Torque of equal magnitude is applied to a solid cylinder and a solid sphere, both having the same mass and radius. Both of them are free to rotate about their axis of symmetry. If `alpha _(1)` and `alpha_(2)` are the angular accelerations of the cylinder and the sphere respectively, then the ratio `(alpha_(1))/(alpha_(2))` will be

Options

A. `(5)/(2)`
B. `(5)/(4)`
C. `(4)/(5)`
D. `(2)/(5)`

Explanation

To find the ratio of angular accelerations \(\frac{\alpha_1}{\alpha_2}\) for a solid cylinder and a solid sphere subjected to the same torque, we can follow these steps:

### Step 1: Identify the moment of inertia for both objects
- For a solid cylinder, the moment of inertia \(I_1\) is given by:
\[ I_1 = \frac{1}{2} m r^2 \]
- For a solid sphere, the moment of inertia \(I_2\) is given by:
\[ I_2 = \frac{2}{5} m r^2 \]

### Step 2: Write the relationship between torque and angular acceleration
The torque \(\tau\) applied to each object is related to its moment of inertia and angular acceleration by the equation:
\[ \tau = I \alpha \]
Thus, for the cylinder and sphere, we have:
- For the cylinder:
\[ \tau = I_1 \alpha_1 \quad \Rightarrow \quad \tau = \frac{1}{2} m r^2 \alpha_1 \]
- For the sphere:
\[ \tau = I_2 \alpha_2 \quad \Rightarrow \quad \tau = \frac{2}{5} m r^2 \alpha_2 \]

### Step 3: Set the torques equal to each other
Since the torque applied to both objects is of equal magnitude:
\[ \frac{1}{2} m r^2 \alpha_1 = \frac{2}{5} m r^2 \alpha_2 \]

### Step 4: Simplify the equation
We can cancel \(m\) and \(r^2\) from both sides (assuming they are non-zero):
\[ \frac{1}{2} \alpha_1 = \frac{2}{5} \alpha_2 \]

### Step 5: Solve for the ratio \(\frac{\alpha_1}{\alpha_2}\)
Rearranging the equation gives:
\[ \alpha_1 = \frac{2}{5} \cdot 2 \alpha_2 = \frac{4}{5} \alpha_2 \]
Thus, the ratio \(\frac{\alpha_1}{\alpha_2}\) is:
\[ \frac{\alpha_1}{\alpha_2} = \frac{4}{5} \]

### Final Answer
The ratio of angular accelerations is:
\[ \frac{\alpha_1}{\alpha_2} = \frac{4}{5} \]