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, we need to determine the height a 60 kg person can climb using the energy obtained from breakfast, given that the efficiency of the person is 30%.

### Step-by-Step Solution:

1. **Convert Calories to Joules:**
The energy obtained from breakfast is given as 5000 calories. We need to convert this energy into joules using the conversion factor \(1 \text{ cal} = 4.2 \text{ J}\).
\[ E = 5000 \text{ cal} \times 4.2 \text{ J/cal} = 21000 \text{ J} \]

2. **Calculate Usable Energy:**
The efficiency of the person is 30%. This means only 30% of the total energy can be used for work (climbing). We calculate the usable energy \(E_u\):
\[ E_u = \text{Efficiency} \times E = \frac{30}{100} \times 21000 \text{ J} = 6300 \text{ J} \]

3. **Use Energy for Potential Energy Change:**
The energy used by the person to climb is converted into gravitational potential energy. The formula for gravitational potential energy is:
\[ \Delta PE = mgh \]
where \(m\) is mass (60 kg), \(g\) is the acceleration due to gravity (approximately \(10 \text{ m/s}^2\)), and \(h\) is the height climbed.

4. **Set Usable Energy Equal to Potential Energy:**
We set the usable energy equal to the change in potential energy:
\[ E_u = mgh \]
Substituting the values:
\[ 6300 \text{ J} = 60 \text{ kg} \times 10 \text{ m/s}^2 \times h \]

5. **Solve for Height \(h\):**
Rearranging the equation to solve for \(h\):
\[ h = \frac{6300 \text{ J}}{60 \text{ kg} \times 10 \text{ m/s}^2} = \frac{6300}{600} = 10.5 \text{ m} \]

### Final Answer:
The height up to which the person can climb is **10.5 meters**.

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