JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
- JEE Mains 2020
- Medium
- MCQ
A bullet of mass \(5\, g\), travelling with a speed of \(210\) \(m / s ,\) strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is \(0.030\, cal /\left( g -{ }^{\circ} C \right)\) \(\left(1\, cal =4.2 \times 10^{7}\, ergs \right)\) close to\(.......^oC\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(87.5\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2} m v^{2} \times \frac{1}{2}=m s \Delta T\)
\(\Delta T=\frac{v^{2}}{4 \times 5}=\frac{210^{2}}{4 \times 30 \times 4.200}\)
\(=87.5^{\circ} C\)
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