Two boys A and B lift 100 bricks through the same height in 5 minutes and 6 minutes respectively. Then

Options

A. A has more power than B
B. B has more power than A
C. both have same power
D. data insufficient.

Explanation

To solve the problem, we need to determine the power exerted by both boys A and B while lifting the bricks. Let's break down the solution step by step.

### Step 1: Understand the Work Done
The work done (W) in lifting the bricks can be calculated using the formula:
\[ W = n \times m \times g \times h \]
where:
- \( n \) = number of bricks (100 bricks)
- \( m \) = mass of a single brick (let's denote it as \( m \))
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height through which the bricks are lifted (let's denote it as \( h \))

### Step 2: Calculate the Time Taken
- Boy A takes 5 minutes to lift the bricks.
- Boy B takes 6 minutes to lift the bricks.
We need to convert these times into seconds for consistency:
- Time taken by A, \( t_A = 5 \times 60 = 300 \, \text{s} \)
- Time taken by B, \( t_B = 6 \times 60 = 360 \, \text{s} \)

### Step 3: Calculate the Power
Power (P) is defined as the work done per unit time:
\[ P = \frac{W}{t} \]
For both boys, the work done is the same since they are lifting the same number of bricks through the same height. Thus, we can express the power for each boy as follows:

For Boy A:
\[ P_A = \frac{W}{t_A} = \frac{n \times m \times g \times h}{300} \]

For Boy B:
\[ P_B = \frac{W}{t_B} = \frac{n \times m \times g \times h}{360} \]

### Step 4: Compare the Powers
Since the work done \( W \) is the same for both boys, we can compare their powers:
- \( P_A = \frac{W}{300} \)
- \( P_B = \frac{W}{360} \)

To compare \( P_A \) and \( P_B \):
- Since \( 300 < 360 \), it follows that \( P_A > P_B \).

### Conclusion
Boy A has more power than Boy B because he completed the work in less time.

### Final Answer
A has more power than B.

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