A lead bullet moving with velocity ' $V$ ' strikes a wall and stops. If $75 \%$ of its energy is converted…

A lead bullet moving with velocity ' $V$ ' strikes a wall and stops. If $75 \%$ of its energy is converted into heat, then the increase in temperature is ( $\mathrm{s}=$ specific heat of lead, $\mathrm{J}=$ mechanical equivalent of heat)
  1. $\frac{3 \mathrm{~V}^2}{8 \mathrm{Js}}$
  2. $\frac{5 \mathrm{~V}^2}{8 \mathrm{Js}}$
  3. $\frac{3 \mathrm{~V}^2}{4 \mathrm{Js}}$
  4. $\frac{5 \mathrm{~V}^2}{4 \mathrm{Js}}$

Solution

$\text {K.E. }=\frac{1}{2} \mathrm{MV}^2$
As $75 \%$ of the kinetic energy is converted to heat, Heat energy, $W=\frac{3}{4} \times \frac{1}{2} \mathrm{MV}^2=\frac{3}{8} \mathrm{MV}^2$ $\begin{aligned} & \mathrm{J}=\frac{\mathrm{W}}{\mathrm{Q}}=\frac{\text { change in } \mathrm{KE}}{\text { Heat energy }} \\ \therefore \quad & \mathrm{J}=\frac{\frac{3}{8} \mathrm{MV}^2}{\Delta \mathrm{~T} \times \mathrm{M} \times \mathrm{s}} \\ \therefore \quad & \Delta \mathrm{~T}=\frac{3 \mathrm{~V}^2}{8 \mathrm{Js}} \end{aligned}$ /

Asked in: MHT CET 2024 (10 May Shift 1)

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