Q161JEE Main 2018MCQ

Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is :[Main 8 April 2018]

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📖 Explanation

Use parallel axis theorem $I_0=6\times \$ [{MR^2}/2]$+$[{MR^2}/2]$+6M(2R)^2=3MR^{2+}{MR^2}/2+24MR^2=55/2MR^2I_P=I_0+7M(3R^2)=55/2MR^{2+}63MR^2=181/2MR^2$

Q162JEE Main 2018MCQ

A force of 40 N acts on a point B at the end of an L-shaped object, as shown in the figure. The angle θ that will produce maximum moment of the force about point A is given by :[Main 15 April 2018 S1]

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📖 Explanation

We can produce maximum moment of force, when line of action of force is perpendicular to line $A B$ . $\therefore \tan \theta =\;\frac{2}{4}=\;\frac{1}{2}$

Q163JEE Main 2017MCQ

In a physical balance working on the principle of moments, when 5 mg weight is placed on the left pan, the beam becomes horizontal. Both the empty pans of the balance are of equal mass. Which of the following statements is correct ?[Main 8 April 2017]

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Q164JEE Main 2017MCQ

A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is :[Main 2017]

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📖 Explanation

Taking torque about pivot $\tau =I\alpha$ mg sin θ $l/2={ml^2}/3\alpha$ $\alpha =\frac{3g}{2l}$ sin θ

Q165JEE Main 2017MCQ

The machine as shown has 2 rods of length 1 m connected by a pivot at the top. The end of one rod is connected to the floor by a stationary pivot and the end of the other rod has a roller that rolls along the floor in a slot. As the roller goes back and forth,a 2 kg weight moves up and down. If the roller is moving towards right at a constant speed, the weight moves up with a :[Main 9 April 2017]

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Q166JEE Main 2017MCQ

A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is :[Main 8 April 2017]

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📖 Explanation

$mg-T=ma$ $RT=I\propto$ $RT={MR^2}/2{.}a/R$ $T={Ma}/2$ $mg-{Ma}/2=ma$ $mg=a(M/2+m)$ $mg=a({M+2m}/2)$ $a=\frac{2mg}{M+2m}$

Q167JEE Main 2017MCQ

Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is $I_o$ as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is :[Main 8 April 2017]

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📖 Explanation

$I_0=kml^2$ $BC=I$ $I_\text{DEF}=Km/4(I/2)^2$ $=k/16ml^2$ $I_\text{DEF}=I_0/16$ $I_\text{remain}=I_0=I_0/16$ $={15I_0}/16$

Q168JEE Main 2017MCQ

A circular hole of radius $R/4$ is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is :[Main 9 April 2017]

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Q169JEE Main 2017MCQ

The moment of inertia of a uniform cylinder of length l and radius R about its perpendicularbisector is I. What is the ratio l/R such that the moment of inertia is minimum ?[Main 2017]

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📖 Explanation

$I={ml^2}/12+{mR^2}/4$ or $I=m/4(l^2/3+R^2)$ ......(1)Also $m=\pi R^2l\rho$ $\Rightarrow R^2 =m{\pi l\rho }$ Put in equation (1) $I=m/4(l^2/3+m/{\pi l\rho })$ For maxima & minima $\frac{dI}{dl}=m/4({2l}/3-m/{\pi l^2\rho })=0$ $\Rightarrow {2l}/3=m/{\pi l^2\rho }\Rightarrow {2l}/3=\frac{\pi R^2l\rho }{\pi l^2\rho }$ or ${2l}/3=R^2/l$ $\Rightarrow l^2/R^2=3/2$ or $l^2/R\Rightarrow \sqrt{3/2}$

Q170JEE Main 2016MCQ

A cubical block of side 30 cm is moving with velocity $2ms^{-1}$ on a smooth horizontal surface. The surface has a bump at a point O as shown in figure. The angular velocity (in rad/s) of the block immediately after it hits the bump, is :[Main 9 Apr 2016]

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📖 Explanation

Using conservation of angular momentum $mva/2=2/3ma^2\omega \Rightarrow \omega =\frac{3v}{4a}=5$ rad/s

Q171JEE Main 2016MCQ

Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute(rpm) to ensure proper mixing is close to :(Take the radius of the drum to be 1.25 mand its axle to be horizontal) :[Main 10 Apr 2016]

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📖 Explanation

$C\le \sqrt{5gr}$ $\omega =v/r\le \sqrt{{5g}/r}=\sqrt{{50\times 8}/10}=\sqrt{400/10}$ $\omega =\sqrt40$ rad/s $={60\sqrt10}/\pi$ rpm

Q172JEE Main 2016MCQ

A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to :[Main 2016]

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Q173JEE Main 2016MCQ

A particle of mass m is moving along the side of a square of side 'a', with auniform speed v in the x-y plane as shown in the figure :Which of the following statements is false for the angular momentum $\vec{L}$ about the origin ?[Main 2016]

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📖 Explanation

$\vec{L}_O=mvR/\sqrt2(-k^{∧})$ [D to A] $\vec{L}_O=mv[R/\sqrt2+a]k^{∧}$ [C to D]

Q174JEE Main 2015MCQ

From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is :[Main 2015]

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📖 Explanation

For maximum possible volume of cube $2R=\sqrt{3a}$ ,a is side of the cubeMoment of inertia about the required axis $=I= \rho a^3a^2/6,a^2/6,$ where $\rho =M/{4/3\pi R^3}$ $I=\frac{3M}{4\pi R^3}1/6({2R}/\sqrt3)^5=\frac{3M}{4\pi R^3}1/6\frac{32R^5}{9\sqrt3}=\frac{4MR^2}{9\sqrt3\pi }=\frac{4MR^2}{9\sqrt3\pi }$

Q175JEE Main 2014MCQ

A mass 'm' is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R.If the string does not slip on the cylinder,with what acceleration will the mass fall on release?[Main 2014]

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📖 Explanation

For the mass m,mg-T=mafor the cylinder,TR $=mR^2q/R$ ⇒T = ma⇒mg = 2ma⇒a = g/2

Q176JEE Main 2014MCQ

A bob of mass m attached to an inextensible string of the length $l$ is suspended from a vertical support.The bob rotates in a horizontal circle with an angular speed $\omega$ rad/s about the vertical.About the point of suspension:[Main 2014]

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📖 Explanation

$\vec{L}$ changes in direction not in magnitude

Q177JEE Main 2013MCQ

A hoop of radius r and mass m rotating with an angular velocity $\omega _0$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?[Main 2013]

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📖 Explanation

From conservation of angular momentum about any fix point on the surface $mr^2\omega _0=2mr^2\omega$ $\therefore \omega ={\omega _0}/2$ $\therefore V_{CM}={\omega _0r}/2$

Q178JEE Main 2011MCQ

A pulley of radius $2 {\text{m}}$ is rotated about its axis by a force $F= (20 t-5 t^2)$ newton (where $t$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is $10 {\text{kg}}- {\text{m}}^2$ the number of rotation made by the pulley before its direction of motion is reversed, is:

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📖 Explanation

Given $F=20 t-5 t^2, {\text{R}}=2 {\text{m}}$ and $I=10 {\text{kg}} {\text{m}}^2$ Torque applied on pulley $\tau =F R$ $\; \therefore \alpha =\;\frac{F R}{I}[\;\text{ as }\; \tau =I \alpha ]$ $\;\Rightarrow \alpha =\;\frac{ (20 t-5 t^2) \times 2}{10}$ $\;\Rightarrow \alpha =4 t-t^2$ $\;\Rightarrow \;\frac{d \omega }{d t}=4 t-t^2$ $\;\Rightarrow ∫_{0}^{\omega } d \omega =∫_{0}^{t} (4 t-t^2) \text{dt}$ $\;\Rightarrow \omega =2 t^{2-}\;\frac{r^3}{3}$ $\;(\;\text{ At }\; t=0,6 s \omega =0)$ $\;\omega =\;\frac{d \theta }{d t}=2 t^{2-}\;\frac{t^3}{3}$ $\;∫_{0}^{\theta } d \theta =∫_{0}^{6} (2 t^{2-}\;\frac{r^3}{3}) \text{dt}$ $\Rightarrow \theta =36 {\text{rad}} \Rightarrow n=\;\frac{36}{2 \pi }< 6$

Q179JEE Main 2011MCQ

A thin horizontal circular disc is rotating about a vertical axis passing through its center. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc.

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📖 Explanation

Here no external force is applied on the disc so Torque $(\tau )=0$ .So angular momentum is conserved.That means $I_1 \omega _1=I_2 \omega _2$ $\Rightarrow \omega _2=\;\frac{I_1 \omega _1}{I_2}$ $\therefore$ Angular speed is inversely proportional to Moment of inertia.For ${\text{disc}} I=\;\frac{1}{2} M R^2$ $\therefore$ Moment of Inertia is proportional to Mass.As insect moves along a diameter, the effective mass of disc first decreases then increases and hence the moment of inertia first decreases then increases so from principle of conservation of angular momentum, angular speed, first increases then decreases.

Q180JEE Main 2011MCQ

A mass $m$ hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass $m$ and radius $R$ . Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass $m$ , if the string does not slip on the pulley, is:

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📖 Explanation

This is the free body diagram of pulley and massFor translation motion of the block, $m g-T=m a$ For rotational motion of the pulley, $T \times R=I \alpha =I \;\frac{a}{R}$ where $\alpha =$ angular acceleration of disc $=\;\frac{a}{R}$ and $I=\;\frac{1}{2} m R^2$ (For circular disc)Solving (1) & (2), $\;a=\;\frac{m g}{ (m+\;\frac{I}{R^2}) }=\;\frac{m g}{m+\;\frac{m R^2}{2 R^2}}$ $\;=\;\frac{2 m g}{3 m}=\;\frac{2 g}{3}$