MHT CET · Physics · Thermal Properties of Matter
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 \(\frac{v^2 s}{2 J}\)
- B \(\frac{\mathrm{v}^2}{4 \mathrm{~sJ}}\)
- C \(\frac{\mathrm{v}^2 \mathrm{~s}}{\mathrm{~J}}\)
- D \(\frac{2 \mathrm{v}^2}{\mathrm{~Js}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{v}^2}{4 \mathrm{~sJ}}\)
Step-by-step Solution
Detailed explanation
Energy converted to heat \(=50 \%\) of K.E.
\(\begin{array}{ll}
\therefore \quad & E=\frac{1}{2} \mathrm{mv}^2 \times\left(\frac{50}{100}\right)=\frac{1}{4} \mathrm{mv}^2 \\
& \text { Heat, } \mathrm{Q}=\mathrm{ms} \Delta \mathrm{~T} \\
& \mathrm{~J}=\text { Mechanical equivalent of heat }=\frac{W}{\mathrm{Q}} \\
\therefore \quad & \mathrm{~J}=\frac{\frac{1}{4} \mathrm{mv}^2}{\mathrm{~ms} \Delta \mathrm{~T}} \\
& \Delta \mathrm{~T}=\frac{\mathrm{v}^2}{4 \mathrm{Js}}
\end{array}\)
\(\begin{array}{ll}
\therefore \quad & E=\frac{1}{2} \mathrm{mv}^2 \times\left(\frac{50}{100}\right)=\frac{1}{4} \mathrm{mv}^2 \\
& \text { Heat, } \mathrm{Q}=\mathrm{ms} \Delta \mathrm{~T} \\
& \mathrm{~J}=\text { Mechanical equivalent of heat }=\frac{W}{\mathrm{Q}} \\
\therefore \quad & \mathrm{~J}=\frac{\frac{1}{4} \mathrm{mv}^2}{\mathrm{~ms} \Delta \mathrm{~T}} \\
& \Delta \mathrm{~T}=\frac{\mathrm{v}^2}{4 \mathrm{Js}}
\end{array}\)
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