Question

A bread gives a boy of mass $40kg$ an energy of $21kJ$. If the efficiency is $28\%$ then the height can be climbed by him using this energy is

& A)22.5m \\\ & B)14.7m \\\ & C)5m \\\ & D)10m \\\ \end{aligned}$$

Answer 1

Here, the bread gives an energy of $21kJ$ to the boy. The efficiency of the boy is given in the question. Using the above two values can find out the efficient energy. The boy uses this energy to climb a height $h$, which is equal to the potential energy at a height $h$. We know that potential energy is related to mass, acceleration due to gravity and height. Therefore, by finding the potential energy we can find out the height.

Complete answer:
Given,
Efficiency of boy, $\eta =28\%$
Energy of one bread, $E=21kJ=21000J$
We have,
Efficient energy, ${{E}_{efficient}}=E\times \eta$
Substituting the values in the above equation we get,
The energy consumed by boy is ${{E}_{efficient}}=\dfrac{28}{100}\times 21000=5880J$ ---------1
To climb a height $h$ the boy utilizes potential energy, $P.E=mgh$
Where,
$m$is the mass of the boy
$g$ is the acceleration due to gravity
Given,
Substituting the values of$\text{m and g}$in the above equation, we get,
$P.E=40\times 10\times h=400h$ -------- 2
The efficient energy is utilized in giving potential energy.
Then,
${{E}_{efficient}}=mgh$
Equating equation 1 and 2, we get,
$5880=400h$
$h=\dfrac{5880}{400}=14.7m$

Therefore, the answer is option B.

Note:
A method of reducing energy consumption by using less energy input to attain the same amount of useful output is known as efficiency. Efficiency can be determined quantitatively by the ratio of useful output to the total input. The ratio of energy transferred in a useful form compared to the total energy supplied is called the efficiency of a device. It is denoted by $\eta$. It has no unit.
$\text{ }\\!\\!\eta\\!\\!\text{ =}\dfrac{\text{Work output}}{\text{Work input}}\times 100\%$