Question

Morning breakfast gives $5000\text{cal}$ to a $60\text{kg}$ person. The efficiency of the person is $30\%$. The height upto which the person can climb by using the energy obtained from breakfast is [Take $g=10{\text{m/s}}^2$]

• A. $5.00\text{m}$
• B. $10.45\text{m}$
• C. $15.00\text{m}$
• D. $16.50\text{m}$

Answer 1

The correct option is B $10.45\text{m}$

Given:
Efficiency of person $\left(\eta \right)=30\%$
Energy supplied to person $\left(H\right)=5000\text{cal}$
As the person is climbing up, the potential energy of the person changes.
Thus, mechanical work done by the person $\left(W\right)=mgh$
We know that,
Efficiency $\left(\eta \right)=\frac{\text{Output}}{\text{Input}}=\frac{\text{Work done}}{\text{Heat supplied}}$
Using this, we can modify $W=JH$ as $W=\eta JH$
$\therefore mgh=\eta JH$
From this,
$h=\frac{\eta JH}{mg}=\frac{\left(\frac{30}{100}\right)\times 4.18\times 5000}{60\times 10}=10.45\text{m}$
Thus, option (b) is the correct answer.