NTA Abhyas JEE Main2020PhysicsThermal Properties of MatterPractice
The figure below shows an isosceles triangular frame ABC made out of two different materials and D is the midpoint of the side AB . The coefficient of thermal expansion of the rod ADB is α 1 and for the rods AC and BC is α 2 . End C is fixed and the whole system is placed on a smooth horizontal surface. When the temperature of the system increases, it is found that the distance remains fixed, then
Options
- Al 1 l 2 = 2 α 2 α 1
- Bl 1 l 2 = 2 α 1 α 2
- Cl 1 l 2 = α 1 α 2
- Dl 1 l 2 = α 2 α 1
Correct answer
A. l 1 l 2 = 2 α 2 α 1
Step-by-step solution
According to condition of the problem, height of the isosceles triangle A B C is unchanged. The dotted lines show configuration After a temperature rise. Increase in length of rod A B , ∆ l 1 = l 1 α 1 ∆ T Thus A A ′ = 1 / 2 l 1 α 1 ∆ T We draw a normal from A to A ′ C (the final length of A C ). Increase in length of A C is A ′ N A ′ N = l 2 α 2 ∆ T Considering increase in angle θ to be very small. A ′ N ≃ A A ′ cos θ Where cos θ = l 1 2 l 2 Thus, we have l 2 α 2 ∆ T = 1 2 l