A spherical shell and a solid cylinder of same radius rolls down an inclined plane. The ratio of their accelerations will be:-

Options

A. `15 : 14`
B. `9 : 10 `
C. `2 :3`
D. `3:5`

Answer

To find the ratio of the accelerations of a spherical shell and a solid cylinder rolling down an inclined plane, we can follow these steps:

### Step 1: Identify the Moment of Inertia
The moment of inertia (I) for each object is given by:
- For a spherical shell: \( I = \frac{2}{3} m r^2 \)
- For a solid cylinder: \( I = \frac{1}{2} m r^2 \)

### Step 2: Write the Formula for Acceleration
The acceleration of the center of mass (a) for a rolling object down an incline can be expressed as:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{m r^2}} \]
where \( g \) is the acceleration due to gravity, \( \theta \) is the angle of the incline, \( m \) is the mass, and \( r \) is the radius.

### Step 3: Calculate the Acceleration for Each Object
**For the spherical shell:**
\[ a_{\text{spherical}} = \frac{g \sin \theta}{1 + \frac{\frac{2}{3} m r^2}{m r^2}} = \frac{g \sin \theta}{1 + \frac{2}{3}} = \frac{g \sin \theta}{\frac{5}{3}} = \frac{3g \sin \theta}{5} \]

**For the solid cylinder:**
\[ a_{\text{cylinder}} = \frac{g \sin \theta}{1 + \frac{\frac{1}{2} m r^2}{m r^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]

### Step 4: Find the Ratio of Accelerations
Now, we need to find the ratio of the accelerations:
\[ \text{Ratio} = \frac{a_{\text{spherical}}}{a_{\text{cylinder}}} = \frac{\frac{3g \sin \theta}{5}}{\frac{2g \sin \theta}{3}} \]

### Step 5: Simplify the Ratio
Cancelling \( g \sin \theta \) from the numerator and denominator:
\[ \text{Ratio} = \frac{3}{5} \times \frac{3}{2} = \frac{9}{10} \]

### Final Result
Thus, the ratio of the accelerations of the spherical shell to the solid cylinder is:
\[ \text{Ratio} = \frac{9}{10} \]