Question

A steel drill making $180rpm$ is used to drill a hole in a block of steel. The mass of the steel block and the drill is $180gm$. If the entire mechanical work is used up in producing heat and the rate of rise in temperature of the block and the drill is ${0.5}^{\circ }C/s$.The torque required to drive the drill in Nm is

Specific heat of steel $=0.1$ and $J=4.2J/cal$. Use: $P=\tau \omega$

Answer 1

Amount of heat used to raise the temp: $Q=180\times 0.1\times 0.5=9cal/s=4.2\times 9J=37.8J/s$
Power: $P=\tau \omega =\tau \times \frac{180\times 2\pi }{60}=\tau \times 6\pi$
$\therefore 37.8=\tau \times 6\pi$
$\Rightarrow \tau =\frac{6.3\times 7}{22}=2Nm$