Verified Solution • 2025–26 Session moderate

QJEEAdv20Ph0186: JEE Advanced 2020 Physics - Class 11 JEEAdv20Ph0186 Solution: A small roller of diameter 20 cm has an axle of diameter 10 cm…

A small roller of diameter 20 cm has an axle of diameter 10 cm (see figure below on the left). It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle (see figure on the right). The scale is now pushed slowly on the axle so that it moves without slipping on the axle, and the roller starts rolling without slipping. After the roller has moved 50 cm, the position of the scale will look like (figures are schematic and not drawn to scale) JEE Advanced 2020 Paper 1 Offline Physics - Rotational Motion Question 42 English

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/ce00090aee0f4ed05e2f0b210a0313ce95aef57173cca46280e5a2aea00ee5d9.webp
  • B https://ik.imagekit.io/novatra/photos/8fb24871ebf0a3861ab52f69e2564cc1c01cb6932c4512c3bab9c34eddd1741a.webp
  • C https://ik.imagekit.io/novatra/photos/05a932f0916c5b77254b53c37f644253156feb7bfd5837d13cd17180d757985a.webp
  • D https://ik.imagekit.io/novatra/photos/fe42786c0bf26c461e009513f6ea914dbd8d38746ba6427a8aff99bece53a97b.webp

Step-by-Step Verified Solution

For no slipping at ground,

vc = ωR

(where, r is the radius of roller)

∴ Velocity of scale = (vc + ωr) (where, r is the radius of axle)

Given, vc . t = 50 cm

∴ Distance moved by scale = (vc + ωr)t = (vc+vcrr)t=3vc2.t=75cm

Therefore, relative displacement (with respect to centre of roller) is (75 − 50) cm = 25 cm.

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