JEE Advance - Physics (2021 - Paper 1 Online - No. 13)
A particle of mass M = 0.2 kg is initially at rest in the xy-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/s2 along the positive x-direction. Its angular momentum and torque with respect to the origin, in SI units, are represented by $$\overrightarrow L $$ and $$\overrightarrow \tau $$ respectively. $$\widehat i$$, $$\widehat j$$ and $$\widehat k$$ are unit vectors along the positive x, y and z-directions, respectively. If $$\widehat k$$ = $$\widehat i$$ $$\times$$ $$\widehat j$$ then which of the following statement(s) is(are) correct?
The particle arrives at the point (x = l, y = $$-$$h) at time t = 2 s
$$\overrightarrow \tau = 2\widehat k$$ when the particle passes through the point (x = l, y = $$-$$h)
$$\overrightarrow L = 4\widehat k$$ when the particle passes through the point (x = l, y = $$-$$h)
$$\overrightarrow \tau = \widehat k$$ when the particle passes through the point (x = 0, y = $$-$$h)
Explanation
Force acting along $\mathrm{X}$-axis $=m a=0.2 \times 10$
$$ =2 \mathrm{~N} $$
Distance covered in $2 \mathrm{~s}$ along $\mathrm{X}$-axis
$$ =\frac{1}{2} a t^2=\frac{1}{2} \times 10 \times 4=20 \mathrm{~m}=2 l $$
Particle reaches at the point $(l,-h)$ at $t=2 \mathrm{~s}$ (Option (A) is correct)
$\tau$ at $(l,-h)=F h=2 \times 1=2 \mathrm{Nm}$
$\vec{\tau}=2(\hat{k}) \quad$ (Option (B) is correct)
$v(t=2 \mathrm{~s})=10 \times 2=20 \mathrm{~m} / \mathrm{s}$
$|\overrightarrow{\mathrm{L}}|$ at $(l,-h)=\mathrm{m} v h=0.2 \times 20 \times 1=4 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}$
$\overrightarrow{\mathrm{L}}=4(\hat{k}) \quad$ (Option $(\mathrm{C})$ is correct)
$\tau$ at $(0,-h)=F h=2 \times 1=2 \mathrm{Nm}$
$\vec{\tau}=2(\hat{k}) \quad$ (Option (D) is incorrect)
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