The weight of a person is `60` kg . If he gets `10` calories of heat through food and the efficiency of his body is `28%`, then upto what height he can climb? Take g = `10m s^(-2)`

Options

A. `100` cm
B. `1.96` cm
C. `400` cm
D. `1000` cm

Explanation

To solve the problem step by step, we need to determine how high a person weighing 60 kg can climb after consuming 10 calories of heat with an efficiency of 28%. We will also use the acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).

### Step 1: Convert calories to joules
The first step is to convert the energy gained from calories to joules. We know that:
1 calorie = 4.2 joules

Given that the person gains 10 calories:
\[ \text{Energy in joules} = 10 \, \text{calories} \times 4.2 \, \text{joules/calorie} = 42 \, \text{joules} \]

### Step 2: Calculate usable energy
Next, we need to calculate the usable energy based on the efficiency of the body. The efficiency is given as 28%, or 0.28 in decimal form. Thus, the usable energy \( U \) is:
\[ U = \text{Efficiency} \times \text{Total Energy} = 0.28 \times 42 \, \text{joules} = 11.76 \, \text{joules} \]

### Step 3: Use the potential energy formula
The potential energy \( U \) gained when climbing to a height \( h \) is given by the formula:
\[ U = m \cdot g \cdot h \]
Where:
- \( m = 60 \, \text{kg} \) (mass of the person)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h \) is the height we need to find.

### Step 4: Rearrange the formula to find height
Rearranging the potential energy formula to solve for height \( h \):
\[ h = \frac{U}{m \cdot g} \]

### Step 5: Substitute the known values
Now we substitute the values we have calculated:
\[ h = \frac{11.76 \, \text{joules}}{60 \, \text{kg} \times 10 \, \text{m/s}^2} = \frac{11.76}{600} \approx 0.0196 \, \text{meters} \]

### Step 6: Convert height to centimeters
To express the height in centimeters:
\[ h \approx 0.0196 \, \text{meters} \times 100 \, \text{cm/m} = 1.96 \, \text{cm} \]

### Final Answer
Thus, the height to which the person can climb is approximately:
\[ \boxed{1.96 \, \text{cm}} \]