MHT CET 2021 Physics Solved Paper

(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., ‌‌‌

3

8

mv2×10−3≥15m×4.2
⇒vmin=410m∕s