1. Set up forces along the incline

Let the block have weight $mg$ on an incline of angle $\theta = 45^\circ$.

• Component of weight down the incline: $mg\sin\theta$
• Normal reaction: $N = mg\cos\theta$
• Limiting friction: $f = \mu N = \mu mg\cos\theta$

We assume the applied force is along the incline.

2. Force required to just push the block up the incline

If the block is about to move up, friction acts down the plane.

So the required force is $F_1 = mg\sin\theta + \mu mg\cos\theta$

3. Force required to just prevent it from sliding down

To just prevent downward sliding, the block tends to move down, so friction acts up the plane.

Hence the required upward force is $F_2 = mg\sin\theta - \mu mg\cos\theta$

4. Use the given condition

Given: $F_1 = 2F_2$

Substitute: $mg\sin\theta + \mu mg\cos\theta = 2\left(mg\sin\theta - \mu mg\cos\theta\right)$

Cancel $mg$: $\sin\theta + \mu \cos\theta = 2\sin\theta - 2\mu \cos\theta$

Rearrange: $3\mu \cos\theta = \sin\theta$

Thus, $\mu = \frac{\sin\theta}{3\cos\theta} = \frac{\tan\theta}{3}$

Since $\theta = 45^\circ$, $\tan 45^\circ = 1$, so $\mu = \frac{1}{3}$

So, $\mu \approx 0.33$

5. Check options

• A: $0.60$ ❌
• B: $0.33$ ✅
• C: $0.25$ ❌
• D: $0.50$ ❌

Therefore, the correct option is B.