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QJEEAdv21Ph07a3: JEE Advanced 2021 Physics - Class 11 JEEAdv21Ph07a3 Solution: A particle of mass M = 0.2 kg is initially at rest in the…

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 L→ and τ→ respectively. i^, j^ and k^ are unit vectors along the positive x, y and z-directions, respectively. If k^ = i^ × j^ then which of the following statement(s) is(are) correct?

Multiple Choice Options

  • A The particle arrives at the point (x = l, y = −h) at time t = 2 s
  • B τ→=2k^ when the particle passes through the point (x = l, y = −h)
  • C L→=4k^ when the particle passes through the point (x = l, y = −h)
  • D τ→=k^ when the particle passes through the point (x = 0, y = −h)

Step-by-Step Verified Solution

JEE Advanced 2021 Paper 1 Online Physics - Impulse & Momentum Question 11 English Explanation
Force acting along X\mathrm{X} -axis =ma=0.2×10=m a=0.2 \times 10

=2 N

Distance covered in 2 s2 \mathrm{~s} along X\mathrm{X} -axis

=12at2=12×10×4=20 m=2l

Particle reaches at the point (l,−h)(l,-h) at t=2 st=2 \mathrm{~s} (Option (A) is correct)

τ\tau at (l,−h)=Fh=2×1=2Nm(l,-h)=F h=2 \times 1=2 \mathrm{Nm}

τ⃗=2(k^)\vec{\tau}=2(\hat{k}) \quad (Option (B) is correct)

v(t=2 s)=10×2=20 m/sv(t=2 \mathrm{~s})=10 \times 2=20 \mathrm{~m} / \mathrm{s}

∣L→∣|\overrightarrow{\mathrm{L}}| at (l,−h)=mvh=0.2×20×1=4 kg m2/s(l,-h)=\mathrm{m} v h=0.2 \times 20 \times 1=4 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}

L→=4(k^)\overrightarrow{\mathrm{L}}=4(\hat{k}) \quad (Option (C)(\mathrm{C}) is correct)

τ\tau at (0,−h)=Fh=2×1=2Nm(0,-h)=F h=2 \times 1=2 \mathrm{Nm}

τ⃗=2(k^)\vec{\tau}=2(\hat{k}) \quad (Option (D) is incorrect)

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