A thin rod of mass `m` and length `l` is hinged at the lower end to a level floor and stands vertically. Then its upper end will strike the floor with a velocity given by:

Options

A. `sqrt(2gl)`
B. `sqrt(3gl)`
C. `sqrt(5gl)`
D. `sqrt(mgl)`

Explanation

To solve the problem of determining the velocity with which the upper end of a thin rod strikes the floor when it is hinged at the lower end and stands vertically, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the System**:
- We have a thin rod of mass \( m \) and length \( l \) that is hinged at one end and initially stands vertically.

2. **Determine the Center of Mass**:
- The center of mass of the rod is located at a distance of \( \frac{l}{2} \) from the hinge (the lower end).

3. **Calculate Initial Potential Energy**:
- The potential energy (PE) of the center of mass when the rod is vertical is given by:
\[ PE = mgh = mg\left(\frac{l}{2}\right) = \frac{mgl}{2} \]

4. **Determine the Moment of Inertia**:
- The moment of inertia \( I \) of the rod about the hinge (lower end) is:
\[ I = \frac{1}{3} ml^2 \]

5. **Relate Kinetic Energy to Angular Velocity**:
- When the rod falls, the potential energy converts into kinetic energy (KE). The kinetic energy of the rod when it is falling is given by:
\[ KE = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{3} ml^2\right) \omega^2 = \frac{ml^2 \omega^2}{6} \]

6. **Apply Conservation of Energy**:
- According to the conservation of energy, the initial potential energy equals the final kinetic energy:
\[ \frac{mgl}{2} = \frac{ml^2 \omega^2}{6} \]

7. **Simplify the Equation**:
- Cancel \( m \) from both sides:
\[ \frac{gl}{2} = \frac{l^2 \omega^2}{6} \]
- Rearranging gives:
\[ gl = \frac{l^2 \omega^2}{3} \]
- Therefore:
\[ \omega^2 = \frac{3g}{l} \]

8. **Relate Angular Velocity to Linear Velocity**:
- The linear velocity \( v \) of the upper end of the rod is related to the angular velocity \( \omega \) by:
\[ v = \omega l \]
- Substituting \( \omega \):
\[ v = l \sqrt{\frac{3g}{l}} = \sqrt{3gl} \]

9. **Final Result**:
- The velocity with which the upper end of the rod strikes the floor is:
\[ v = \sqrt{3gl} \]