A small object of uniform density rolls up a curved surface with an initial velocity v'. It reaches upto to maximum height of `(3v^(2))/(4g)` with respect to the initial position. The object is

Options

A. ring
B. solid sphere
C. hollow sphere
D. disc

Explanation

To solve the problem, we need to analyze the energy conservation of the small object as it rolls up the curved surface. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial and Final Energy States
- Initially, the object has kinetic energy due to its velocity \( v' \) and no potential energy since we consider the starting height as the datum (h = 0).
- The total initial energy \( E_i \) can be expressed as:
\[ E_i = KE_i + PE_i = \frac{1}{2} mv'^2 + 0 = \frac{1}{2} mv'^2 \]

### Step 2: Determine the Final Energy at Maximum Height
- At the maximum height \( h = \frac{3v'^2}{4g} \), the object momentarily comes to rest, so its kinetic energy becomes zero.
- The potential energy at this height is given by:
\[ PE_f = mgh = mg \left(\frac{3v'^2}{4g}\right) = \frac{3}{4} mv'^2 \]
- Therefore, the total final energy \( E_f \) is:
\[ E_f = KE_f + PE_f = 0 + \frac{3}{4} mv'^2 = \frac{3}{4} mv'^2 \]

### Step 3: Apply Conservation of Energy
- According to the conservation of energy principle, the total initial energy must equal the total final energy:
\[ E_i = E_f \]
Substituting the expressions we derived:
\[ \frac{1}{2} mv'^2 = \frac{3}{4} mv'^2 \]

### Step 4: Consider the Rolling Motion
- Since the object is rolling without slipping, we need to account for both translational and rotational kinetic energy. The total kinetic energy when rolling is:
\[ KE = \frac{1}{2} mv'^2 + \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For pure rolling, we have:
\[ \omega = \frac{v'}{r} \]
Thus, \( \omega^2 = \frac{v'^2}{r^2} \).

### Step 5: Substitute and Rearrange
- The total kinetic energy can be rewritten as:
\[ KE = \frac{1}{2} mv'^2 + \frac{1}{2} I \left(\frac{v'^2}{r^2}\right) \]
- Setting this equal to the potential energy at maximum height:
\[ \frac{1}{2} mv'^2 + \frac{1}{2} I \left(\frac{v'^2}{r^2}\right) = \frac{3}{4} mv'^2 \]

### Step 6: Solve for Moment of Inertia
- Factor out \( \frac{v'^2}{2} \):
\[ \frac{v'^2}{2} \left( m + \frac{I}{r^2} \right) = \frac{3}{4} mv'^2 \]
- Dividing both sides by \( v'^2 \) (assuming \( v' \neq 0 \)):
\[ \frac{1}{2} \left( m + \frac{I}{r^2} \right) = \frac{3}{4} m \]
- Rearranging gives:
\[ m + \frac{I}{r^2} = \frac{3}{2} m \]
\[ \frac{I}{r^2} = \frac{3}{2} m - m = \frac{1}{2} m \]
- Therefore, we find:
\[ I = \frac{1}{2} m r^2 \]

### Step 7: Identify the Object
- The moment of inertia \( I = \frac{1}{2} m r^2 \) corresponds to a solid disk. Hence, the object is a disk.

### Conclusion
The object is a **disk** (Option 4).

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