Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Answer:

‘A’ option is correct 100%

chahe shart lga le agar galat to main tumhe inst pe follow krunga nhi to tu krega

Answer 2

A student skates up at a ramp that makes an angle 30° with the horizontal. He/she starts (as shown in figure) at the bottom of the ramp with the speed v₀ and wants to turn around a semicircular path xyz of radius R during which he/she reaches a maximum height h (at any point y) from the ground as shown in 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..

solution : from energy conservation between the bottom point and point Y.

kinetic energy at bottom point = potential energy at point Y + kinetic energy at point Y

⇒1/2 mv₀² = mgh + 1/2 mv₁²

⇒v₁² = v₀² - 2gh ......(1)

at point Y, centripetal force is balanced by component of weight along plane.

i.e., mgsin30° = mv₁²/R

⇒v₁² = gR/2 , putting it in equation (1),

v₀² - 2gh = gR/2

therefore option (A) is correct choice.

at point X and Z of circular path, the points are at same height but less than h. so the velocity more than a point Y.

so the required centripetal force must be more because centripetal force = mv²/R is more ..

so option (D) is also correct that the centripetal force required is maximum at points x and z.

Therefore options (A) and (D) are correct choices.