A solid sphere of mass 2 kg rolls up a `30^(@)` incline with an initial speed of `10 ms^(-1)`. The maximum height reached by the sphere is `(g = 10 ms^(-2))`

Options

A. 3.5 m
B. 7 m
C. 10.5 m
D. 14 m

Explanation

To solve the problem of a solid sphere rolling up a 30-degree incline with an initial speed of 10 m/s, we will use the principles of conservation of energy.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Mass of the sphere, \( m = 2 \, \text{kg} \)
- Initial speed, \( v = 10 \, \text{m/s} \)
- Angle of incline, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Initial Kinetic Energy:**
The total kinetic energy (KE) of the rolling sphere is the sum of translational kinetic energy and rotational kinetic energy.
- Translational KE:
\[ KE_{\text{trans}} = \frac{1}{2} mv^2 = \frac{1}{2} \times 2 \times (10)^2 = 100 \, \text{J} \]
- Moment of inertia \( I \) for a solid sphere about its center of mass is given by:
\[ I = \frac{2}{5} m r^2 \]
- The angular velocity \( \omega \) is related to the translational velocity \( v \) by \( \omega = \frac{v}{r} \).
- Rotational KE:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) = \frac{1}{5} mv^2 \]
- Thus, the total kinetic energy is:
\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = 100 + \frac{1}{5} \times 2 \times (10)^2 = 100 + 40 = 140 \, \text{J} \]

3. **Set Up the Energy Conservation Equation:**
At the maximum height \( h \), all kinetic energy is converted into potential energy (PE):
\[ KE_{\text{total}} = PE \]
\[ 140 = mgh \]

4. **Substitute Values into the Equation:**
\[ 140 = 2 \times 10 \times h \]
\[ 140 = 20h \]

5. **Solve for Height \( h \):**
\[ h = \frac{140}{20} = 7 \, \text{m} \]

### Final Answer:
The maximum height reached by the sphere is \( h = 7 \, \text{m} \).