(b) Let mass of the bullet be m gram, then total heat required for bullet to just melt down Q1‌‌=mc∆T+mL ‌‌=m×(0.03)(327−27)+m×6 ‌‌=15m−cal ‌‌=(16m×4⋅2)J Now, when bullet is struck by obstacles, the loss in its mechanical energy =‌
1
2
(m×10−3)v2 The energy absorbed by bullet, Q2‌‌=‌
72
100
×‌
1
2
mv2×10−3 ‌‌=‌
3
8
mv2×10−3J
Now, the bullet will melt if Q2≥Q1 ‌ i.e., ‌‌‌