If a solid sphere of mass 1 kg and radius 0.1 m rolls without slipping at a uniform velocity of 1 m/s along a straight line on a horizontal floor, the kinetic energy is

Options

A. a.`(7)/(5) J`
B. b.`(2)/(5)J`
C. c.`(7)/(10)J`
D. d.1 J

Explanation

To find the total kinetic energy of a solid sphere rolling without slipping, we need to consider both its translational and rotational kinetic energy.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the sphere, \( m = 1 \, \text{kg} \)
- Radius of the sphere, \( r = 0.1 \, \text{m} \)
- Velocity of the sphere, \( v = 1 \, \text{m/s} \)

2. **Calculate the translational kinetic energy (TKE):**
The formula for translational kinetic energy is:
\[ TKE = \frac{1}{2} mv^2 \]
Substituting the values:
\[ TKE = \frac{1}{2} \times 1 \, \text{kg} \times (1 \, \text{m/s})^2 = \frac{1}{2} \times 1 \times 1 = 0.5 \, \text{J} \]

3. **Calculate the moment of inertia (I) of the sphere:**
The moment of inertia for a solid sphere is given by:
\[ I = \frac{2}{5} m r^2 \]
Substituting the values:
\[ I = \frac{2}{5} \times 1 \, \text{kg} \times (0.1 \, \text{m})^2 = \frac{2}{5} \times 1 \times 0.01 = \frac{2}{500} = 0.004 \, \text{kg m}^2 \]

4. **Relate angular velocity (\(\omega\)) to linear velocity (v):**
Since the sphere rolls without slipping, we have:
\[ \omega = \frac{v}{r} \]
Substituting the values:
\[ \omega = \frac{1 \, \text{m/s}}{0.1 \, \text{m}} = 10 \, \text{rad/s} \]

5. **Calculate the rotational kinetic energy (RKE):**
The formula for rotational kinetic energy is:
\[ RKE = \frac{1}{2} I \omega^2 \]
Substituting the values:
\[ RKE = \frac{1}{2} \times 0.004 \, \text{kg m}^2 \times (10 \, \text{rad/s})^2 = \frac{1}{2} \times 0.004 \times 100 = 0.2 \, \text{J} \]

6. **Calculate the total kinetic energy (TKE + RKE):**
\[ \text{Total Kinetic Energy} = TKE + RKE = 0.5 \, \text{J} + 0.2 \, \text{J} = 0.7 \, \text{J} \]

### Final Answer:
The total kinetic energy of the rolling solid sphere is \( 0.7 \, \text{J} \).