Verified Solution • 2025–26 Session moderate

QJEEM20Ph0406: JEE Main 2020 Physics - Class 11 JEEM20Ph0406 Solution: A bullet of mass 5 g, travelling with a speed of 210 m/s,…

A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030 cal/(g – oC) (1 cal = 4.2 × 107 ergs) close to :

Multiple Choice Options

  • A 87.5 oC
  • B 83.3 oC
  • C 38.4 oC
  • D 119.2 oC

Step-by-Step Verified Solution

12mv2×12=msΔT{1 \over 2}m{v^2} \times {1 \over 2} = ms\Delta T

⇒\Rightarrow ΔT=v24×5=21024×30×4.200\Delta T = {{{v^2}} \over {4 \times 5}} = {{{{210}^2}} \over {4 \times 30 \times 4.200}}

=87.5∘C= 87.5^\circ C

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