A conductor of mass `m=5kg` and length `l=4m` is placed on horizontal surface and a uniform magnetic field `B=2` Tesla exist parallel to horizontal surface but perpendicular to the conductor. Suddenly, a certain amount of charge is passed through it, then it is found to jump to a height `h=5m`. Then, find the amount of charge that passes through the conductor in coulomb. (take `g=10m//s^(2)`)

Explanation

To solve the problem, we need to find the amount of charge that passes through the conductor when it jumps to a height of 5 meters in a magnetic field. We will use the principles of force, momentum, and energy conservation.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Conductor:**
The conductor experiences a magnetic force when a current flows through it in the presence of a magnetic field. The force \( F \) on the conductor can be expressed as:
\[ F = I \cdot L \cdot B \]
where \( I \) is the current, \( L \) is the length of the conductor, and \( B \) is the magnetic field strength.

2. **Relate Force to Change in Momentum:**
The force also relates to the change in momentum of the conductor. If \( \Delta t \) is the time duration for which the current flows, the change in momentum \( \Delta p \) can be expressed as:
\[ F \cdot \Delta t = m \cdot v_1 \]
where \( m \) is the mass of the conductor and \( v_1 \) is the final velocity after the force has acted.

3. **Use Conservation of Energy:**
When the conductor jumps to a height \( h \), its potential energy at the peak is equal to the kinetic energy it had just before jumping:
\[ mgh = \frac{1}{2} mv_1^2 \]
From this equation, we can solve for \( v_1 \):
\[ v_1 = \sqrt{2gh} \]

4. **Substituting \( v_1 \) into the Momentum Equation:**
Substitute \( v_1 \) back into the momentum equation:
\[ F \cdot \Delta t = m \cdot \sqrt{2gh} \]

5. **Substituting the Expression for Force:**
Substitute \( F = I \cdot L \cdot B \) into the equation:
\[ I \cdot L \cdot B \cdot \Delta t = m \cdot \sqrt{2gh} \]

6. **Express Current in Terms of Charge:**
The current \( I \) can be expressed in terms of charge \( Q \) and time \( \Delta t \):
\[ I = \frac{Q}{\Delta t} \]
Substitute this into the equation:
\[ \frac{Q}{\Delta t} \cdot L \cdot B \cdot \Delta t = m \cdot \sqrt{2gh} \]
This simplifies to:
\[ Q \cdot L \cdot B = m \cdot \sqrt{2gh} \]

7. **Solve for Charge \( Q \):**
Rearranging gives:
\[ Q = \frac{m \cdot \sqrt{2gh}}{L \cdot B} \]

8. **Substituting Known Values:**
Now, substitute the known values:
- \( m = 5 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( h = 5 \, \text{m} \)
- \( L = 4 \, \text{m} \)
- \( B = 2 \, \text{T} \)

\[ Q = \frac{5 \cdot \sqrt{2 \cdot 10 \cdot 5}}{4 \cdot 2} \]

9. **Calculating \( \sqrt{2gh} \):**
\[ \sqrt{2gh} = \sqrt{2 \cdot 10 \cdot 5} = \sqrt{100} = 10 \]

10. **Final Calculation:**
Substitute back into the equation for \( Q \):
\[ Q = \frac{5 \cdot 10}{8} = \frac{50}{8} = \frac{25}{4} \, \text{C} \]

### Final Answer:
The amount of charge that passes through the conductor is \( \frac{25}{4} \, \text{C} \).