Rotational Motion JEE Advanced previous year questions with solutions

4 solved JEE Advanced questions on Rotational Motion, free to read — no sign-in needed. The full chapter has 52 questions; sign in to attempt the remaining 48 in the exam simulator.

1. Q1JEE Advanced Adv 2018 (Paper 1)
   Consider a body of mass $1.0kg$ at rest at the origin at time $t=0$. A force $\vec{F}=\left(\alpha t\hat{i}+\beta \hat{j}\right)$ is applied on the body, where $\alpha =1.0N{s}^{-1}and\beta =1.0N$ . The torque acting on the body about the origin at time $=1.0sis\vec{\tau }$ . Which of the following statements is (are) true?
   1. A.$\left|\vec{\tau }\right|=\dfrac{1}{3}Nm$
   2. B.The torque $\vec{\tau }$ is in the direction of the unit vector $+\vec{k}$
   3. C.The velocity of the body at $t=1sis\vec{v}=\dfrac{1}{2}\left(\hat{i}+2\hat{j}\right)m{s}^{-1}$
   4. D.The magnitude of displacement of the body at $t=1sis\dfrac{1}{6}m$
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   Answer: A,C

   $\vec{F}=\left(\alpha t\right)\hat{i}+\beta \hat{j}$ $\left[att=0,v=0,\vec{r}=\vec{0}\right]$ $\alpha =1,\beta =1$ $\vec{F}=t\hat{i}+\hat{j}$ $m\dfrac{d\vec{v}}{dt}=t\hat{i}+\hat{j}$ On integrating $m\vec{v}=\dfrac{{t}^{2}}{2}\hat{i}+t\hat{j}$ $\left[m=1kg\right]$ $\dfrac{d\vec{v}}{dt}\dfrac{{t}^{2}}{2}\hat{i}+t\hat{j}$ $\left[\vec{r}=\vec{0}att=0\right]$ On integrating $\vec{r}=\dfrac{{t}^{3}}{6}\hat{i}+\dfrac{{t}^{2}}{2}\hat{j}$ $\vec{\tau }=-\dfrac{1}{3}\hat{k}$ $\vec{v}=\dfrac{{t}^{2}}{2}\hat{i}+t\hat{j}$ At $t=1\vec{v}=\left(\dfrac{1}{2}\hat{i}+\hat{j}\right)=\dfrac{1}{2}\left(\hat{i}+2\hat{j}\right)m/sec$ At $t=1\vec{s}=\vec{{r}_{1}}-\vec{{r}_{0}}$ $=\left[\dfrac{1}{6}\hat{i}+\dfrac{1}{2}\hat{j}\right]-\left[\vec{0}\right]$ $\vec{s}=\dfrac{1}{6}\hat{i}+\dfrac{1}{2}\hat{j}$ $\left|\vec{s}\right|=\sqrt{{\left(\dfrac{1}{6}\right)}^{2}+{\left(\dfrac{1}{2}\right)}^{2}}=\dfrac{\sqrt{10}}{6}m$
2. Q2JEE Advanced Adv 2008 (Paper 1)
   Statement 1 Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first. and Statement 2 By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.
   1. A.Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
   2. B.Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
   3. C.Statement 1 is true, Statement 2 is false.
   4. D.Statement 1 is false, Statement 2 is true
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   Answer: (D)

   In case of pure rolling on inclined plane, $$\begin{aligned} & a=\frac{g \sin \theta}{1+I / m R^2} \\ & I_{\text {solid }} < I_{\text {hollow }} \\ \therefore \quad & a_{\text {solid }}>a_{\text {hollow }} \end{aligned}$$ $\therefore$ solid cylinder will reach the bottom first. Further, in case of pure rolling on stationary ground, work done by friction is zero. Therefore, mechanical energy of both the cylinders will remain constant. $$\begin{aligned} & \therefore(\mathrm{KE})_{\text {Hollow }}=(\mathrm{KE})_{\text {solid }}=\text { decrease in } \mathrm{PE}=m g h \\ & \therefore \text { correct option is }(\mathrm{d}) \end{aligned}$$
3. Q3JEE Advanced Adv 2007 (Paper 1)
   Paragraph: $\mathbf{P}_{17 \text { - } 19}$ : Paragraph for Questions Nos. 17 to 19 Two discs $A$ and $B$ are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I, respectively about the common axis. Disc $A$ is imparted an initial angular velocity $2 \omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc $B$ is imparted an angular velocity $\omega$ by a spring having the same spring constant and compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.Question: The loss of kinetic energy during the above process is
   1. A.$\frac{I \omega^2}{2}$
   2. B.$\frac{i \omega^2}{3}$
   3. C.$\frac{I \omega^2}{4}$
   4. D.$\frac{I \omega^2}{6}$
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   Answer: (B)

   Loss of kinetic energy $=K_i-K_f$ $$\begin{aligned} & =\left\{\frac{1}{2} I(2 \omega)^2+\frac{1}{2}(2 I)(\omega)^2\right\}-\frac{1}{2}(3 I)\left(\frac{4}{3} \omega\right)^2 \\ & =\frac{1}{3} I \omega^2 \end{aligned}$$ $\therefore$ Option (b) is correct.
4. Q4JEE Advanced Adv 2007 (Paper 1)
   Paragraph: $\mathbf{P}_{17 \text { - } 19}$ : Paragraph for Questions Nos. 17 to 19 Two discs $A$ and $B$ are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I, respectively about the common axis. Disc $A$ is imparted an initial angular velocity $2 \omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc $B$ is imparted an angular velocity $\omega$ by a spring having the same spring constant and compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.Question: The ratio $\frac{x_1}{x_2}$ is
   1. A.2
   2. B.$\frac{1}{2}$
   3. C.$\sqrt{2}$
   4. D.$\frac{1}{\sqrt{2}}$
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   Answer: (C)

   $$\begin{aligned} \frac{1}{2} I(2 \omega)^2 & =\frac{1}{2} k x_1^2 \\ \frac{1}{2}(2 I)(\omega)^2 & =\frac{1}{2} k x_2^2 \end{aligned}$$ From Eqs. (i) and (ii), we have $$\frac{x_1}{x_2}=\sqrt{2}$$ $\therefore$ Option (c) is correct.