JEE Advanced
Physics
Laws of Motion
2020
JEE Advanced 2020 (Paper 2)
JEE Advanced Physics Question (2020) — Solution
Question
A student skates up a ramp that makes an angle 30 ° with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed v 0 and wants to turn around over a semicircular path x y z of radius R during which he/she reaches a maximum height h (at point y ) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then ( g is the acceleration due to gravity)
Options
- A. v 0 2 - 2 g h = 1 2 g R
- B. v 0 2 - 2 g h = 3 2 g R
- C. the centripetal force required at points x and z is zero
- D. the centripetal force required is maximum at points x and z
Answer
D. the centripetal force required is maximum at points x and z
Step-by-step solution
Speed at x and z is equal and also maximum in circular track. From energy conversation, 1 2 m v 0 2 = m g h + 1 2 m v 2 .......(1) at y ,   mg sin 30 ° = m v 2 R ⇒ v 2 = R g 2 ∴ From equation (1) v 0 2 - 2 g h = R g 2
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Related: Physics — Laws of Motion · All PYQ Banks