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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) $74.6J$

Last updated date: 14th Aug 2024
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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.

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.