Question

A lead bullet moving with velocity '\(v\)' strikes a wall and stops. \(75\%\) of its energy is converted into heat. If the increase in the temperature is \(\frac{v^{2}}{nJs}\), then the value of \(n\) is \(\{s=\text{specific heat of lead},J=\text{mechanical equivalent of heat}\}\)

• A. \(\frac{8}{3}\)
• B. \(\frac{4}{3}\)
• C. \(\frac{3}{8}\)
• D. \(\frac{3}{4}\)

Hint:

Equate 75 percent of the kinetic energy to the heat gained, converting calories to joules with the mechanical equivalent J.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The bullet stops, so its kinetic energy \(\frac{1}{2}mv^{2}\) is lost. Of this, 75 percent appears as heat.

Step 2: Energy balance
Heat energy in joules \(=0.75\times\frac{1}{2}mv^{2}=\frac{3}{8}mv^{2}\).

Step 3: Calculation
The heat absorbed is \(ms\Delta T\) in heat units, which is \(J\,ms\Delta T\) in work units. So \(J\,ms\Delta T=\frac{3}{8}mv^{2}\), giving \(\Delta T=\frac{3v^{2}}{8Js}\).

Step 4: Compare with the given form
Comparing with \(\Delta T=\frac{v^{2}}{nJs}\) gives \(n=\frac{8}{3}\), which is option (A).

Final Answer:
The value of \(n\) is \(\frac{8}{3}\). \[ \boxed{\text{(A) }\dfrac{8}{3}} \]