A particle of mass M = 0 . 2kg is initially at rest in the xy -plane at a point (x = - l, y = - h) , where l = 10m and h = 1m . The particle is accelerated at time t = 0 with a constant acceleration a = 10m s ^ - 2

Correct answer: A. The particle arrives at the point (x = l, y = - h) at time t = 2s . At& (t = 2s): displacement along + X ~axis = (1)/(2) × (10) × t^ 2 = (1)/(2) × 10 × 4 = 20m Hence, particle arrives at~(10, - 1) At~(x = l, y = - h) , → τ = → r × → F = (10 ^ i - ^ j ) × \ 0 .2 × 10\ ^ i = (2 ^ k )Nm → L = → r × → p = (10 ^ i - ^ j ) × \ 0 .2 × v ^ i \ and~ v(att = 2s) = 0 + 10 × 2 = 20ms^ - 1 ∴ → L = (10 ^ i - ^ j ) × 4 ^ i = (4 ^ k )Nms At~(x = 0, y = - h): → r = 0 ^ i - ^ j → F = 2 ^ i ∴ → τ = → r × → F = (2 ^ k )Nm

JEE Advanced 2021Paper 1

Exam: JEE Advanced
Year: 2021
Paper: Paper 1
Subject: Physics
Chapter: Rotational Motion

JEE Advanced 2021 · Paper 1 · Physics · Rotational Motion

A particle of mass M = 0 . 2kg is initially at rest in the xy...

A particle of mass

M = 0 . 2kg

is initially at rest in the

xy

-plane at a point \left(x = - l, y = - h\right)

, where

l = 10m and h = 1m

. The particle is accelerated at time

t = 0

with a constant acceleration

a = 10m{s}^{- 2}

along the positive

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

\overset{\to}{L} and \overset{\to}{τ}

, respectively.

\overset{^}{i}, \overset{^}{j} and

\overset{^}{k}

are unit vectors along the positive

x, y and z

-directions, respectively. If

\overset{^}{k} = \overset{^}{i} \times \overset{^}{j}

then which of the following statement(s) is (are) correct? A particle of mass M = 0 . 2kg is initially at rest in the xy -plane at a point (x = - l, y = - h) , where l = 10m and

Options

AThe particle arrives at the point $\left(x = l, y = - h\right)$ at time $t = 2s$ .
B $\overset{\to}{τ} = 2\overset{^}{k}$ when the particle passes through the point \left(x = l, y = - h\right).
C $\overset{\to}{L} = 4\overset{^}{k}$ when the particle passes through the point \left(x = l, y = - h\right).
D $\overset{\to}{τ} = \overset{^}{k}$ when the particle passes through the point \left(x = 0, y = - h\right).

Correct answer

A. The particle arrives at the point $\left(x = l, y = - h\right)$ at time $t = 2s$ .

Step-by-step solution

A particle of mass M = 0 . 2kg is initially at rest in the xy -plane at a point (x = - l, y = - h) , where l = 10m and

At& $\left(t = 2s\right):$

displacement along $+ X$ ~axis

$= \dfrac{1}{2} \times \left(10\right) \times t^{2}$

$= \dfrac{1}{2} \times 10 \times 4 = 20m$

Hence, particle arrives at~\left(10, - 1\right) $At~\left(x = l, y = - h\right)$ ,

$\overset{→}{τ} = \overset{→}{r} \times \overset{→}{F}$

$= \left(10\overset{^}{i} - \overset{^}{j}\right) \times \left\{0 .2 \times 10\right\}\overset{^}{i}= \left(2\overset{^}{k}\right)Nm$

$\overset{→}{L} = \overset{→}{r} \times \overset{→}{p}= \left(10\overset{^}{i} - \overset{^}{j}\right) \times \left\{0 .2 \times v\overset{^}{i}\right\}$

and~ $v\left(att = 2s\right) = 0 + 10 \times 2 = 20ms^{- 1}$

$∴ \overset{→}{L} = \left(10\overset{^}{i} - \overset{^}{j}\right) \times 4\overset{^}{i} = \left(4\overset{^}{k}\right)Nms$

At~\left(x = 0, y = - h\right): $$ \overset{→}{r} = 0\overset{^}{i} - \overset{^}{j} $$ \overset{→}{F} = 2\overset{^}{i} $$ ∴ \overset{→}{τ} = \overset{→}{r} \times \overset{→}{F} = \left(2\overset{^}{k}\right)Nm

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