A particle starts from rest from the top of an inclined plane and again comes to rest on reaching the bottom most point of the inclined plane. If the coefficient of friction on some part of inclined plane is `3tantheta` and rest portion is smooth then the ratio of the rough length of the inclined to smooth length is `1 : n_(1)` then value of n is .....?

Explanation

To solve the problem, we need to analyze the motion of the particle on the inclined plane, considering the effects of friction and the smooth portion of the incline. Here’s the step-by-step solution:

### Step 1: Understand the Setup
The particle starts from rest at the top of an inclined plane and comes to rest at the bottom. There are two sections of the incline: a rough section with a coefficient of friction of \(3 \tan \theta\) and a smooth section.

### Step 2: Define Variables
Let:
- \(L\) = length of the rough portion
- \(nL\) = length of the smooth portion (where \(n\) is the ratio we need to find)
- \(g\) = acceleration due to gravity
- \(\theta\) = angle of the incline

### Step 3: Forces Acting on the Particle
On the rough portion:
- The gravitational force component along the incline: \(mg \sin \theta\)
- The frictional force opposing the motion: \(f = \mu N = 3 \tan \theta \cdot mg \cos \theta\)

Where \(N = mg \cos \theta\) is the normal force.

### Step 4: Net Force on the Rough Portion
The net force acting on the particle while it is on the rough portion can be expressed as:
\[ F_{\text{net}} = mg \sin \theta - f = mg \sin \theta - 3 \tan \theta \cdot mg \cos \theta \]
\[ F_{\text{net}} = mg \sin \theta - 3mg \frac{\sin \theta}{\cos \theta} \cos \theta = mg \sin \theta - 3mg \sin \theta = -2mg \sin \theta \]

### Step 5: Acceleration on the Rough Portion
Using Newton's second law, \(F = ma\):
\[ ma = -2mg \sin \theta \]
\[ a = -2g \sin \theta \]

### Step 6: Motion on the Rough Portion
Using the equation of motion \(v^2 = u^2 + 2as\) for the rough portion where initial velocity \(u = 0\):
\[ 0 = 0 + 2(-2g \sin \theta)L \]
This gives us the final velocity at the end of the rough portion.

### Step 7: Motion on the Smooth Portion
On the smooth portion, the only force acting is the component of gravity:
\[ F = mg \sin \theta \]
The acceleration on this portion is:
\[ a = g \sin \theta \]

### Step 8: Using the Work-Energy Principle
The work done by gravity on the entire incline must equal the work done against friction on the rough portion:
\[ \text{Work done by gravity} = mg(h_{\text{total}}) = mg(L \sin \theta + nL \sin \theta) \]
\[ \text{Work done against friction} = f \cdot L = 3 \tan \theta \cdot mg \cos \theta \cdot L \]

### Step 9: Setting Up the Equation
Equating the work done by gravity and work done against friction:
\[ mg(L + nL) \sin \theta = 3 \tan \theta \cdot mg \cos \theta \cdot L \]
Cancelling \(mg\) from both sides:
\[ (L + nL) \sin \theta = 3 \tan \theta \cdot \cos \theta \cdot L \]
Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\[ (L + nL) \sin \theta = 3 \sin \theta \cdot L \]

### Step 10: Solving for \(n\)
Dividing both sides by \(\sin \theta\) (assuming \(\sin \theta \neq 0\)):
\[ L + nL = 3L \]
\[ 1 + n = 3 \]
\[ n = 2 \]

### Final Answer
The value of \(n\) is \(2\).

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