A body takes `1(1/3)` times as much time to slide down a rough inclined plane as it takes to slide down an identical bust smooth inclined plane. If the angle of inclination is 45°, find the coefficient of friction.

Options

A. `1/16`
B. `3/16`
C. `5/16`
D. `7/16`

Answer

To solve the problem, we need to find the coefficient of friction (μ) for a body sliding down a rough inclined plane compared to a smooth inclined plane. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between time taken on both planes
Let \( t_s \) be the time taken to slide down the smooth inclined plane and \( t_r \) be the time taken to slide down the rough inclined plane. According to the problem, we have:
\[ t_r = \frac{4}{3} t_s \]

### Step 2: Analyze the motion on the smooth inclined plane
For a smooth inclined plane, the acceleration \( a_s \) of the body can be calculated using the formula:
\[ a_s = g \sin \theta \]
where \( g \) is the acceleration due to gravity and \( \theta \) is the angle of inclination. For \( \theta = 45^\circ \):
\[ a_s = g \sin 45^\circ = g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}} \]

### Step 3: Calculate the time taken on the smooth inclined plane
Using the equation of motion \( s = \frac{1}{2} a t^2 \), we can express the time taken on the smooth plane:
\[ s = \frac{1}{2} a_s t_s^2 \implies t_s^2 = \frac{2s}{a_s} = \frac{2s \sqrt{2}}{g} \]
Thus, the time taken on the smooth inclined plane is:
\[ t_s = \sqrt{\frac{2s \sqrt{2}}{g}} \]

### Step 4: Analyze the motion on the rough inclined plane
For the rough inclined plane, the net acceleration \( a_r \) is given by:
\[ a_r = g \sin \theta - g \mu \cos \theta \]
Substituting \( \theta = 45^\circ \):
\[ a_r = g \cdot \frac{1}{\sqrt{2}} - g \mu \cdot \frac{1}{\sqrt{2}} = \frac{g(1 - \mu)}{\sqrt{2}} \]

### Step 5: Calculate the time taken on the rough inclined plane
Using the same distance \( s \):
\[ s = \frac{1}{2} a_r t_r^2 \implies t_r^2 = \frac{2s}{a_r} = \frac{2s \sqrt{2}}{g(1 - \mu)} \]
Thus, the time taken on the rough inclined plane is:
\[ t_r = \sqrt{\frac{2s \sqrt{2}}{g(1 - \mu)}} \]

### Step 6: Set up the equation using the time relationship
From the relationship \( t_r = \frac{4}{3} t_s \), we can substitute the expressions for \( t_r \) and \( t_s \):
\[ \sqrt{\frac{2s \sqrt{2}}{g(1 - \mu)}} = \frac{4}{3} \sqrt{\frac{2s \sqrt{2}}{g}} \]

### Step 7: Square both sides and simplify
Squaring both sides gives:
\[ \frac{2s \sqrt{2}}{g(1 - \mu)} = \frac{16}{9} \cdot \frac{2s \sqrt{2}}{g} \]
Cancelling \( \frac{2s \sqrt{2}}{g} \) from both sides leads to:
\[ \frac{1}{1 - \mu} = \frac{16}{9} \]

### Step 8: Solve for the coefficient of friction \( \mu \)
Rearranging gives:
\[ 1 - \mu = \frac{9}{16} \implies \mu = 1 - \frac{9}{16} = \frac{7}{16} \]

### Final Answer
The coefficient of friction \( \mu \) is:
\[ \mu = \frac{7}{16} \]