A hollow sphere and a solid sphere both having the same mass of 5 kg and radius 10 m are initially at rest. If they are made to roll down the same inclined plane without slipping, the ratio of their speeds when they reach the bottom of the plane `(V_("hollow"))/(V_("solid"))`, will be

Options

A. 1
B. `(7)/(sqrt(10))`
C. `(7)/(sqrt(12))`
D. `(21)/(sqrt(25))`

Answer

To solve the problem of finding the ratio of the speeds of a hollow sphere and a solid sphere when they roll down an inclined plane, we can follow these steps:

### Step 1: Understand the Problem
Both spheres have the same mass (5 kg) and radius (10 m) and are initially at rest. We need to find the ratio of their speeds when they reach the bottom of the inclined plane.

### Step 2: Apply Conservation of Energy
When the spheres roll down the inclined plane, their potential energy is converted into kinetic energy. The potential energy at the top is given by:
\[ PE = mgh \]
where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height of the incline.

At the bottom of the incline, the total kinetic energy (KE) is the sum of translational and rotational kinetic energy:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
where \( v \) is the linear speed, \( I \) is the moment of inertia, and \( \omega \) is the angular speed. Since the spheres roll without slipping, we have:
\[ v = \omega r \]
Thus, \( \omega = \frac{v}{r} \).

### Step 3: Write the Moment of Inertia
For the hollow sphere:
\[ I_{hollow} = \frac{2}{3} m r^2 \]
For the solid sphere:
\[ I_{solid} = \frac{2}{5} m r^2 \]

### Step 4: Set Up the Energy Equation
Using conservation of energy:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \left(\frac{v}{r}\right)^2 \]

Substituting for \( I \):
- For the hollow sphere:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{3} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 \]
\[ mgh = \left(\frac{1}{2} + \frac{1}{3}\right) mv^2 \]
\[ mgh = \frac{5}{6} mv^2 \]

- For the solid sphere:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]
\[ mgh = \left(\frac{1}{2} + \frac{1}{5}\right) mv^2 \]
\[ mgh = \frac{7}{10} mv^2 \]

### Step 5: Solve for Velocity
Now, solving for \( v \):
- For the hollow sphere:
\[ gh = \frac{5}{6} v^2 \]
\[ v_{hollow}^2 = \frac{6gh}{5} \]
\[ v_{hollow} = \sqrt{\frac{6gh}{5}} \]

- For the solid sphere:
\[ gh = \frac{7}{10} v^2 \]
\[ v_{solid}^2 = \frac{10gh}{7} \]
\[ v_{solid} = \sqrt{\frac{10gh}{7}} \]

### Step 6: Find the Ratio of Speeds
Now, we find the ratio of the speeds:
\[ \frac{v_{hollow}}{v_{solid}} = \frac{\sqrt{\frac{6gh}{5}}}{\sqrt{\frac{10gh}{7}}} \]
\[ = \sqrt{\frac{6gh}{5} \cdot \frac{7}{10gh}} \]
\[ = \sqrt{\frac{42}{50}} \]
\[ = \sqrt{\frac{21}{25}} \]
\[ = \frac{\sqrt{21}}{5} \]

### Final Answer
Thus, the ratio of their speeds when they reach the bottom of the inclined plane is:
\[ \frac{V_{hollow}}{V_{solid}} = \sqrt{\frac{21}{25}} \]