A ball of mass `1g` and charge `10^(-8) C` moves from a point `A`. Where potential is `600` volt to the point `B` where potential is zero. Velocity of the ball at the point `B` is `20 cm//s`. The velocity of the ball at the point `A` will be

Options

A. `22.8 " cms"^(-1)`
B. `228 " cms"^(-1)`
C. `16.8 " ms"^(-1)`
D. `168 " ms"^(-1)`

Answer

To find the velocity of the ball at point A, we can use the principle of conservation of energy, which states that the change in kinetic energy of the ball will be equal to the change in potential energy as it moves from point A to point B.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the ball, \( m = 1 \, \text{g} = 1 \times 10^{-3} \, \text{kg} \)
- Charge of the ball, \( q = 10^{-8} \, \text{C} \)
- Potential at point A, \( V_A = 600 \, \text{V} \)
- Potential at point B, \( V_B = 0 \, \text{V} \)
- Velocity at point B, \( v_B = 20 \, \text{cm/s} = 0.2 \, \text{m/s} \)

2. **Calculate Change in Potential Energy:**
\[ \Delta V = V_A - V_B = 600 \, \text{V} - 0 \, \text{V} = 600 \, \text{V} \]
The change in potential energy (\( \Delta PE \)) is given by:
\[ \Delta PE = q \cdot \Delta V = (10^{-8} \, \text{C}) \cdot (600 \, \text{V}) = 6 \times 10^{-6} \, \text{J} \]

3. **Calculate Change in Kinetic Energy:**
The kinetic energy at point B (\( KE_B \)) is:
\[ KE_B = \frac{1}{2} m v_B^2 = \frac{1}{2} (1 \times 10^{-3} \, \text{kg}) (0.2 \, \text{m/s})^2 = \frac{1}{2} (1 \times 10^{-3}) (0.04) = 2 \times 10^{-5} \, \text{J} \]

4. **Apply Conservation of Energy:**
The change in kinetic energy (\( \Delta KE \)) is equal to the change in potential energy:
\[ \Delta KE = \Delta PE \]
Therefore,
\[ KE_A - KE_B = \Delta PE \]
Rearranging gives:
\[ KE_A = KE_B + \Delta PE \]

5. **Substituting Values:**
\[ KE_A = 2 \times 10^{-5} \, \text{J} + 6 \times 10^{-6} \, \text{J} = 2.6 \times 10^{-5} \, \text{J} \]

6. **Calculate Velocity at Point A:**
\[ KE_A = \frac{1}{2} m v_A^2 \]
Rearranging gives:
\[ v_A^2 = \frac{2 KE_A}{m} \]
Substituting the values:
\[ v_A^2 = \frac{2 (2.6 \times 10^{-5})}{1 \times 10^{-3}} = \frac{5.2 \times 10^{-5}}{1 \times 10^{-3}} = 0.052 \]
Taking the square root:
\[ v_A = \sqrt{0.052} \approx 0.228 \, \text{m/s} = 22.8 \, \text{cm/s} \]

### Final Answer:
The velocity of the ball at point A is approximately \( 22.8 \, \text{cm/s} \).

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