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 tocket, 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 suface such that the mass does not move is:

Options

A. `2tan theta`
B. `3tan theta`
C. `tan theta`
D. `tan2theta`

Answer

To solve the problem, we need to determine the minimum coefficient of friction (μ_min) required to prevent a mass m from sliding down an inclined plane inside a rocket that is accelerating upward with an acceleration of 2g.

### Step-by-Step Solution:

1. **Identify the Effective Acceleration**:
The rocket is accelerating upward with an acceleration of \(2g\). Therefore, the effective gravitational acceleration acting on the mass inside the rocket is:
\[ g_{\text{effective}} = g + 2g = 3g \]

2. **Analyze Forces on the Mass**:
The mass m experiences two forces:
- The gravitational force acting downward, which is \(F_g = mg_{\text{effective}} = 3mg\).
- The normal force (N) exerted by the inclined plane, which acts perpendicular to the surface.

3. **Resolve Forces Along the Incline**:
The gravitational force can be resolved into two components:
- Along the incline: \(F_{\parallel} = 3mg \sin \theta\)
- Perpendicular to the incline: \(F_{\perpendicular} = 3mg \cos \theta\)

4. **Determine the Normal Force**:
The normal force (N) acting on the mass is equal to the perpendicular component of the gravitational force:
\[ N = 3mg \cos \theta \]

5. **Frictional Force**:
The frictional force (F_f) that opposes the motion of the mass down the incline is given by:
\[ F_f = \mu N = \mu (3mg \cos \theta) \]

6. **Set Up the Equation for Equilibrium**:
For the mass to remain at rest (not sliding down), the frictional force must balance the component of gravitational force acting down the incline:
\[ F_f = F_{\parallel} \]
Therefore, we have:
\[ \mu (3mg \cos \theta) = 3mg \sin \theta \]

7. **Simplify the Equation**:
We can cancel \(3mg\) from both sides (assuming \(m \neq 0\)):
\[ \mu \cos \theta = \sin \theta \]

8. **Solve for the Minimum Coefficient of Friction**:
Rearranging the equation gives:
\[ \mu = \frac{\sin \theta}{\cos \theta} = \tan \theta \]

Thus, the minimum coefficient of friction required to prevent the mass from sliding down the incline is:
\[ \mu_{\text{min}} = \tan \theta \]