A boy standing on a weighing machine observes his weight as 200 N. When he suddenly jumpes upwards, his friend notices that the reading increased to 400 N. The acceleration by which the boy jumped will be-

Options

A. (a)`9.8 m//s^(2)`
B. (b)`29.4 m//s^(2)`
C. (c)`4.9m//s^(2)`
D. (d)`14.7 m//s^(2)`

Answer

To solve the problem, we need to analyze the forces acting on the boy when he jumps and how they relate to the readings on the weighing machine.

### Step-by-Step Solution:

1. **Understanding the Initial Condition:**
- The boy is standing on the weighing machine, and the reading shows 200 N. This indicates that the weight of the boy (W) is 200 N.
- Weight (W) is given by the formula:
\[ W = mg \]
- Here, \( m \) is the mass of the boy and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

2. **Calculating the Mass of the Boy:**
- From the weight equation, we can express the mass of the boy as:
\[ m = \frac{W}{g} = \frac{200 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 20.41 \, \text{kg} \]

3. **Understanding the Jump:**
- When the boy jumps, the reading on the weighing machine increases to 400 N. This is the apparent weight when he is accelerating upwards.
- The apparent weight (normal force, \( N \)) when the boy jumps can be expressed as:
\[ N = mg + ma \]
- Here, \( a \) is the upward acceleration of the boy.

4. **Setting Up the Equation:**
- We know that when he jumps, the normal force (N) becomes 400 N. Therefore, we can write:
\[ 400 \, \text{N} = mg + ma \]
- Substituting \( mg \) with 200 N (the boy's weight):
\[ 400 \, \text{N} = 200 \, \text{N} + ma \]

5. **Solving for Acceleration (a):**
- Rearranging the equation gives us:
\[ ma = 400 \, \text{N} - 200 \, \text{N} = 200 \, \text{N} \]
- Now, substituting \( m \) (which we calculated as approximately 20.41 kg):
\[ 20.41 \, \text{kg} \cdot a = 200 \, \text{N} \]
- Solving for \( a \):
\[ a = \frac{200 \, \text{N}}{20.41 \, \text{kg}} \approx 9.8 \, \text{m/s}^2 \]

6. **Conclusion:**
- The acceleration by which the boy jumped is approximately \( 9.8 \, \text{m/s}^2 \).

### Final Answer:
The acceleration by which the boy jumped is \( 9.8 \, \text{m/s}^2 \).