NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice
The system is released form rest with both the springs in unstretched position. Mass of each block is 1 kg and force constant of each spring is 10 N/m. Assume pulleys and strings are massless and all contacts are smooth.Then which of the following is incorrect. (g = 10 m/s 2 )
Options
- AExtension of horizontal spring in equilibrium is 2/5 m
- BExtension of vertical spring in equilibrium is 1/5 m
- CMaximum speed of the block A is 8 5 m/s
- DMaximum speed of the block A is 1 4 m/s
Correct answer
D. Maximum speed of the block A is 1 4 m/s
Step-by-step solution
The system is released... T = 2 kx ⇒ 5 kx = mg x = mg 5 k = 1 × 1 0 5 × 1 0 = 1 5 (ii) 1 g x = 1 2 1 v 2 + 1 2 1 2 v 2 x + 1 2 1 0 0.2 2 + 1 2 1 0 0.4 2 2 = V 2 2 1 + 4 + 1 2 0.4 + 1.6 ⇒ v = 2 5 v A = 2 v = 2 5 2 = 8 5 So only wrong option is (iv)