Q141JEE Main 2019MCQ

A thin disc of mass M and radius R has mass per unit area $\sigma (r) =kr^2$ where r is the distance from its center . Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is :

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📖 Explanation

${I}=∑{dmr}^{2}$ ${I}=∑(2 \pi {x}) {d} {x} ( {kx}^{2} ) {x}^{2}$ $=2 \pi {k} ∫{x}^{5} {d} {x}=2 \pi {k} \$ [ \frac{ {x}^{6}}{6} ]_{0}^{ {r}}=2 \pi {k} \frac{ {R}^{6}}{6}= \frac{2 \pi {kR}^{6} × 4}{6× 4}= \frac{2 {mR}^{2}}{3}$

Q142JEE Main 2019MCQ

A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of θ , where θ is the angle by which it has rotated, is given as 2 $k\theta ^{2}$ . If its moment of inertia is I then the angular accelerationof the disc is:

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📖 Explanation

$K E=K \theta ^{2}$ $\Rightarrow\ \frac{1}{2} I W^{2}=K \theta$ Differentiating on both sides w.r.t to θ $\ \frac{1}{2} I \ \frac{d}{d_{0}}\ (\omega ^{2} )=k \ \frac{d}{d \theta }\ (\theta ^{2}\ )$ $I \alpha =2 \theta (K)\Rightarrow \alpha =\ \frac{2 k}{I}=\theta$

Q143JEE Main 2019MCQ

Two particles A, B are moving on two concentric circles of radii $R_1$ and $R_2$ with equal angular speed ω. At t = 0, their positions and direction of motion are shown in the figure : The relative velocity $υ_\vec{A}- υ_\vec{B}$ at t = $\pi /{2\omega }$ is given by :

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📖 Explanation

θ = ωt = ω $\pi /{2\omega }$ = $\pi /2$ $V_\vec{A}-V_\vec{S}$ = ω $R_1(-\hat{i})$ - ω $R_2(-i)$

Q144JEE Main 2019MCQ

Two Coaxial discs, having moments of inertia $I_1$ and $I_1 /2$ are rotating with respective angular velocity $\omega _1$ and $\omega _1/2$ , about their common axis.They are brought in contact with each other and thereafter they rotate with a common angular velocity. If $E_f$ and $E_i$ , are the final and initial total energies, then $(E_{f}-E_{i} )$ is

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Q145JEE Main 2019MCQ

An L-shaped object, made of thin rods ofuniform mass density, is suspended with astring as shown in figure. If AB = BC, and theangle made by AB with downward vertical isθ, then : [9-Jan-2019 Shift 1]

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📖 Explanation

Let mass of one rod is m. Balancing torque about hinge point mg $(C_1P)$ = mg $(C_2N)$ mg $(L/2 sin\theta )$ = mg $(L/2 cos\theta - Lsin\theta )$ \Rightarrow $3/2$ mgL sin θ = ${mgL}/2$ cos θ \Rightarrow tan θ = $1/3$

Q146JEE Main 2019MCQ

A metal of mass 5 g and radius 1 cm is fixed to a thin stick AB if negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25 rotations per second in 5s, is close to:

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📖 Explanation

$w=w_{0}+\alpha t$ $(25)(2 \pi )=0+\alpha (5)$ $\alpha =10 \pi \ \;{rad}$ / ${sec} ^{2}$ $\tau =I \alpha$ $=\ (\ \frac{M R^{2}}{4}+M R^{2}\ )(10 \pi )$ $= (\ \frac{5}{4} M R^{2}\ )(10 \pi )$ $= \frac{5}{4} \\times 5 \\times 10^{-3} \\times (1 \times 10^{-2}\ )^{2} \\times 10 \times \pi$ $=2 \\times 10^{-5} N m$

Q147JEE Main 2019MCQ

A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of $7M/ 8$ and is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere. Let $I_1$ be the moment of inertia of the disc about its axis and $I_2$ be the moment of inertia of the new sphere about its axis. The ratio $I_1$ / $I_2$ is given by:

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📖 Explanation

$I_{1}=\ \frac{1}{2} m_{1} r_{2}, I_{2}= \frac{2}{5} m_{2} r_{2}^{2}$ $\Rightarrow\frac{I_{1}}{I_{2}}= \frac{\ \frac{1}{2} \ \frac{7 m}{8}(2 R)^{2}}{\ \frac{2}{5} \ \frac{m}{8}\ (\ \frac{R}{2}\ )^{2}}$ $=\ \frac{1}{2} \times \frac{7}{8} \\times 4 \times \frac{5}{2} \times \frac{8}{1} \\times \frac{4}{1}$ $=140$

Q148JEE Main 2019MCQ

An electric dipole is formed by two equal and opposite charges q with separation d. The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency $\omega$ is :

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📖 Explanation

$\vec{T}= {P} ^{\to }\times \vec{E}$ $2 (m ( \frac{d}{2} )^{2} ) \alpha =(q d)(E) \sin \theta$ $\Rightarrow \omega =\sqrt{ \frac{2 q E d}{m d^{2}}}$ $\Rightarrow \omega =\sqrt{ \frac{2 q E}{m d}}$

Q149JEE Main 2019MCQ

Moment of inertia of a body about a given axis is $1. 5 kgm ^{2}$ . Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration of ${20 rad}/s^{2}$ must be applied about the axis for a duration of:

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📖 Explanation

$I=1.5 {kgm}^{2}$ Rot. $K \epsilon =1200 \ {J}$ $1200= \frac{1}{2} \\times 1.5 \\times {\omega }^{2}$ $\omega =40 \ {rad} / {s}$ $\ \therefore \;\;\;\; t=\ \frac{\omega }{\alpha }=\ \frac{40}{20}=2 s$

Q150JEE Main 2019MCQ

The moment of inertia of a solid sphere, about anaxis parallel to its diameter and at a distance of xfrom it, is I(x)'. Which one of the graphs representsthe variation of I(x) with x correctly ?

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📖 Explanation

I = $2/5 mR^2 + mx^2$

Q151JEE Main 2019MCQ

A slob is subjected to two forces $F_\vec{1}$ and $F_\vec{2}$ of same magnitude F as shown in the figure. Force $F_\vec{2}$ is in XY-plane while force $F_\vec{1}$ acts along z-axis at the point $(2\hat{i}+3\hat{j})$ . The moment of these forces about point O will be :

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📖 Explanation

Torque for $F_\vec{1}$ force $F_\vec{1}$ = $F/2(-\hat{i})$ + ${F\sqrt3}/2 (-\hat{j})$ $r_\vec{1}$ = $0\hat{i}+6\hat{j}$ $\tau _\vec{F_1}$ = $r_\vec{1}\times F_\vec{1}$ = $3F\hat{k}$ Torque for $F_\vec{2}$ force $F_\vec{2}$ = $F\hat{k}$ $r_\vec{2}$ = $2\hat{i}+3\hat{j}$ $\tau _\vec{F_2}$ = $r_\vec{2}\times F_\vec{2}$ = $3F\hat{i}+2F(-\hat{j})$ $\tau _\vec{\text{net}}$ = $\tau _\vec{F_1}+\tau _\vec{F_2}$ = $3F\hat{i}+2F(-\hat{j})+3F(\hat{k})$

Q152JEE Main 2019MCQ

A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration ofthe cylinder will be (Neglect the mass and thickness of the string) :-

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📖 Explanation

40 + f = m(Rα) .....(i) 40 × R - f × R = m $R^2$ α 40 - f = mRα ...... (ii) From (i) and (ii) α = $\frac{40}{mR}$ = 16

Q153JEE Main 2019MCQ

A particle of mass m is moving along a trajectory given by $x=x_{0}+a \cos \omega _{1} t$ $\;\; y= y_{0}+b \sin \omega _{2} t$ The torque,acting on the particle about the origin, at t=0 is :

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📖 Explanation

${\ \vec{\text{r}}}=\ (\ {x}_{\ {o}}+\ {a} \cos \omega _{1} {\text{t}}\ ) {\text{i}}+\ (\ {\text{y}}_{0}+\ {\text{b}} \sin \omega _{2} {\text{t}}\ ) \ {\text{j}};$ at t=0 $\ \vec{r}=\ (x_{0}+a\ ) i+y_{0} \ \text{j}$ $\ \vec{a}=\ \frac{d^{2} x}{d t^{2}} \ \hat{i}+\ \frac{d^{2} y}{d t^{2}} \ \hat{j}$ $\ \vec{r}=\ (-a \omega _{1}^{2} \cos \omega _{1} t\ ) \ \hat{i}+\ (-b \omega _{2}^{2} \sin \omega _{2} t\ ) \text{j};$ at t=0 $\ {\ \vec{a}}=-\ {a} \omega _{1}^{2} \ {\ \hat{i}}$ ${T}=\ {\ \vec{r}} \times \ {\ \vec{F}}$ $=\ {m}(\ {\ \vec{r}} \times {\ \vec{a}})$ $=\ {m} (\ {y}_{\ {o}}\ (-\ {a} \omega _{1}^{2}\ )-\ {k} )$ $=\ {my}_{0} a\omega \hat{i}\hat{k}$

Q154JEE Main 2019MCQ

A circular disc $D_1$ of mass M and radius R has two identical discs $D_2$ and $D_3$ of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO', passing through the centre of $D_1$ , asshown in the figure, will be:-

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📖 Explanation

I = ${MR^2}/2+2({MR^2}/4+MR^2)$ = ${MR^2}/2+{MR^2}/2+2MR^2$ = 3 $MR^2$

Q155JEE Main 2019MCQ

A thin circular plate of mass M and radius R has its density varying as $\rho (r)=\rho _{0} r$ with $\rho _0$ as constant and r is the distance from its center. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is $I= a MR^2$ . The value of the coefficient a is:

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📖 Explanation

$∫ d I_{m}=∫_{{0}} ^{{R} }d m r^{2}$ $=2 \pi ∫ \rho _{0} r^{4} d r$ $= {2 \pi \rho _{0} R^{5}}{5}= \frac{3 M R^{2}}{5}$ $M=∫ d m=∫_{0}^ {R} \rho _{0} (2 \pi r^{2} d r )$ $M= \frac{\rho _{0}(2 \pi ) R^{3}}{3}$ $I_{ {edge}}=I_{c m}+M R^{2}= \frac{8}{5} M R^{2}$

Q156JEE Main 2018MCQ

A thin ${\text{rod}} MN$ , free to rotate in the vertical plane about the fixed end N , is held horizontal. When the end $M$ is released the speed of this end, when the rod makes an angle a with the horizontal, will be proportional to : (see figure)

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Q157JEE Main 2018MCQ

A thin uniform bar of length $L$ and mass 8 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in the same horizontal plane from opposite sides of the bar with speeds 2 v and $v$ respectively. The masses stick to the bar after collision at a distance $\;\frac{ L }{3}$ and $\;\frac{ L }{6}$ respectively from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be :

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Q158JEE Main 2018MCQ

From a uniform circular disc of radius R and mass $9M$ , a small disc of radius $R/3$ is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is :[Main 8 April 2018]

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📖 Explanation

$I_\text{full}=I_\text{scooped}+I_\text{remaining}$ ${9M.R^2}/2=\$ [\frac{MR^2}{2×9}+{4MR^2}/9]$+I_\text{rem}{9MR^2}/2=9/18MR^{2+}I_\text{rem}I_\text{rem}=4MR^2$

Q159JEE Main 2018MCQ

A thin circular disk is in the xy plane as shown in the figure. The ratio of its moment of inertia about z and z' axes will be :[Main 16 April 2018 S1]

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Q160JEE Main 2018MCQ

A uniform rod AB is suspended from a point X, at a variable distance x from A,as shown. To make the rod horizontal, a mass m is suspended from its end A. Aset of (m,x) values is recorded. The appropriate variables that give a straight line, when plotted, are :[Main 15 April 2018 S1]

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