A uniform hollow sphere of mass 400 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 10 cm/s. Its kinetic energy is:

Options

A. (a)`4 xx10^(-3)J`
B. (b)`8xx10^(-3)J`
C. (c)`3.33 xx 10^(-3) J`
D. (d)`3.5 xx 10^(-3) J`

Explanation

To find the kinetic energy of a uniform hollow sphere rolling without slipping, we can follow these steps:

### Step 1: Understand the components of kinetic energy
The total kinetic energy (KE) of a rolling object is the sum of its translational kinetic energy and its rotational kinetic energy. The formula for total kinetic energy is given by:
\[ KE = KE_{\text{translational}} + KE_{\text{rotational}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
where:
- \(m\) is the mass of the sphere,
- \(v\) is the linear velocity of the center of mass,
- \(I\) is the moment of inertia of the sphere,
- \(\omega\) is the angular velocity.

### Step 2: Identify the moment of inertia for a hollow sphere
The moment of inertia \(I\) of a uniform hollow sphere about its center is given by:
\[ I = \frac{2}{3} m r^2 \]
where \(r\) is the radius of the sphere.

### Step 3: Relate angular velocity to linear velocity
For a rolling object without slipping, the relationship between linear velocity \(v\) and angular velocity \(\omega\) is:
\[ \omega = \frac{v}{r} \]

### Step 4: Substitute the values into the kinetic energy formula
Substituting the moment of inertia and the angular velocity into the kinetic energy formula:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{3} m r^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{3} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
\[ KE = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 \]
\[ KE = \left(\frac{1}{2} + \frac{1}{3}\right) mv^2 \]

### Step 5: Find a common denominator and simplify
To combine the fractions:
\[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]
Thus, the total kinetic energy becomes:
\[ KE = \frac{5}{6} mv^2 \]

### Step 6: Plug in the values
Given:
- Mass \(m = 400 \text{ g} = 0.4 \text{ kg}\)
- Speed \(v = 10 \text{ cm/s} = 0.1 \text{ m/s}\)

Now substituting these values into the kinetic energy formula:
\[ KE = \frac{5}{6} \times 0.4 \text{ kg} \times (0.1 \text{ m/s})^2 \]
Calculating this:
\[ KE = \frac{5}{6} \times 0.4 \times 0.01 = \frac{5 \times 0.4 \times 0.01}{6} = \frac{0.02}{6} = \frac{1}{300} \approx 0.00333 \text{ J} \]

### Final Answer
Thus, the kinetic energy of the hollow sphere is approximately:
\[ KE \approx 3.33 \times 10^{-3} \text{ J} \]