JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 9)
A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of bullet is :
(Given : initial temperature of the bullet = 127$$^\circ$$C, Melting point of the bullet = 327$$^\circ$$C, Latent heat of fusion of lead = 2.5 $$\times$$ 104 J kg$$-$$1, Specific heat capacity of lead = 125 J/kg K)
125 ms$$-$$1
500 ms$$-$$1
250 ms$$-$$1
600 ms$$-$$1
Explanation
$${2 \over 5} \times {1 \over 2}m{v^2} = mL + ms\Delta T$$
$$ \Rightarrow {{{v^2}} \over 5} = 2.5 \times {10^4} + 125 + 200$$
$$ \Rightarrow {{{v^2}} \over 5} = 5 \times {10^4}$$
$$ \Rightarrow v = 500$$ m/s
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