JEE Advanced 2022 — Paper 1 — PHY-I solution

JEE Advanced2022Paper 1PHY-I

At time $t = 0$ , a disk of radius 1 m starts to roll without slipping on a horizontal plane with an angular acceleration of $α = \frac{2}{3} rad s^{- 2}$ . A small stone is stuck to the disk. At $t = 0$ , it is at the contact point of the disk and the plane. Later, at time $t = π$ s, the stone detaches itself and flies off tangentially from the disk. The maximum height (in m) reached by the stone measured from the plane is $\frac{1}{2} + \frac{x}{10}$ . The value of $x$ is _____. [Take $g = 10 ms^{- 2}$ .]

Your answer

Solution

Kinematics at detachment ( $t = π$ s):

Angle rotated: $θ = \frac{1}{2} α t^{2} = \frac{1}{2} \cdot \frac{2}{3} \cdot π = \frac{π}{3}$ rad.

Angular velocity: $ω = α t = \frac{2 π}{3}$ rad/s. Velocity of CM: $v_{c m} = ω R = \frac{2 π}{3}$ m/s.

Velocity of stone: The stone is at angular position $π /3$ from the bottom. Its velocity is the vector sum of $v_{c m}$ (horizontal) and the tangential velocity $ω R$ at angle $12 0^{\circ}$ relative to $v_{c m}$ .

$v = v_{c m}^{2} + (ω R)^{2} + 2 v_{c m} (ω R) cos 12 0^{\circ} = v_{c m} = \frac{2 π}{3}$ m/s.

$tan θ_{proj} = \frac{ω R sin 12 0 ^{\circ}}{v _{c m} + ω R cos 12 0 ^{\circ}} = 3 \Rightarrow θ_{proj} = \frac{π}{3}$ .

Maximum height of trajectory above release point:

$h_{1} = \frac{v ^{2} sin ^{2} 6 0 ^{\circ}}{2 g} = \frac{( 4 π /9 ) ( 3/4 )}{20} = \frac{π}{60}$ m.

Height of release point above the plane: $R (1 - cos 6 0^{\circ}) = \frac{1}{2}$ m.

Total: $H_{m a x} = \frac{1}{2} + \frac{π}{60} = \frac{1}{2} + \frac{x}{10}$ , giving $x = \frac{π}{6} \approx 0.52$ .

← Back to full paper

Practice more PHY-I

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →