Question

A quarter horsepower motor runs at a speed of 600 rpm. Assuming 40% efficiency, the work done by the motor in one rotation would be
(A) $ 7.46J $
(B) $ 7400J $
(C) $ 7.46erg $
(D) $ 74.6J $

Answer 1

Hint
The angular speed 1 rpm implies that a body completes one revolution in one minute. We should convert all units to the same system. The work done will be equal to the product of the energy required for rotation and its efficiency.
Formula used: In this solution we will be using the following formula,
 $ \Rightarrow E = Pt $ where $ E $ is the energy consumed through time $ t $ and $ P $ is the power.
 $ \Rightarrow W = E \times \eta $ where $ W $ is the work done, and $ \eta $ (pronounced nu) is the efficiency.
 $ \Rightarrow \omega (rpm){\text{ = }}\dfrac{{\text{N}}}{t} $ where $ \omega (rpm) $ is the angular speed in revolution per minute, $ {\text{N}} $ is angular distance given in number of revolution and $ t $ is time in minutes.

Complete step by step answer
A motor is a device that converts electrical energy to mechanical energy.
The motor in the question runs at a quarter horsepower. i.e. $ P = \dfrac{{1hp}}{4} $.
Let’s convert the power to SI units. To do so we can use the factor,
 $ \Rightarrow 1hp = 746{\text{W}} $,
Hence we get the power as,
 $ \Rightarrow P{\text{ = }}\dfrac{{{\text{746W}}}}{4} = 186.5{\text{W}} $
Now, at a quarter horsepower, the motor runs at 600 rpm, meaning 600 revolutions are made by the motor per minute. This is the angular speed.
Angular speed is given by
 $ \Rightarrow \omega (rpm){\text{ = }}\dfrac{{\text{N}}}{t} $ where $ \omega (rpm) $ is angular speed in revolution per minute,
Hence, the time taken for one revolution is
 $ \Rightarrow t = \dfrac{N}{\omega } = \dfrac{1}{{600}}\min $
60 seconds make one minute, hence we can write
 $ \Rightarrow t = \dfrac{1}{{600}} \times 60s = 0.1s $
Energy consumed by a process or machine for a period of time $ t $ is given by
 $ \Rightarrow E = Pt $
So energy for one revolution corresponding to a time of $ 0.1s $ is given by
 $ \Rightarrow E = 186.5 \times 0.1 = 18.65J $
The work done is given by the formula
 $ \Rightarrow W = E \times \eta $ where $ \eta $ (pronounced nu) is the efficiency.
Hence on substituting we get,
 $ \Rightarrow W = 18.65 \times 40\% = 18.65 \times 0.4 = 7.46J $
 $ \therefore W = 7.46J $
Hence the correct answer is option (A).

Note
Alternatively, the angular speed can be converted to radians per second and the angular distance to radians.
One revolution completes a circle which subtends angle $ 2\pi rad $
Hence 1 rpm is $ 2\pi rad/\min $
Furthermore, $ \omega = 600rpm = 600 \times 2\pi rad/\min $
But $ 1\min = 60s $
Hence,
 $ \omega = 600rpm = \dfrac{{600 \times 2\pi rad}}{{60s}} = (10 \times 2\pi )rad/s $
So
 $ t = \dfrac{N}{\omega } = \dfrac{{2\pi }}{{10 \times 2\pi }}s = 0.1s $
This is identical to the time calculated in the solution.