JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Rotational Mechanics

Consider a large disk of radius R and two smaller disks, each of radius

r = R /50

, lying on its circumference, as shown in the figure. The smaller disks are initially in contact with each other, with an angular separation

Δ θ

between their centers. They are made to roll without slipping in opposite directions, with constant angular velocities

ω

and

2 ω

while the large disk is held stationary. The time

τ

at which the smaller disks are again in contact is :
[Use

sin (Δ θ) = Δ θ

and ignore gravity.]

Answer: (C)

Solution

The two smaller disks roll with angular velocities

ω

and

2 ω

in opposite directions. The relative angular velocity is

3 ω

. The disks start with an angular separation

Δ θ

between their centers.

For the disks to be in contact again, they must travel a combined angular distance of

2 π - Δ θ

(the remaining arc of the large circle).

Using the small angle approximation:

\frac{Δ θ}{2} = \frac{R /50}{R + \frac{R}{50}} = \frac{1}{51}

, so

Δ θ = \frac{2}{51}

rad.

The arc length traveled is

(π - Δ θ) (R + \frac{R}{50}) = \frac{51 R}{50} (π - Δ θ)

.

This equals the sum of distances rolled by both disks:

ω t \cdot \frac{R}{50} + 2 ω t \cdot \frac{R}{50} = 3 ω t \cdot \frac{R}{50}

.

Solving:

t = \frac{51}{3 ω} (π - \frac{2}{51}) = \frac{51}{3 ω} (2 π - \frac{4}{51})

sec.

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Single correct option