Question

A man weighing 60kg lifts a body of 15kg to the top of a building 10m high in 30 minutes. His efficiency is:
A. 10%
B. 20%
C. 30%
D. 40%

Answer 1

Hint: The efficiency is the ratio of output power to input power. The input power is the power produced by the man to reach the top of the building along with the body and the output power is the power required to take the body to the top of the building alone. Calculate input power and output using the formula of power.

Formula used:
The formula of power is given by,
$Power = \dfrac{{mgh}}{t}$
where,
$m$ represents the mass of the body,
$g$ represents the acceleration due to gravity,
And $h$ represents the height at which the body is placed.

Complete step by step answer:
The efficiency of a man is given by the ratio of his output power to input power.
Thus,
Efficiency = $\eta = \dfrac{{{\text{Output power}}}}{{{\text{Input power}}}} \times 100\% $
Here, in this case, input power is the power produced by the man to reach the top of the building along with the body and the output power is the power required to take the body to the top of the building alone.
The power is defined as the work done in a given time.
Thus,
$Power = \dfrac{{Workdone}}{{Time}}$
We know,
${\text{Workdone = Force x displacement}}$
Therefore,
${\text{Power = }}\dfrac{{{\text{Force x displacement}}}}{{{\text{Time}}}}$
Where the force is the product of the mass of the body and its acceleration.
Let, M be the mass of a man, m be the mass of the body, h be total displacement which is the height of the building, and t be the time taken to reach the top of the building.
Therefore,

M=60kg, m=15kg, h=10m, t=30min
Thus, the input power can be calculated as
${\text{Input Power}} = \dfrac{{(m + M)g \times h}}{t}$
The output power can be calculated as
${\text{Output Power}} = \dfrac{{mgh}}{t}$
The efficiency of man can be calculated as

$\eqalign{
  & \eta = \dfrac{{\dfrac{{mgh}}{t}}}{{\dfrac{{(m + M)g \times h}}{t}}} \times 100\% \cr
  & \eta = \dfrac{m}{{m + M}} \times 100\% \cr} $
$\eqalign{
  & \eta = \dfrac{{15}}{{15 + 60}} \times 100\% \cr
  & \eta = 20\% \cr} $
Thus, the efficiency of a man is 20%.
Hence, the correct answer is option B.

Note:
Power is the rate at which the work is done or work done per unit time. It has nothing to do with how much energy is spent to perform a task. If you carefully observe the equation, you will see that the efficiency is independent of the displacement of the body and the time taken to do a certain task. It only depends on the mass of the man and the mass of the body he is carrying.