Most Important Selected Qs for JEE AdvancedPhysicsWork, Power and Energy
In the figure shown, ADB & BEF are two fixed circular paths in a vertical plane. A block of mass m enters in the frictionless tube ADB through point A with minimum velocity to reach point B . From there it moves on another circular path BEF of radius R ^ . And it is just able to complete the circle.
Options
- AVelocity at A must be 4 Rg .
- BVelocity at A must be 2 Rg .
- CR^ R = 2 3
- DThe normal reaction at point E is 6 mg.
Correct answer
D. The normal reaction at point E is 6 mg.
Step-by-step solution
For minimum velocity at A ; 1 2 m V_A ^2=m g R V_A= 2 g R Now, 1 2 mv _ B ^2+ mgR^ = 1 2 ~m v _ E ^2 As, V_B= 2 g R For looping the loop; aligned & V_E= 5 g R^ & 1 2 m 2 g R+m g R^ = 1 2 m 5 g R^ & R^ R = 2 3 aligned And also N-m g= m v_E^2 R^ N - mg = m 5 ~g R ^ R ^ N =6 mg