1. Choose the rocket frame

Since the rocket is accelerating upward with acceleration $2g$, we analyze the block in the non-inertial frame of the rocket.

In this frame, a pseudo force acts on the block downward of magnitude $m(2g).$

Also, the real gravitational force acts downward: $mg.$

So the block experiences an effective downward force equal to $mg + 2mg = 3mg.$

Hence the effective gravity inside the rocket is $g_{\text{eff}} = 3g$ downward.

2. Resolve forces along and perpendicular to the incline

The incline makes an angle $\theta$ with the horizontal.

Under effective gravity $3g$ downward:

• Component along the plane: $3mg\sin\theta$
• Component perpendicular to the plane: $3mg\cos\theta$

Therefore the normal reaction is $N = 3mg\cos\theta.$

3. Condition for no slipping

To prevent the block from sliding down the plane, friction must balance the component of effective gravity along the plane.

Required friction: $f = 3mg\sin\theta.$

Maximum static friction: $f_{\max} = \mu N = \mu(3mg\cos\theta).$

For the block to remain at rest, $3mg\sin\theta \le \mu(3mg\cos\theta).$

Cancelling $3mg$, $\tan\theta \le \mu.$

Thus the minimum coefficient of friction is $\mu_{\min} = \tan\theta.$

4. Check options

• A: $\tan\theta$ ✅
• B: $2\tan\theta$ ❌
• C: $3\tan\theta$ ❌
• D: $\tan 2\theta$ ❌

So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

Derived answer: A

They agree.