A solid cylinder rolls down an inclined plane of height `3m` and reaches the bottom of plane with angular velocity of `2sqrt2 rad//s`. The radius of cylinder must be [take `g=10m//s^(2)`]

Options

A. `5 cm`
B. `0.5 cm`
C. `sqrt10 cm`
D. `sqrt5 m`

Explanation

To solve the problem, we need to find the radius of the solid cylinder given that it rolls down an inclined plane of height \( h = 3 \, \text{m} \) and reaches the bottom with an angular velocity \( \omega = 2\sqrt{2} \, \text{rad/s} \). We will use the principles of energy conservation and the relationship between linear and angular velocity.

### Step-by-Step Solution:

1. **Calculate the potential energy at the top of the incline:**
The potential energy (PE) of the cylinder at the height \( h \) is given by:
\[
PE = mgh
...