Q90 / 180 · AIIMS 2015 · Physics

An air compressor is powered by a $200 \mathrm{rad} / \mathrm{s}$ electric motor using a V-belt drive. The motor pulley is $8 \mathrm{~cm}$ in radius, and the tension in the $\mathrm{V}$-belt is $135 \mathrm{~N}$ on one side and $45 \mathrm{~N}$ on the other. The power of the motor will be

  • Correct answer
    A
    $1.44 \mathrm{~kW}$
  • B
    $14.4 \mathrm{~kW}$
  • C
    $2.88 \mathrm{~kW}$
  • D
    $28.8 \mathrm{~kW}$

Solution

Since the net force on the belt is the difference between the two tensions, the torque exerted by the motor on the belt is, $$ \begin{aligned} \tau & =F r=(135 \mathrm{~N}-45 \mathrm{~N})(0.08 \mathrm{~m}) \\ & =7.2 \mathrm{~N} \mathrm{~m} \end{aligned} $$ The power output of the motor is therefore $$ \begin{aligned} P & =\tau \omega=(72 \mathrm{~N} \cdot \mathrm{m})\left(200 \mathrm{rad} \mathrm{s}^{-1}\right) \\ & =1440 \mathrm{~W}=1.44 \mathrm{~kW} \end{aligned} $$

Hence, the correct answer is option: 'A'.

More Rotational Motion Questions