An object takes 1 second to slide down a rough `45^(@)` inclined plane. The time taken to slide down a smooth `30^(@)` inclined plane having the same slope length is (`mu = 0.5`)

Options

A. `sqrt2`s
B. `(1)/(sqrt2)` s
C. `(1)/(2 sqrt2) s`
D. `2^(-1//4) s`

Answer

To solve the problem, we will analyze the motion of an object sliding down two inclined planes: one rough with a 45-degree angle and the other smooth with a 30-degree angle. We will find the time taken to slide down the smooth incline.

### Step 1: Analyze the rough incline (45 degrees)
1. **Identify forces**: The gravitational force acting down the incline is given by \( mg \sin(45^\circ) \), and the normal force is \( mg \cos(45^\circ) \).
2. **Frictional force**: The frictional force \( f \) can be expressed as:
\[ f = \mu \cdot N = \mu \cdot mg \cos(45^\circ) = \mu \cdot mg \cdot \frac{1}{\sqrt{2}} \]
where \( \mu = 0.5 \).

3. **Net force and acceleration**: The net force acting on the object along the incline is:
\[ mg \sin(45^\circ) - f = ma \]
Substituting for \( f \):
\[ mg \sin(45^\circ) - \mu mg \cos(45^\circ) = ma \]
Simplifying this gives:
\[ mg \cdot \frac{1}{\sqrt{2}} - 0.5 \cdot mg \cdot \frac{1}{\sqrt{2}} = ma \]
\[ mg \cdot \frac{1}{\sqrt{2}} \left(1 - 0.5\right) = ma \]
\[ mg \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = ma \]
Canceling \( m \) from both sides, we find:
\[ a = \frac{g}{4\sqrt{2}} \]

### Step 2: Calculate distance using time
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \) and \( t = 1 \) second:
\[ s = 0 + \frac{1}{2} \cdot \frac{g}{4\sqrt{2}} \cdot (1^2) \]
\[ s = \frac{g}{8\sqrt{2}} \]

### Step 3: Analyze the smooth incline (30 degrees)
1. **Forces on the smooth incline**: The only force acting down the incline is \( mg \sin(30^\circ) \) since there is no friction.
2. **Acceleration**: The acceleration \( a \) is given by:
\[ a = g \sin(30^\circ) = g \cdot \frac{1}{2} = \frac{g}{2} \]

### Step 4: Calculate time for the smooth incline
Using the same distance \( s \):
\[ s = ut + \frac{1}{2} a t^2 \]
Again, \( u = 0 \):
\[ s = 0 + \frac{1}{2} \cdot \frac{g}{2} \cdot t^2 \]
\[ s = \frac{g}{4} t^2 \]
Setting the distances equal:
\[ \frac{g}{8\sqrt{2}} = \frac{g}{4} t^2 \]
Canceling \( g \) from both sides:
\[ \frac{1}{8\sqrt{2}} = \frac{1}{4} t^2 \]
Multiplying both sides by 4:
\[ \frac{1}{2\sqrt{2}} = t^2 \]
Taking the square root:
\[ t = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} \text{ seconds} \]

### Final Answer
The time taken to slide down the smooth 30-degree incline is \( \frac{1}{2} \) seconds.

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