Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A steel ball of mass m1 = 1 kg moving with velocity 50 m s-1 collides with another ball of mass m2 = 200 g lying on the ground. Due to collision, the KE is lost and their internal energies change equally and T1 and T2 are the temperature changes of masses m1 and m2, respectively. If the specific heat of steel is unity and J = 4.18 J cal-1, then
PhysicsThermodynamicsJEE Main
Options:
  • A T1 = 7.1 °C and T2 = 1.47 °C
     
  • B T1 = 1.47 °C and T2 = 7.1 °C
  • C T1 = 3.4 °C and T2 = 17.0 °C
  • D T1 = 17.0 °C and T2 = 3.4 °C
Solution:
2734 Upvotes Verified Answer
The correct answer is: T1 = 3.4 °C and T2 = 17.0 °C
Half of KE is attained as heat by each ball

       1 2 KE = m 1 s 1 T 2

      1 2 × 1 × 5 0 2   = 1 × 0.105 × 4 1 8 × 1 0 3 × 1 0 3 × T 1

           T 1 = 5 0 × 5 0 2 × 0.105 × 4.18 × 1 0 3

               = 2 5 2.1 × 4.18 = 2 5 8.778 3.4 K

As      m 2 = m 1 5 , so T2 = 5T1 = 17 K
 

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.