JEE Advanced2011PhysicsThermal Properties of MatterActual
Steel wire of length L at 40^ C is suspended from the ceiling and then a mass m is hung from its free end. The wire is cooled down from 40^ C to 30^ C to regain its original length L . The coefficient of linear thermal expansion of the steel is 10⁻⁵ / ^ C , Young's modulus of steel is 10¹¹ ~N / m ^2 and radius of the wire is 1 ~mm . Assume that L> diameter of the wire. Then, the value of m in kg is nearly
Correct answer
0
Step-by-step solution
l₁= F L A Y = m g L r^2 Y = Increase in length l₂=L = Decrease in length To regain its original length, aligned & l₁ & = l₂ & m g L r^2 Y & =L & m & = ( r^2 Y g ) aligned Substituting the values, we get m 3 ~kg Answer is 3 . Analysis of Question Question is very simple. In my opinion any student who have gone through the syllabus once can solve this problem easily.