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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Thermal Properties of Matter

JIPMER 2016 · Physics · Thermal Properties of Matter

A lead bullet penetrates into a solid object and melts. Assuming...

A lead bullet penetrates into a solid object and melts. Assuming that 50%50 \% of its kinetic energy was used to heat it, the initial speed of the bullet is (the initial temperature of the bullet is 25∘C25^{\circ} \mathrm{C} and its melting point is 300∘C300^{\circ} \mathrm{C}. Latent heat of fusion of lead =2.5×104 J/kg=2.5 \times 10^4 \mathrm{~J} / \mathrm{kg} and specific heat capacity of lead =125 J/kg−K=125 \mathrm{~J} / \mathrm{kg}-\mathrm{K}

Options

  1. A100 m/s
  2. B490 m/s
  3. C520 m/s
  4. D360 m/s

Correct answer

B. 490 m/s

Step-by-step solution

Given, S=125 J/kg−KS=125 \mathrm{~J} / \mathrm{kg}-\mathrm{K} θ=25∘C,θ=300∘C,m=2.5×104 J/kg\theta=25^{\circ} \mathrm{C}, \theta=300^{\circ} \mathrm{C}, m=2.5 \times 10^4 \mathrm{~J} / \mathrm{kg}, then 12mv2×12=mSΔθ+mL\frac{1}{2} m v^2 \times \frac{1}{2}=m S \Delta \theta+m L ⇒v24=125(300−25)+2.5×104⇒v24=59375⇒v2=59375×4⇒v=237500⇒v=487.3 m/s⇒≈490 m/s\begin{array}{rlrl}\Rightarrow & & \frac{v^2}{4} & =125(300-25)+2.5 \times 10^4 \\ \Rightarrow & & \frac{v^2}{4} & =59375 \\ \Rightarrow & & v^2 & =59375 \times 4 \\ \Rightarrow & & v & =\sqrt{237500} \Rightarrow v=487.3 \mathrm{~m} / \mathrm{s} \\ \Rightarrow & & \approx 490 \mathrm{~m} / \mathrm{s}\end{array}
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