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*Simulator*

**Previous Years JEE Advanced Questions**

(A) at $\theta=30^{\circ}$, the block will start sliding down the plane

(B) the block will remain at rest on the plane up to certain $\theta$ and then it will topple

(C) at $\theta=60^{\circ}$, the block will start sliding down the plane and continue to do so at higher angles

(D) at $\theta=60^{\circ}$, the block will start sliding down the plane and on further increasing q, it will topple at certain q

**[IIT-JEE 2009]**

**Sol.**(B)

(A) linear momentum of the system does not change in time

(B) kinetic energy of the system does not change in time

(C) angular momentum of the system does not change in time

(D) potential energy of the system does not change in time

**[IIT-JEE 2009]**

**Sol.**(A)

(A) $\vec{v}_{C}-\vec{v}_{A}=2\left(\vec{v}_{B}-\vec{v}_{C}\right)$

(B) $\vec{v}_{C}-\vec{v}_{B}=\vec{v}_{B}-\vec{v}_{A}$

(C) $\left|\vec{v}_{C}-\vec{v}_{A}\right|=2\left|\vec{v}_{B}-\vec{v}_{C}\right|$

(D) $\left|\vec{v}_{C}-\vec{v}_{A}\right|=4\left|\vec{v}_{B}\right|$

**[IIT-JEE 2009]**

**Sol.**(B,C)

**[IIT-JEE 2011]**

**Sol.**4

corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is $\mathrm{N} \times 10^{-4} \mathrm{kg}-\mathrm{m}^{2}$, then N is

**[IIT-JEE 2011]**

**Sol.**9

(A) the ring has pure rotation about its stationary CM

(B) the ring comes to a complete stop

(C) friction between the ring and the ground is to the left

(D) there is no friction between the ring and the ground

**[IIT-JEE 2011]**

**Sol.**(A,C or C)

**[IIT-JEE 2012]**

**Sol.**(B)

(A) $\vec{L}_{o}$ and $\vec{L}_{p}$ do not vary with time

(B) $\vec{L}_{o}$ varies with time while $\vec{L}_{p}$ remains constant

(C) $\vec{L}_{o}$ remains constant while $\vec{L}_{P}$ varies with time

(D) $\vec{L}_{o}$ and $\vec{L}_{P}$ both vary with time

**[IIT-JEE 2012]**

**Sol.**(C)

**[IIT-JEE 2012]**

**Sol.**3

$\mathrm{I}_{0}=\mathrm{I}_{\text {biggle }}-\mathrm{I}_{\text {mall }}$

$=\frac{4 \mathrm{m}(2 \mathrm{R})^{2}}{2}-\left(\frac{\mathrm{mR}^{2}}{2}+\mathrm{mR}^{2}\right)$

$=8 \mathrm{mR}^{2}-\frac{3}{2} \mathrm{mR}^{2}=\frac{13}{2} \mathrm{mR}^{2}$

$\mathrm{I}_{\mathrm{P}}=\left(\frac{4 \mathrm{m}(2 \mathrm{R})^{2}}{2}+4 \mathrm{m}(2 \mathrm{R})^{2}\right)-\left(\frac{\mathrm{mR}^{2}}{2}+\mathrm{m} 4 \mathrm{R}^{2}\right)=20-\frac{1}{2}=\frac{39}{2}$

$\therefore$ ratio $=\frac{39}{13}=3$

**Paragraph for Questions 10 and 11**

The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed , the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed $\omega$ in this case.

Now consider two similar systems as shown in the figure : case (A) the disc with its face vertical and parallel to x-z plane; Case (B) the disc with its face making an angle of $45^{\circ}$ with x-y plane and its horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed about the z-axis.

(A) It is $\sqrt{2 \omega}$ for both the cases.

(B) It is $\omega$ for case $(a) ;$ and $\frac{\omega}{\sqrt{2}}$ for case $(b)$

(C) It is $\omega$ for case $(a) ;$ and $\sqrt{2 \omega}$ for case (b).

(D) It is $\omega$ for both the cases.

**[IIT-JEE 2012]**

**Sol.**(D)

$\omega$ for both the cases.

$\therefore$ in same t $2 \pi$ angle.

(A) It is vertical for both the cases (a) and (b).

(B) It is vertical for case (a); and is at $45^{\circ}$ to the x-z plane and lies in the plane of the disc for case (b).

(C) It is horizontal for case (a); and is at $45^{\circ}$ to the x-z plane and is normal to the plane of the disc for case (b).

(D) It is vertical for case (a); and is at $45^{\circ}$ to the x-z plane and is normal to the plane of the disc for case (b)

**[IIT-JEE 2012]**

**Sol.**(A)

vertical for both.

(A) the point O has a linear velocity $3 R \omega \hat{i}$

(B) the point $P$ has a linear velocity $\frac{11}{4} R \omega \hat{i}+\frac{\sqrt{3}}{4} R \omega \hat{k}$

(C) the point P has a linear velocity $\frac{13}{4} R \omega \hat{i}-\frac{\sqrt{3}}{4} R \omega \hat{k}$

(D) the point P has a linear velocity $\left(3-\frac{\sqrt{3}}{4}\right) R \omega \hat{i}+\frac{1}{4} R \omega \hat{k}$

**[IIT-JEE 2012]**

**Sol.**(A,B)

(A) Both cylinders P and Q reach the ground at the same time.

(B) Cylinder P has larger acceleration than cylinder Q.

(C) Both cylinders reach the ground with same translational kinetic energy.

(D) Cylinder Q reaches the ground with larger angular speed.

**[IIT-JEE 2012]**

**Sol.**(D)

10 rad $\mathrm{s}^{-1}$ about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s–1) of the system is

**[IIT-JEE 2013]**

**Sol.**8

by angular momentum conservation

$\mathrm{I}_{\text {disc }} \omega_{0}=\left(\mathrm{I}_{\text {disc }}+2 \mathrm{I}_{\text {ring }}\right) \omega$

**[JEE Advanced-2014]**

**Sol.**4

Since no external torque acts therefore angular momentum remains conserved.

Angular momentum of ball = Angular momentum of plateform

$0.05 \times 9 \times 0.25 \times 2=\frac{1}{2} \times 0.45 \times 0.5 \times 0.5 \times \omega$

$\omega=4 \mathrm{rad} / \mathrm{s}$

**[JEE Advanced-2014]**

**Sol.**2

Angular impulse = change in angular momentum

$\tau \Delta \mathrm{t}=\mathrm{I} \omega$

$3 \times \mathrm{F} \times \mathrm{R} \sin 30 \times \Delta \mathrm{t}=\mathrm{I} \omega$

$3 \times 0.5 \times 0.5 \times \frac{1}{2} \times 1=\frac{1}{2} \times 1.5 \times 0.5 \times 0.5 \times \omega$

$\omega=2 \mathrm{rad} / \mathrm{s}$

**[JEE Advanced-2015]**

**Sol.**7

(A) $\frac{2}{3} \mathrm{R}$

(B) $\frac{1}{3} \mathrm{R}$

(C) $\frac{3}{5} \mathrm{R}$

(D) $\frac{4}{5} \mathrm{R}$

**[JEE Advanced-2015]**

**Sol.**(C,D)

**[JEE Advanced-2015]**

**Sol.**6

(A) $\frac{\mathrm{h}}{\ell}=\frac{\sqrt{3}}{16}, \mathrm{f}=\frac{16 \sqrt{3}}{3} \mathrm{N}$

(B) $\frac{\mathrm{h}}{\ell}=\frac{3}{16}, \mathrm{f}=\frac{16 \sqrt{3}}{3} \mathrm{N}$

(C) $\frac{\mathrm{h}}{\ell}=\frac{3 \sqrt{3}}{16}, \mathrm{f}=\frac{8 \sqrt{3}}{3} \mathrm{N}$

(D) $\frac{\mathrm{h}}{\ell}=\frac{3 \sqrt{3}}{16}, \mathrm{f}=\frac{16 \sqrt{3}}{3} \mathrm{N}$

**[JEE Advanced-2016]**

**Sol.**(D)

where $\alpha=\frac{10}{3} \mathrm{ms}^{-3}, \beta=5 \mathrm{ms}^{-2}$ and $\mathrm{m}=0.1 \mathrm{kg} .$ At $\mathrm{t}=1 \mathrm{s},$ which of the following statement(s)

is(are) true about the particle?

(A) The velocity $\overrightarrow{\mathrm{v}}$ is given by $\overrightarrow{\mathrm{v}}=(10 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}) \mathrm{ms}^{-1}$

(B) The angular momentum $\overrightarrow{\mathrm{L}}$ with respect to the origin is given by $\overrightarrow{\mathrm{L}}=-\left(\frac{5}{3}\right) \hat{\mathrm{k}} \mathrm{Nms}$

(C) The force $\overrightarrow{\mathrm{F}}$ is given by $\overrightarrow{\mathrm{F}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}) \mathrm{N}$

(D) The torque $\vec{\tau}$ with respect to the origin is given by $\vec{\tau}=-\left(\frac{20}{3}\right) \hat{\mathrm{k}} \mathrm{Nm}$

**[JEE Advanced-2016]**

**Sol.**(A,B,D)

$\overrightarrow{\mathrm{r}}=\alpha \mathrm{t}^{3} \hat{\mathrm{i}}+\beta \mathrm{t}^{2} \hat{\mathrm{j}}$

$\overrightarrow{\mathrm{v}}=\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=3 \alpha \mathrm{t}^{2} \hat{\mathrm{i}}+2 \beta \hat{\mathrm{t} \hat{\mathrm{j}}}$

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d}^{2} \overrightarrow{\mathrm{r}}}{\mathrm{dt}^{2}}=6 \alpha t \hat{\mathrm{i}}+2 \beta \hat{\mathrm{j}}$

At t = 1

(A) $\overrightarrow{\mathrm{v}}=3 \times \frac{10}{3} \times 1 \hat{\mathrm{i}}+2 \times \mathrm{J} \times 1 \hat{\mathrm{j}}$

$=10 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}$

$(\mathrm{B}) \quad \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}$

$=\left(\frac{10}{3} \times 1 \hat{i}+5 \times 1 \hat{j}\right) \times 0.1(10 \hat{i}+10 \hat{j})$

$=-\frac{5}{3} \hat{\mathrm{k}}$

(C) $\overrightarrow{\mathrm{F}}=\mathrm{m} \times\left(6 \times \frac{10}{3} \times 1 \hat{\mathrm{i}}+2 \times 5 \hat{\mathrm{j}}\right)=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}$

$(\mathrm{D}) \vec{\tau}=\mathrm{r} \times \overrightarrow{\mathrm{F}}$

$=\left(\frac{10}{3} \hat{\mathrm{i}}+5 \hat{\mathrm{j}}\right) \times(2 \hat{\mathrm{i}}+\hat{\mathrm{j}})$

$=+\frac{10}{3} \hat{\mathrm{k}}+10(-\hat{\mathrm{k}})$

$=-\frac{20}{3} \hat{\mathrm{k}}$

(A) The magnitude of angular momentum of the assembly about its center of mass is $17 \mathrm{ma}^{2} \omega / 2$

(B) The magnitude of the z-component $\overrightarrow{\mathrm{L}}$ is $55 \mathrm{ma}^{2} \omega$

(C) The magnitude of angular momentum of center of mass of the assembly about the point O is 81 $\mathrm{ma}^{2} \mathrm{\omega}$

(D) The center of mass of the assembly rotates about the z-axis with an angular speed of $\omega$/ 5

**[JEE Advanced-2016]**

**Sol.**(A,D)

By no slip condition, here $\omega^{\prime}$ is angular velocity about z axis

$\omega^{\prime} x=\omega r$

$\omega^{\prime}=\frac{\omega r}{x}=\omega \sin \theta=\frac{\omega}{5}$

**Paragraph 1**

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity is an example of a non-inertial frame of reference. The relationship between the force $\overrightarrow{\mathrm{F}}_{\mathrm{rot}}$ experienced by a particle of mass m moving on the rotating disc and the force $\overrightarrow{\mathrm{F}}_{\mathrm{in}}$ experienced by the particle in an inertial frame of reference is

$\overrightarrow{\mathrm{F}}_{\mathrm{rot}}=\overrightarrow{\mathrm{F}}_{\mathrm{in}}+2 \mathrm{m}\left(\overrightarrow{\mathrm{v}}_{\mathrm{rot}} \times \vec{\omega}\right)+\mathrm{m}(\vec{\omega} \times \overrightarrow{\mathrm{r}}) \times \vec{\omega}$

where $\vec{v}_{\text {rot }}$ is the velocity of the particle in the rotating frame of reference and $\overrightarrow{\mathrm{r}}$ is the position

vector of the particle with respect to the centre of the disc.

Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed about its vertical axis through its center. We assign a coordinate system with the origin at the centre of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis $(\vec{\omega}=\omega \hat{k})$ A small block of mass m is gently placed in the slot at $\overrightarrow{\mathrm{r}}=(\mathrm{R} / 2) \hat{\mathrm{i}}$ at $\mathrm{t}=0$ and is constrained to move only along the slot.

(A) $\frac{\mathrm{R}}{4}\left(\mathrm{e}^{2 \omega \mathrm{t}}+\mathrm{e}^{-2 \omega \mathrm{t}}\right)$

(B) $\frac{\mathrm{R}}{2} \cos 2 \omega \mathrm{t}$

(C) $\frac{\mathrm{R}}{2} \mathrm{cos} \omega \mathrm{t}$

(D) $\frac{\mathrm{R}}{4}\left(\mathrm{e}^{\mathrm{ot}}+\mathrm{e}^{-\mathrm{ot}}\right)$

**[JEE Advanced-2016]**

**Sol.**

**(D)**

Force on block along slot $=m \omega^{2} \mathrm{r}=\mathrm{ma}=\mathrm{m}\left(\frac{\mathrm{vdv}}{\mathrm{dr}}\right)$

$\int_{0}^{\mathrm{v}} \mathrm{vd} \mathrm{v}=\int_{\mathrm{R} / 2}^{\mathrm{r}} \omega^{2} \mathrm{rdr}$

$\frac{\mathrm{v}^{2}}{2}=\frac{\omega^{2}}{2}\left(\mathrm{r}^{2}-\frac{\mathrm{R}^{2}}{4}\right) \Rightarrow \mathrm{v}=\omega \sqrt{\mathrm{r}^{2}-\frac{\mathrm{R}^{2}}{4}}=\frac{\mathrm{dr}}{\mathrm{dt}}$

$\Rightarrow \int_{\mathrm{R} / 4}^{\mathrm{r}} \frac{\mathrm{dr}}{\sqrt{\mathrm{r}^{2}-\frac{\mathrm{R}^{2}}{4}}}=\int_{0}^{\mathrm{t}} \omega \mathrm{dt}$

$\ell_{\mathrm{n}}\left(\frac{\mathrm{r}+\sqrt{\mathrm{r}^{2}-\frac{\mathrm{R}^{2}}{4}}}{\frac{\mathrm{R}}{2}}\right)-\ell \mathrm{n}\left(\frac{\mathrm{R} / 2+\sqrt{\frac{\mathrm{R}^{2}}{4}-\frac{\mathrm{R}^{2}}{4}}}{\frac{\mathrm{R}}{2}}\right)=\omega \mathrm{t}$

$\Rightarrow \mathrm{r}+\sqrt{\mathrm{r}^{2}-\frac{\mathrm{R}^{2}}{4}}=\frac{\mathrm{R}}{2} \mathrm{e}^{\omega \mathrm{t}}$

$\Rightarrow \mathrm{r}^{2}-\frac{\mathrm{R}^{2}}{4}=\frac{\mathrm{R}^{2}}{4} \mathrm{e}^{2 \omega \mathrm{t}}+\mathrm{r}^{2}-2 \mathrm{r} \frac{\mathrm{R}}{2} \mathrm{e}^{\omega \mathrm{t}}$

$\Rightarrow \mathrm{r}=\frac{\frac{\mathrm{R}^{2}}{4} \mathrm{e}^{2 \omega \mathrm{t}}+\frac{\mathrm{R}^{2}}{4}}{\mathrm{Re}^{\mathrm{ot}}}=\frac{\mathrm{R}}{4}\left(\mathrm{e}^{\omega \mathrm{t}}+\mathrm{e}^{-\omega \mathrm{t}}\right)$

(A) $-\mathrm{m\omega}^{2} \mathrm{R}$ cos $\omega$ tì $\hat{\mathrm{j}}-\mathrm{mg} \hat{\mathrm{k}}$

(B) $\mathrm{m\omega}^{2} \mathrm{R} \sin \omega \hat{\mathrm{j}}-\mathrm{mg} \hat{\mathrm{k}}$

(C) $\frac{1}{2} \mathrm{m\omega}^{2} \mathrm{R}\left(\mathrm{e}^{\omega \mathrm{t}}-\mathrm{e}^{-\omega \mathrm{t}}\right) \hat{\mathrm{j}}+\mathrm{mg} \hat{\mathrm{k}}$

(D) $\frac{1}{2} \mathrm{m\omega}^{2} \mathrm{R}\left(\mathrm{e}^{2 \omega \mathrm{t}}-\mathrm{e}^{-2 \omega \mathrm{t}}\right) \hat{\mathrm{j}}+\mathrm{mg} \hat{\mathrm{k}}$

**[JEE Advanced-2016]**

**Sol.**(C)

(A) When the bar makes an angle $\theta$ with the vertical, the displacement of its midpoint from the initial position is proportional to $(1-\cos \theta)$

(B) The midpoint of the bar will fall vertically downward

(C) Instantaneous torque about the point in contact with the floor is proportional to $\sin \theta$

(D) The trajectory of the point $A$ is a parabola

**[JEE Advanced-2017]**

**Sol.**(A,B,C)

When the bar makes an angle $\theta ;$ the height of its COM (mid point) is $\frac{\mathrm{L}}{2} \cos \theta$

$\therefore$ displacement $=\mathrm{L}-\frac{\mathrm{L}}{2} \cos \theta=\frac{\mathrm{L}}{2}(1-\cos \theta)$

Since force on COM is only along the vertical direction, hence COM is falling vertically downward.

Instantaneous torque about point of contact is

$\mathrm{Mg} \times \frac{\mathrm{L}}{2} \sin \theta$

(A) If the force is applied normal to the circumference at point X then $\tau$ is constant

(B) If the force is applied tangentially at point S then $\tau \neq 0$ but the wheel never climbs the step

(C) If the force is applied normal to the circumference at point P then $\tau$ is zero

(D) If the force is applied at point P tangentially then t decreases continuously as the wheel climbs

**[JEE Advanced-2017]**

**Sol.**(C)

(A) is incorrect.

if force is applied at P tangentially the

$\tau=\mathrm{F} \times 2 \mathrm{R}=\mathrm{constant}$

**Paragraph for Questions 27 and 28**

One twirls a circular ring (of mass M and radius R) near the tip of one’s finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity $\omega_{0}$. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is µ and the acceleration due to gravity is g.

(A) $\mathrm{M} \omega_{0}^{2} \mathrm{R}^{2}$

(B) $\operatorname{Mo}_{0}^{2}(\mathrm{R}-\mathrm{r})^{2}$

(C) $\frac{1}{2} \mathrm{M} \omega_{0}^{2}(\mathrm{R}-\mathrm{r})^{2}$

(D) $\frac{3}{2} \mathrm{M} \omega_{0}^{2}(\mathrm{R}-\mathrm{r})^{2}$

**[JEE Advanced-2017]**

**Sol.**Bonus

(A) $\sqrt{\frac{3 \mathrm{g}}{2 \mu(\mathrm{R}-\mathrm{r})}}$

(B) $\sqrt{\frac{\mathrm{g}}{\mu(\mathrm{R}-\mathrm{r})}}$

(C) $\sqrt{\frac{2 \mathrm{g}}{\mu(\mathrm{R}-\mathrm{r})}}$

(D) $\sqrt{\frac{\mathrm{g}}{2 \mu(\mathrm{R}-\mathrm{r})}}$

**[JEE Advanced-2017]**

**Sol.**(B)

$\mu \mathrm{M} \omega_{0}^{2}(\mathrm{R}-\mathrm{r})=\mathrm{Mg}$

$\omega_{0}=\sqrt{\frac{\mathrm{g}}{\mu(\mathrm{R}-\mathrm{r})}}$

(A) $v=\sqrt{\frac{\mathrm{k}}{2 \mathrm{m}}} \mathrm{R}$

(B) $v=\sqrt{\frac{k}{m}} R$

(C) $\mathrm{L}=\sqrt{\mathrm{mk}} \mathrm{R}^{2}$

**[JEE Advanced-2018]**

**Sol.**(B,C)

applied on the body, where $\alpha=1.0 \mathrm{Ns}^{-1}$ and $\beta=1.0 \mathrm{N}$. The torque acting on the body about

the origin at time $\mathrm{t}=1.0 \mathrm{s}$ is $\vec{\tau} .$ Which of the following statements is (are) true?

(A) $|\vec{\tau}|=\frac{1}{3} \mathrm{Nm}$

(B) The torque $\vec{\tau}$ is in the direction of the unit vector $+\hat{\mathrm{k}}$

(C) The velocity of the body at $\mathrm{t}=1 \mathrm{s}$ is $\overrightarrow{\mathrm{v}}=\frac{1}{2}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}) \mathrm{ms}^{-1}$

(D) The magnitude of displacement of the body at $t=1$ s is $\frac{1}{6} m$

**[JEE Advanced-2018]**

**Sol.**(A,C)

**[JEE Advanced-2018]**

**Sol.**0.75

**[JEE Advanced-2018]**

**Sol.**(A)

Thanks Saransh sir