JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 13)
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 :
87.5 oC
83.3 oC
38.4 oC
119.2 oC
Forklaring
$${1 \over 2}m{v^2} \times {1 \over 2} = ms\Delta T$$
$$ \Rightarrow $$ $$\Delta T = {{{v^2}} \over {4 \times 5}} = {{{{210}^2}} \over {4 \times 30 \times 4.200}}$$
$$ = 87.5^\circ C$$
$$ \Rightarrow $$ $$\Delta T = {{{v^2}} \over {4 \times 5}} = {{{{210}^2}} \over {4 \times 30 \times 4.200}}$$
$$ = 87.5^\circ C$$