MHT CET · Physics · Thermodynamics
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)
- A \(\frac{3 \mathrm{~V}^2}{8 \mathrm{Js}}\)
- B \(\frac{5 \mathrm{~V}^2}{8 \mathrm{Js}}\)
- C \(\frac{3 \mathrm{~V}^2}{4 \mathrm{Js}}\)
- D \(\frac{5 \mathrm{~V}^2}{4 \mathrm{Js}}\)
Answer & Solution
Correct Answer
(A) \(\frac{3 \mathrm{~V}^2}{8 \mathrm{Js}}\)
Step-by-step Solution
Detailed explanation
\(\text {K.E. }=\frac{1}{2} \mathrm{MV}^2\)
As \(75 \%\) of the kinetic energy is converted to heat,
Heat energy, \(W=\frac{3}{4} \times \frac{1}{2} \mathrm{MV}^2=\frac{3}{8} \mathrm{MV}^2\)
\(\begin{aligned}
& \mathrm{J}=\frac{\mathrm{W}}{\mathrm{Q}}=\frac{\text { change in } \mathrm{KE}}{\text { Heat energy }} \\
\therefore \quad & \mathrm{J}=\frac{\frac{3}{8} \mathrm{MV}^2}{\Delta \mathrm{~T} \times \mathrm{M} \times \mathrm{s}} \\
\therefore \quad & \Delta \mathrm{~T}=\frac{3 \mathrm{~V}^2}{8 \mathrm{Js}}
\end{aligned}\)
As \(75 \%\) of the kinetic energy is converted to heat,
Heat energy, \(W=\frac{3}{4} \times \frac{1}{2} \mathrm{MV}^2=\frac{3}{8} \mathrm{MV}^2\)
\(\begin{aligned}
& \mathrm{J}=\frac{\mathrm{W}}{\mathrm{Q}}=\frac{\text { change in } \mathrm{KE}}{\text { Heat energy }} \\
\therefore \quad & \mathrm{J}=\frac{\frac{3}{8} \mathrm{MV}^2}{\Delta \mathrm{~T} \times \mathrm{M} \times \mathrm{s}} \\
\therefore \quad & \Delta \mathrm{~T}=\frac{3 \mathrm{~V}^2}{8 \mathrm{Js}}
\end{aligned}\)
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