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ExamsJEE AdvancedPhysics

A steel drill turning at 180 rpm bores a hole in a steel block. The combined mass of block and drill is 180 g, and all the mechanical work converts to heat, raising the temperature at 0.5 C/s. Given specific heat of steel = 0.1 (cal/g-C) and J = 4.2 J/cal, find (a) the power input of the drill and (b) the driving torque. Use P = tau * omega.

  1. (a) 3.78 W, (b) 0.20 N.m
  2. (a) 37.8 W, (b) 2.0 N.m
  3. (a) 378 W, (b) 20 N.m
  4. (a) 0.378 W, (b) 0.02 N.m

Correct answer: (a) 37.8 W, (b) 2.0 N.m

Solution

P = m*c*(dT/dt)*J = 180*0.1*0.5*4.2 = 37.8 W; omega = 2*pi*180/60 = 6*pi ~ 18.85 rad/s; tau = 37.8/18.85 ~ 2.0 N.m.

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