Verified Solution • 2025–26 Session moderate

QJEEAdv20Ph2c27: JEE Advanced 2020 Physics - Class 11 JEEAdv20Ph2c27 Solution: A student skates up a ramp that makes an angle 30 with the…

A student skates up a ramp that makes an angle 30

^\circ

with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed v₀ and wants to turn around over a semicircular path xyz 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)

JEE Advanced 2020 Paper 2 Offline Physics - Work Power & Energy Question 8 English

Multiple Choice Options

A $v_0^2 - 2gh = {1 \over 2}gR$
B $v_0^2 - 2gh = {{\sqrt 3 } \over 2}gR$
C the centripetal force required at points x and z is zero
D the centripetal force required is maximum at points x and z

Step-by-Step Verified Solution

Apply energy conservation,

{1 \over 2}mv_0^2 = mgh + {1 \over 2}m{v^2}

{1 \over 2}m{v^2} = {1 \over 2}mv_0^2 - mgh

..... (i)

At top point y, the centripetal force provided by the component of mg,

mg sin

\theta

= centripetal force =

{{m{v^2}} \over R}

{{m{v^2}} \over R} = mg\sin \theta

Putting this value in Eq. (i), we get

{{mgR\sin \theta } \over 2} = {1 \over 2}mv_0^2 - mgh

{{mgR} \over 2}{1 \over 2} = {1 \over 2}mv_0^2 - mgh

{{gR} \over 2} = v_0^2 - 2gh

At point x and a of circular path, the points are at same height but less than h. So the velocity is more than point y.

So required centripetal force = $\frac{mv^{2}}{r}$ is more.

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank

Browse year-wise authentic previous years' questions with solutions

Authentic CBT Mock Tests

Simulate official exam environment with live timer and percentile scoring