JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2026
- Easy
- MCQ
In a perfectly inelastic collision, two spheres made of the same material with masses 15 kg and 25 kg , moving in opposite directions with speeds of \(10 m / s\) and \(30 m / s\), respectively, strike each other and stick together. The rise in temperature (in \({ }^{\circ} C\) ), if all the heat produced during the collision is retained by these spheres, is :
(specific heat of sphere material \(31 cal / kg .{ }^{\circ} C\) and \(1 cal =4.2 J)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 1.44
Step-by-step Solution
Detailed explanation
\(( K . E )_{\text {lost }}=\frac{1}{2} \mu V_{ rel }^2\left(1- e ^2\right)\)
\(=\frac{1}{2}\left(\frac{m_1 m_2}{m_1+ m _2}\right)(10+30)^2(1-0)\)
\(=\frac{1}{2}\left[\frac{(15)(25)}{40}\right][40]^2\)
\(=7500 J\)
\(( K . E )_{ loss }=\left( m _1+ m _2\right)( S )(\Delta T )\)
\(\left[ S =31 \times 4.2 J / kg -{ }^{\circ} C \right]\)
\(7500=(40)(31)(\Delta T )\)
\(\Delta T =\frac{7500}{40 \times 31 \times 4.2}=1.44^{\circ} C\)
\(=\frac{1}{2}\left(\frac{m_1 m_2}{m_1+ m _2}\right)(10+30)^2(1-0)\)
\(=\frac{1}{2}\left[\frac{(15)(25)}{40}\right][40]^2\)
\(=7500 J\)
\(( K . E )_{ loss }=\left( m _1+ m _2\right)( S )(\Delta T )\)
\(\left[ S =31 \times 4.2 J / kg -{ }^{\circ} C \right]\)
\(7500=(40)(31)(\Delta T )\)
\(\Delta T =\frac{7500}{40 \times 31 \times 4.2}=1.44^{\circ} C\)
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