A body is released from the top of a smooth inclined plane of inclination `theta`. It reaches the bottom with velocity `v`. If the angle of inclina-tion is doubled for the same length of the plane, what will be the velocity of the body on reach ing the ground .

Options

A. v
B. 2v
C. `[2 cos theta]^(1//2)v`
D. `[2 sin theta]^(1//2)v`

Explanation

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the problem
We have a body released from the top of a smooth inclined plane with an angle of inclination \( \theta \). It reaches the bottom with a velocity \( v \). We need to find the velocity when the angle of inclination is doubled to \( 2\theta \) while keeping the length of the incline the same.

### Step 2: Analyze the first scenario (angle \( \theta \))
Using the equation of motion, we know that:
\[ v^2 = u^2 + 2as \]
where:
- \( u = 0 \) (initial velocity)
- \( a = g \sin \theta \) (acceleration down the incline)
- \( s \) is the length of the incline.

Substituting these values, we get:
\[ v^2 = 0 + 2(g \sin \theta)s \]
Thus, we can write:
\[ v^2 = 2gs \sin \theta \]

### Step 3: Analyze the second scenario (angle \( 2\theta \))
For the second scenario where the angle is \( 2\theta \), the acceleration down the incline becomes:
\[ a' = g \sin(2\theta) \]
Using the double angle identity, we can express \( \sin(2\theta) \) as:
\[ \sin(2\theta) = 2 \sin \theta \cos \theta \]
Now, substituting this into the equation of motion:
\[ v'^2 = u^2 + 2a's \]
Again, \( u = 0 \), so:
\[ v'^2 = 0 + 2(g \sin(2\theta))s \]
Substituting for \( \sin(2\theta) \):
\[ v'^2 = 2g(2 \sin \theta \cos \theta)s \]
This simplifies to:
\[ v'^2 = 4gs \sin \theta \cos \theta \]

### Step 4: Relate \( v' \) to \( v \)
From the first scenario, we know:
\[ v^2 = 2gs \sin \theta \]
Now, substituting this into the equation for \( v' \):
\[ v'^2 = 4 \cdot \frac{v^2}{2} \cdot \cos \theta = 2v^2 \cos \theta \]
Taking the square root gives:
\[ v' = \sqrt{2v^2 \cos \theta} = v \sqrt{2 \cos \theta} \]

### Final Answer
Thus, the velocity of the body when the angle of inclination is doubled is:
\[ v' = v \sqrt{2 \cos \theta} \]