Q120 / 150 · MHT CET 2024 (04 May Shift 2) · Physics

A lead bullet moving with velocity ' $v$ ' strikes a wall and stops. If $50 \%$ of its energy is converted into heat, then the increase in temperature is ( $\mathrm{s}=$ specific heat of lead)

  • A
    $\frac{v^2 s}{2 J}$
  • Correct answer
    B
    $\frac{\mathrm{v}^2}{4 \mathrm{~sJ}}$
  • C
    $\frac{\mathrm{v}^2 \mathrm{~s}}{\mathrm{~J}}$
  • D
    $\frac{2 \mathrm{v}^2}{\mathrm{~Js}}$

Solution

Energy converted to heat $=50 \%$ of K.E. $\begin{array}{ll} \therefore \quad & E=\frac{1}{2} \mathrm{mv}^2 \times\left(\frac{50}{100}\right)=\frac{1}{4} \mathrm{mv}^2 \\ & \text { Heat, } \mathrm{Q}=\mathrm{ms} \Delta \mathrm{~T} \\ & \mathrm{~J}=\text { Mechanical equivalent of heat }=\frac{W}{\mathrm{Q}} \\ \therefore \quad & \mathrm{~J}=\frac{\frac{1}{4} \mathrm{mv}^2}{\mathrm{~ms} \Delta \mathrm{~T}} \\ & \Delta \mathrm{~T}=\frac{\mathrm{v}^2}{4 \mathrm{Js}} \end{array}$

Hence, the correct answer is option: 'B'.

Relevant Questions