work done by gravity over distance $\dfrac{L}{2}$ is

$W_1 = mg\sin\phi \cdot \frac{L}{2}$

3. Work done in the rough lower half

Over the lower half, gravity still does positive work:

$W_{g,2} = mg\sin\phi \cdot \frac{L}{2}$

Normal reaction is

$N = mg\cos\phi$

So friction force is

$f = \mu N = \mu mg\cos\phi$

This friction opposes motion, so its work is negative:

$W_f = -\mu mg\cos\phi \cdot \frac{L}{2}$

Therefore net work on the lower half is

$W_2 = mg\sin\phi \cdot \frac{L}{2} - \mu mg\cos\phi \cdot \frac{L}{2}$

4. Total work over the whole incline

$W_{\text{total}} = W_1 + W_2$

So,

$W_{\text{total}} = mg\sin\phi \cdot \frac{L}{2} + \left( mg\sin\phi \cdot \frac{L}{2} - \mu mg\cos\phi \cdot \frac{L}{2} \right)$

$W_{\text{total}} = mgL\sin\phi - \mu mg\cos\phi \cdot \frac{L}{2}$

Since the body starts and ends at rest,

$W_{\text{total}}=0$

Therefore,

$mgL\sin\phi - \mu mg\cos\phi \cdot \frac{L}{2}=0$

Cancel $mgL$:

$\sin\phi - \frac{\mu}{2}\cos\phi = 0$

$\frac{\mu}{2}\cos\phi = \sin\phi$

$\mu = \frac{2\sin\phi}{\cos\phi} = 2\tan\phi$

5. Check options

• A: $2\cos\phi$ ❌
• B: $2\sin\phi$ ❌
• C: $\tan\phi$ ❌
• D: $2\tan\phi$ ✅

6. Conclusion

The required coefficient of friction is

$\boxed{2\tan\phi}$