Question

A steel drill is making $180$ Revolutions per minute, under a constant torque of $5N-m$. If it drills a hole in $7s$. in a steel block of mass $600gm$, rise in temperature of the block is $(s=0.1cal/gm/$${}^{\circ }C$)

• A. ${2.6}^{\circ }C$
• B. ${1.3}^{\circ }C$
• C. ${5.2}^{\circ }C$
• D. $3^{\circ }C$

Answer 1

The correct option is A ${2.6}^{\circ }C$

Given,

$m=600g$

$\tau =5Nm$

$N=180rev/min=\frac{180}{60}=3s^{-1}$

$t=7sec$

$s=0.1cal/g{/}^{0}C=0.1\times 4.2=0.42J/g{/}^{0}C$

According to question

$W=Q$

$\tau \omega t=ms\Delta T$

$\tau .2\pi Nt=ms\Delta T$

$\Delta T=\frac{2\pi \tau Nt}{ms}$

$\Delta T=\frac{2\times 3.14\times 5\times 3\times 7}{600\times 0.42}$

$\Delta T={2.6}^{0}C$

The correct option is A.