NTA Abhyas JEE Main2020PhysicsThermodynamicsPractice
A lead bullet strikes a target with a velocity of 480 ms - 1 . After the impact, the bullet falls dead and the heat produced in the process gets equally shared between the bullet and the target. If the specific heat capacity of lead is c = 0 .03 cal / g / ° C , then the rise in temperature of the bullet is
Options
- A557 ℃
- B457 ℃
- C857 ℃
- D754 ℃
Correct answer
B. 457 ℃
Step-by-step solution
∆ Q = 1 2 K . E . ⇒ m c ∆ T = 1 2 1 2 m v 2 J ∆ T = v 2 4 J c = 480 × 480 4 × 4.2 × ( 0.03 × 10 3 ) = 457 ℃