A body of mass `m` slides down an incline and reaches the bottom with a velocity `v`. If the same mass were in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been

Options

A. v
B. `sqrt2v`
C. `v/sqrt2`
D. `sqrt(2/5)v`

Answer

To solve the problem, we need to analyze the motion of both the sliding block and the rolling ring down the incline. Here's the step-by-step solution:

### Step 1: Analyze the sliding block
1. When the block of mass `m` slides down the incline from a height `H`, it converts its potential energy into kinetic energy.
2. The potential energy (PE) at the top is given by:
\[ PE = mgh \]
3. At the bottom, all this potential energy is converted into kinetic energy (KE), which is given by:
\[ KE = \frac{1}{2} mv^2 \]
4. Setting the potential energy equal to the kinetic energy:
\[ mgh = \frac{1}{2} mv^2 \]
5. Canceling `m` from both sides:
\[ gh = \frac{1}{2} v^2 \]
6. Rearranging gives:
\[ v^2 = 2gh \]
7. Therefore, the velocity `v` of the block at the bottom is:
\[ v = \sqrt{2gh} \]
This is our **Equation 1**.

### Step 2: Analyze the rolling ring
1. When the ring rolls down the incline, it also has potential energy at the top:
\[ PE = mgh \]
2. However, when it reaches the bottom, its energy is divided into translational kinetic energy and rotational kinetic energy:
\[ KE_{trans} = \frac{1}{2} mv_1^2 \]
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]
3. For a ring, the moment of inertia `I` is given by:
\[ I = mr^2 \]
4. The relationship between linear velocity `v_1` and angular velocity `\omega` is:
\[ v_1 = r\omega \quad \Rightarrow \quad \omega = \frac{v_1}{r} \]
5. Substituting `I` and `\omega` into the rotational kinetic energy:
\[ KE_{rot} = \frac{1}{2} (mr^2) \left(\frac{v_1}{r}\right)^2 = \frac{1}{2} mv_1^2 \]
6. Therefore, the total kinetic energy at the bottom is:
\[ KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2} mv_1^2 + \frac{1}{2} mv_1^2 = mv_1^2 \]
7. Setting the potential energy equal to the total kinetic energy:
\[ mgh = mv_1^2 \]
8. Canceling `m` from both sides:
\[ gh = v_1^2 \]
9. Rearranging gives:
\[ v_1^2 = gh \]
10. Now, substituting the expression for `h` from **Equation 1**:
\[ h = \frac{v^2}{2g} \]
11. Thus, we have:
\[ v_1^2 = g \left(\frac{v^2}{2g}\right) = \frac{v^2}{2} \]
12. Therefore, the velocity `v_1` of the ring at the bottom is:
\[ v_1 = \frac{v}{\sqrt{2}} \]

### Final Result
The velocity of the ring at the bottom of the incline is:
\[ v_1 = \frac{v}{\sqrt{2}} \]