Morning breakfast gives 5000 cal to a 60 kg. person. The efficiency of person is 30%. The height upto which the person can climb up by using energy obtained from breakfast is

Options

A. 5m
B. 10.5m
C. 15 m
D. 16.5 m

Answer

To solve the problem step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Understand the Energy Supplied
The person receives 5000 calories from breakfast.

### Step 2: Calculate the Work Done by the Person
The efficiency of the person is given as 30%. Efficiency is defined as the ratio of useful work output to the total energy input. Therefore, the work done (W) can be calculated as:
\[ W = \text{Efficiency} \times \text{Total Energy} \]
Substituting the values:
\[ W = 0.3 \times 5000 \text{ cal} = 1500 \text{ cal} \]

### Step 3: Convert Calories to Joules
We need to convert the work done from calories to joules. The conversion factor is:
\[ 1 \text{ cal} = 4.2 \text{ J} \]
Thus, the work done in joules is:
\[ W = 1500 \text{ cal} \times 4.2 \text{ J/cal} = 6300 \text{ J} \]

### Step 4: Apply the Work-Energy Principle
The work done by the person while climbing is equal to the gravitational potential energy gained, which can be expressed as:
\[ W = mgh \]
Where:
- \( m = 60 \text{ kg} \) (mass of the person)
- \( g = 10 \text{ m/s}^2 \) (acceleration due to gravity)
- \( h \) is the height climbed.

### Step 5: Rearranging the Equation to Find Height
We can rearrange the equation to solve for height \( h \):
\[ h = \frac{W}{mg} \]
Substituting the known values:
\[ h = \frac{6300 \text{ J}}{60 \text{ kg} \times 10 \text{ m/s}^2} \]

### Step 6: Calculate the Height
Calculating the height:
\[ h = \frac{6300}{600} = 10.5 \text{ m} \]

### Final Answer
The height up to which the person can climb using the energy obtained from breakfast is **10.5 meters**.

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