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Physics · Rotational Dynamics

JEE Advanced 2018 — Paper 1 — Question 2

Consider a body of mass 1.0 kg at rest at the origin at time t=0t=0. A force F⃗=(αti^+βj^)\vec{F}=(\alpha t \hat{i}+\beta \hat{j}) is applied on the body, where α=1.0Ns−1\alpha=1.0 \mathrm{Ns}^{-1} and β=1.0 N\beta=1.0 \mathrm{~N}. The torque acting on the body about the origin at time t=1.0 st=1.0 \mathrm{~s} is τ⃗\vec{\tau}. Which of the following statements is (are) true?

  1. Option A:

    ∣τ⃗∣=13Nm|\vec{\tau}|=\frac{1}{3} \mathrm{Nm}

    Correct
  2. Option B:

    The torque τ⃗\vec{\tau} is in the direction of the unit vector +k^+\hat{k}

  3. Option C:

    The velocity of the body at t=1st=1 s is v⃗=12(i^+2j^)ms−1\vec{v}=\frac{1}{2}(\hat{i}+2 \hat{j}) \mathrm{ms}^{-1}

    Correct
  4. Option D:

    The magnitude of displacement of the body at t=1 st=1 \mathrm{~s} is 16m\frac{1}{6} m

Answer: A, C

Step-by-step solution

a⃗=ti^+j^ m/s2\vec{a}=t \hat{i}+\hat{j} \mathrm{~m} / \mathrm{s}^{2}

⇒v⃗=t22i^+tj^ m/s\Rightarrow \vec{v}=\frac{t^{2}}{2} \hat{i}+t \hat{j} \mathrm{~m} / \mathrm{s}

⇒r⃗=t36i^+t22j^m\Rightarrow \vec{r}=\frac{t^{3}}{6} \hat{i}+\frac{t^{2}}{2} \hat{j} m

so, τ⃗=r⃗×F⃗=(t32−t36)(−k^)=t33(−k^)Nm\vec{\tau}=\vec{r} \times \vec{F}=\left(\frac{t^{3}}{2}-\frac{t^{3}}{6}\right)(-\hat{k})=\frac{t^{3}}{3}(-\hat{k}) \mathrm{Nm}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 1
Subject
Physics
Chapter
Rotational Dynamics
Topic
Torque, Equation of Motion and Toppling

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