Get Answers to all your Questions

header-bg qa

A bullet of mass 5g, 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(g-^{\circ}C)\; (1\; cal=4.2\times 10^{7}\; ergs)
Option: 1 87.5^{\circ}C  
Option: 2 83.3^{\circ}C
Option: 3 119.2^{\circ}C
Option: 4 38.4^{\circ}C

Answers (1)

best_answer

\begin{array}{l} \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} \mathrm{C} \end{array}

Posted by

Deependra Verma

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE