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A quarter horse power motor runs at a speed of 600 rpm. Assuming, \[40\% \]efficiency, the work done by the motor in one rotation will be:
(A) \[7.46J\]
(B) \[7400J\]
(C) \[7.46erg\]
(D) \[74.6J\]

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint It is given that the motor runs at a given speed. Now, at a reduced efficiency, we need to calculate work done. Find out the angular velocity using RPM and find out the equation for torque. Use a work done formula with respect to torque and angle of rotation.

Complete step by step answer
It is given that a quarter horse power motor runs at a specified speed. We know that horse power is one of the standard units of power and it is equal to 746 watts. Now the overall power is a quarter of one horse power.
\[ \Rightarrow P = \dfrac{{1Hp}}{4} = \dfrac{{746}}{4} = 186.5Watt\]
Now, it is given that the motor runs at a speed of 600 rotations per minute. We can use this to calculate angular velocity of the motor, using the formula,
\[ \Rightarrow \omega = \dfrac{{2\pi N}}{{60}}\] where N is the speed of motor in RPM
Substituting the values we get,
\[ \Rightarrow \omega = \dfrac{{2\pi \times 600}}{{60}} = 20\pi rad/s\]
Now, work done for a rotary motion can be defined as the product of the torque of the motor and the angle the motor rotates per revolution. Mathematically, we can give it as
\[ \Rightarrow W = \tau \times \theta \]
Torque of a motor is given as the ratio between the power of the motor and the angular velocity at which the motor rotates per second. We can represent this as,
\[\tau = \dfrac{{Power}}{\omega }\]
Substituting this in the work done equation, we get
\[ \Rightarrow W = \dfrac{{Power}}{\omega } \times \theta \]
From the given data and the identified values, substitute for power, angular velocity and \[\theta \]. Since, it is a complete rotation, \[\theta \] will be equal to \[2\pi \]radians. Now,
\[ \Rightarrow W = \dfrac{{186.5}}{{20\pi }} \times 2\pi \]
Cancelling the common terms, we get
\[ \Rightarrow W = \dfrac{{186.5}}{{10}}\]
\[ \Rightarrow W = 18.65J\]
This is work done at \[100\% \] motor efficiency. It is given that the motor rotates at \[40\% \]efficiency. Thus, work done at the given efficiency is
\[ \Rightarrow W = 18.65J \times 0.4\]
\[ \Rightarrow W = 7.46J\]

Thus, option (a) is the right answer.

Note Torque is defined as the moment of force that carries a required tendency of force that can rotate the body at a specified direction when applied. The direction of force is perpendicular to that of the central rotation axis.