JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Rotational Mechanics

A particle of mass

M = 0.2

kg is initially at rest in the

x y

-plane at a point (

x = - l, y = - h

), where

l = 10

m and

h = 1

m. The particle is accelerated at time

t = 0

with a constant acceleration

a = 10

m/s

^{2}

along the positive x-direction. Its angular momentum and torque with respect to the origin, in SI units, are represented by

L

and

τ

, respectively.

\hat{i}, \hat{j}

and

\hat{k}

are unit vectors along the positive x, y and z-directions, respectively. If

\hat{k} = \hat{i} \times \hat{j}

then which of the following statement(s) is(are) correct?

(A)The particle arrives at the point $(x = l, y = - h)$ at time $t = 2$ s.✓ Correct
(B) $τ = 2 \hat{k}$ when the particle passes through the point $(x = l, y = - h)$ ✓ Correct
(C) $L = 4 \hat{k}$ when the particle passes through the point $(x = l, y = - h)$ ✓ Correct
(D) $τ = \hat{k}$ when the particle passes through the point $(x = 0, y = - h)$

Answer: (A), (B), (C)

Solution

The particle moves from

(- 10, - 1)

(10, - 1)

, a distance of 20 m.

Using

s = \frac{1}{2} a t^{2}

20 = \frac{1}{2} \times 10 \times t^{2}

, giving

t = 2

s. ✓ (A)

Force:

F = ma \hat{i} = 0.2 \times 10 \hat{i} = 2 \hat{i}

N.

At point B

(10, - 1)

r_{B} = 10 \hat{i} - \hat{j}

.

Torque:

τ = r_{B} \times F = (10 \hat{i} - \hat{j}) \times 2 \hat{i} = - 2 \hat{j} \times \hat{i} = 2 \hat{k}

N·m. ✓ (B)

Velocity at t = 2 s:

v = a t \hat{i} = 10 \times 2 \hat{i} = 20 \hat{i}

m/s.

Angular momentum:

L = r_{B} \times m v = 0.2 (10 \hat{i} - \hat{j}) \times 20 \hat{i} = 0.2 \times (- 20) \hat{j} \times \hat{i} = 4 \hat{k}

kg·m²/s. ✓ (C)

At point A

(0, - 1)

r_{A} = - \hat{j}

.

Torque:

τ = r_{A} \times F = (- \hat{j}) \times 2 \hat{i} = - 2 \hat{j} \times \hat{i} = 2 \hat{k}

N·m, not

\hat{k}

. ✗ (D)

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · One or more correct options