A rocket is fired vertically from the earth with an acceleration of 2g, where g is the gravitational acceleration. On an inclined plane inside the rocket, making an angle $$\theta$$ with the horizontal, a point object of mass m is kept. The minimum coefficient of friction $$\mu_{min}$$ between the mass and the inclined surface such that the mass does not move is:

Let us examine the situation from the frame of reference of the rocket itself. Because the rocket is accelerating upward with an acceleration $$2g$$, an observer inside the rocket has to introduce a pseudo-force to apply Newton’s laws in the usual way.

Real weight acts downward: $$\vec W = m\vec g\,.$$ Pseudo-force (opposite to the rocket’s upward acceleration) is also downward and equals $$m(2g)\,.$$

Therefore the total effective force that behaves like “weight” for the mass inside the rocket is the vector sum of the two forces, both pointing downward:

$$\vec W_{\text{eff}} = m\vec g + m(2g)\,\hat j = m(3g)\,\hat j\,.$$

In other words, the block feels an effective gravitational acceleration

$$g_{\text{eff}} = 3g\,.$$

The inclined plane makes an angle $$\theta$$ with the horizontal. For such an incline the usual resolution of the weight into components gives:

Component of the effective weight perpendicular to the plane: $$N = mg_{\text{eff}}\cos\theta = m(3g)\cos\theta = 3mg\cos\theta\,.$$

Component of the effective weight parallel to (down) the plane: $$F_{\parallel} = mg_{\text{eff}}\sin\theta = m(3g)\sin\theta = 3mg\sin\theta\,.$$

The static friction force that can oppose this tendency to slide has a maximum value given by the formula

$$f_{\text{max}} = \mu_s N = \mu_s(3mg\cos\theta)\,.$$

For the mass to remain at rest relative to the inclined surface, the frictional force must be at least as large as the downslope component of the effective weight. Hence we demand

$$\mu_s(3mg\cos\theta) \ge 3mg\sin\theta\,.$$

Cancel the common factor $$3mg$$ from both sides:

$$\mu_s\cos\theta \ge \sin\theta\,.$$

Solving for $$\mu_s$$ gives

$$\mu_s \ge \frac{\sin\theta}{\cos\theta} = \tan\theta\,.$$

Thus the least (minimum) value of the coefficient of friction that will just prevent motion is

$$\mu_{\text{min}} = \tan\theta\,.$$

Hence, the correct answer is Option B.