Question

The weight of a person is 60 kg. If he gets ${{10}^{5}}$ calories heat through food and the efficiency of his body is 28% then upto how much height he can climb when he uses the entire energy gained in climbing.
A. 100m
B. 200m
C. 400m
D. 1000m

Answer 1

Hint: To solve this question first convert the energy from calorie to joule. Then find the energy consumed by the man. Let the height the man climbs is h. find the potential energy needed to climb this height. Equate this energy with the obtained energy to find the required answer.

Complete step by step answer:
We have a person whose weight $m=60kg$
He gets ${{10}^{5}}$ calories heat through food.
Converting this heat energy from calories to the unit of joule. We get that
$H={{10}^{5}}\text{calories}=4.2\times {{10}^{5}}\text{joule}$
Now, the efficiency of his body is 28%.
So, the energy obtained by the man is,
$\begin{align}
  & E=\dfrac{28}{100}\times 4.2\times {{10}^{5}}\text{joule} \\
 & E=1.176\times {{10}^{5}}\text{joule} \\
\end{align}$
Now, let the man climb up to a height of h.
So, the potential energy of the man at the height of h will be, $PE=mgh$
Where, g is the acceleration due to gravity.
Again, the potential energy is equal to the energy lost by the while climbing the height h.
So, the potential energy will be equal to the entire energy obtained by the man.
So, we can write,
\[\begin{align}
  & mgh=1.176\times {{10}^{5}}J \\
 & \Rightarrow 60\times 9.8\times h=1.176\times {{10}^{5}} \\
 & \Rightarrow h=\dfrac{1.176\times {{10}^{5}}}{60\times 9.8} \\
 & \Rightarrow h=2\times {{10}^{2}}m \\
 & \Rightarrow h=200m \\
\end{align}\]
So, the man can climb up to a height of 200 m when he uses his entire energy.

The correct option is (B).

Note:
Potential is the energy of a body relative to a reference point. The work done by the man in climbing the height will convert to the potential energy of the man. Since, the man lost his entire energy in climbing the height, the energy obtained by the man will be equal to the potential energy of the man at height h.