Question

A boy blowing a whistle sends in air at 2 gm/sec with a speed of 150m/s His lung power is
A) $2.25 W$
B) $22.5 W$
C) $225 W$
D) $0.225 W$

Answer 1

Hint
As simple here, we have to find the power of his lung when A boy blowing a whistle sends in air $\dfrac{gm}{sec}$ with a speed of $150 \dfrac {m}{s} $.
So find it out we use the formula of the power i.e.
Power = $\dfrac{1}{2}\dfrac{m}{t}{V^2} = \dfrac{{KE}}{T}$
KE = kinetic energy of the sound.

Solution step by step
So here we have to find the power the lung where power of the lung can be defined as
It is the value which is equal to the kinetic energy per unit time
Or it can also be written as
$Power = \dfrac{{KE}}{T}$
Here KE = kinetic energy of the sound
T =time
$Power = \dfrac{1}{2}\dfrac{{m{V^2}}}{t}$
Or for simplicity it can also be written as
$Power = \dfrac{1}{2}\dfrac{m}{t}{V^2}$
Here $m/t = 2 gm/sec = 2 \times 10^{-3} kg/sec$
and $V = 150 m/sec$.
Now on putting the values
$Power = \dfrac{1}{2} \times 2 \times {10}^{-3} \times {(150)}^{2}$
$Power = 22.5W$
Option (B) is correct answer.

Note:
Speed of light is given by the $3 \times 108 \dfrac{m}{s}$
Speed of sound is given by the $342 \dfrac{m}{s}$
And speed of light can also be written as $C^2$ = ${μ}_{0}{ε}_{0}$
SI unit of the heat is $joule$ And SI unit of work done is also $joule$.