A disc rolls down a plane of length L and inclined at angle `theta`, without slipping. Its velocity on reaching the bottom will be :-

Options

A. `sqrt((4gL sin theta)//3)`
B. `sqrt((2gLsin theta)//3)`
C. `sqrt((10gLsin theta)//7)`
D. `sqrt((4gLsin theta))`

Answer

To solve the problem of a disc rolling down an inclined plane without slipping, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Identify the energies at the top and bottom of the incline
At the top of the incline (point A), the disc has potential energy and no kinetic energy (since it starts from rest). At the bottom of the incline (point B), the disc has kinetic energy due to both translation and rotation.

- **Potential Energy at A**: \( PE_A = mgh \)
- **Kinetic Energy at B**: \( KE_B = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \)

### Step 2: Determine the height (h) in terms of L and θ
The height (h) can be expressed in terms of the length of the incline (L) and the angle (θ):
\[ h = L \sin(\theta) \]

### Step 3: Write the energy conservation equation
According to the conservation of energy:
\[ PE_A = KE_B \]
Substituting the expressions for potential and kinetic energy:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

### Step 4: Substitute for I and ω
For a disc, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} m r^2 \]
In pure rolling motion, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ \omega = \frac{v}{r} \]
Substituting \( \omega \) into the kinetic energy equation:
\[ KE_B = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ KE_B = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

### Step 5: Substitute h into the energy equation
Now substituting \( h \) into the energy conservation equation:
\[ mg(L \sin(\theta)) = \frac{3}{4} mv^2 \]

### Step 6: Cancel mass and solve for v
Canceling \( m \) from both sides:
\[ gL \sin(\theta) = \frac{3}{4} v^2 \]
Now, solving for \( v^2 \):
\[ v^2 = \frac{4gL \sin(\theta)}{3} \]

### Step 7: Take the square root to find v
Taking the square root gives:
\[ v = \sqrt{\frac{4gL \sin(\theta)}{3}} \]

### Final Answer
Thus, the velocity of the disc on reaching the bottom of the incline is:
\[ v = \frac{2\sqrt{gL \sin(\theta)}}{\sqrt{3}} \]