Question

A steel drill is making 180 revolutions per minute under a constant couple of 5 Nm. If it drills a hole in 7 seconds in a steel block of mass 600 gm, the rise in temperature of the block is: (S=0.1 cal/gm/K)

• A. ${46}^{0}C$
• B. ${1.3}^{0}C$
• C. ${5.2}^{0}C$
• D. $3^{0}C$

Answer 1

Answer: (A)

${46}^{0}C$

Work done by the couple in 7 second $=\tau \ast \theta =5\ast 2\pi \ast 3\ast 7$
This will make a rise in temp = msT= 210 $\pi$
S= 0.1/4.2 J/g/K
t comes 46K or degree celcius as it's only change in temp