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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
A uniform rod `AB` of length `l` and mass `m` hangs from point `A` in a car moving with velocity `v_(0)` on an inclined plane as shown in Fig. The rod can rotate in vertical plane about the axis at point `A`. if the car suddenly stops, the angular speed with which the rod starts rotating is A. `3/2(v_(0))/l cos theta`B. `(v_(0))/2 (costheta)/l`C. `3/2(v_(0))/lsintheta`D. `5/2(v_(0))/lsintheta` |
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Answer» Correct Answer - A When the car is moving with the constant speed, the rod will be vertical. The centre of mass of the rod is moving with a velocity `v_(0)` parallel to the plane. Conserving agular momentum about `A`, we get `mv_(0)l/2costheta=(ml^(2))/3omegaimplies omega=3/2v_(0)/lcostheta` |
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| 102. |
The device shown in Fig. rotates on the vertical axle as shown. The frame has negligible mass as compared to the four masses each of mass `m`. Initial angular velocity of the system is `omega_(0)`. Due to an internal mechanism the spokes in the frame lengthen so that the radii of the masses become `2a`. Initially, it was a. What will be the new angular velocity of the system? |
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Answer» The will be no torque on the system if we ignore frictional torque at the axle. Therefore, the angular momentum of the system is conserved. `vecL_("initial")=vecL_("final")` `I_("initial")omega_(0)=I_("final")omega` `4ma^(2) omega_(0)=4m(2a)^(2)omega` from which we have `omega=(omega_(0))/4` |
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| 103. |
A wheel `A` is connected to a second wheel `B` by means of inextensible string, passing over a pulley `C`, which rotates about a fixed horizontal axle `O`, as shown in the figure. The system is released from rest. The wheel `A` rolls down the inclined plane `OK` thus pulling up wheel `B` which rolls along the inclined plane `ON`. Determine the velocity (in m/s) of the axle of wheel `A`, when it has travelled a distance `s = 3.5 m` down the to be slope. Both wheels and the pulley are assumed homogeneous disks of identical weight and radius. Neglect the weight of the string. The string does not over `C`. [Take `alpha =53^@` and `beta=37^@`] |
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Answer» Correct Answer - 2 Applying conservation of energy we get `mgs(sinalpha-sinbeta)=1/2Iomega^(2)+2(1/2mv^(2)+1/2Iomega^(2))` where `omega=v/r` and `I=mr^(2)//2` putting values and solving we get `v=2m//s` |
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| 104. |
Many great rivers flow toward the equator. The sediments that they carry, increases the time of rotation of the earth about its own axis. The angular momentum of the earth about its rotation axis is conserved.A. Statement I is True, Statement II is True, Statement ll is a correct explanation for Statement I.B. Statement I is True, Statement II is True, Statement II is NOT a correct explanation for Statement IC. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True. |
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Answer» Correct Answer - A Sediment deposited at the equator (away from the axis of rotation) increases the moment of inertia (not mass) of the Earth. Since `Iomega=`constant, `omega` decreases and thus `I=2pi//omega` increases. |
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| 105. |
A smooth sphere A is moving on a frictionless horizontal plane with angular speed `omega` and centre of mass velocity `upsilon`. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are `omega_A and omega_B` respectively. ThenA. `omega_(A)ltomega_(B)`B. `omega_(A)=omega_(B)`C. `omega_(A)=omega`D. `omega_(B)=omega` |
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Answer» Correct Answer - C There is no torque about C.O.M on sphere A so `omega_(A)=omega` |
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| 106. |
A smooth sphere A is moving on a frictionless horizontal plane with angular speed `omega` and centre of mass velocity `upsilon`. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are `omega_A and omega_B` respectively. ThenA. `omega_(A) ltomega_(B)`B. `omega_(A)=omega_(B)`C. `omega_(A)=omega`D. `omega_(B)=omega` |
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Answer» Correct Answer - C As the spheres are smooth, there will be no transfer of anglar momentum. Thus , `A` after collision, will remain with its initial angular momentum. |
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| 107. |
A soldi sphere a hollow sphere and a disc, all haing same mass and radius are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. Least time will be taken in reaching the bottom byA. th solid sphereB. the hollow sphereC. the discD. all will take the same time |
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Answer» Correct Answer - D Since linear acceleration is same for all `(a=Mgsintheta-mumgcostheta)` as they have same mass `M` and same `mu`. Hence all will reach the bottom simultaneously. |
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| 108. |
Three identical solid spheres move down three incline `A, B` and `C` are all of the same dimensions. A is without friction, the friction between `B` and a sphere is sufficient to cause rolling without slipping, the friction between `C` and a sphere causes rolling with slipping. The kinetic energies, of `A, B, C` at the bottom of the inclines are `E_(A),E_(B),E_(C)`.A. `E_(A)=E_(B)-E_(C)`B. `E_(A)=E_(B)gtE_(C)`C. `E_(A)gtE_(B)gtE_(C)`D. `E_(A)gtE_(B)=E_(C)` |
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Answer» Correct Answer - B For `A` and `B` energy will remains conserved but energy of `C` will dissipate due to action friction force. |
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| 109. |
Two uniform solid spheres having unequal rdii are released from rest from the same height on a rough incline. Ilf the spheres roll without slippingA. The heavier sphere reaches the bottom firstB. The bigger sphere reaches the bottom firstC. The two spheres reach the bottom togetherD. The information given is not sufficient to tell which sphere will reach the bottom first |
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Answer» Correct Answer - C `a=(gsintheta)/(1+(K^(2))/(R^(2))),` For any sphere `(K^(2))/(R^(2))=2/5` So, acceleration is independent of mass of radius. |
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| 110. |
A small solid marble of mass `M` and radius `r` rolls down along the loop track, without slipping. Find the height `h` above the base, from where it has to start rolling down the incline such that the sphere just completes the vertical circular loop of radius `R`. |
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Answer» Here the centre of mass of the marble will move in a circle of radius `(R-r)` so for just looping the loop, at `H` `v=sqrt(g(R-r))`……i Now as in roling `K=K_(T)+K_(R)=1/2Mv^(2)+1/2omega^(2)` and here `I=(2/5)Mr^(2)` with `v=romega` so `K=1/2Mv^(2)+1/2[2/5Mr^(2)][v/r]^(2)` i.e., `=1/2Mv^(2)+1/5Mv^(2)=7/10 Mv^(2)`.........ii So in the light of eqn i eqn ii becomes `K=7/10Mg(R-r)`.....iii As this kinetic energy is provided by loss in `PE`, applying conservation of mechanical energy between `P` and `H`. `0+Mgh=7/10mg(R-r)+Mg(2R_(r))` `h=1/10(27R-17r)` |
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| 111. |
A cylindrical drum, pushed along by a board rolls forward on the ground. There is no slipping at any contact. The distance moved by the man who is pushing the board, when axis of the cylinder covers a distance `L` will be. |
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Answer» Let `v_(0)` be the linear speed of the axis of the cylinder and `omega` be its angular speed about the axis. As it does not slip on the ground hence `omega=v_(0)//R`. Where `R` is the radius of the cylinder. Speed time taken by the axis of move a distance `L` is equal to `t=L//v_(0)`. In the same interval of time distance moved by the topmost point is `S=2v_(0)xxL/(v_(0))=2L` As there is no slipping between any point of contact hence distance moved by the man is `2L`. |
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| 112. |
Two cylinders having radii `2R` and `R` and moment of inertia `4I` and `I` about their central axes are supported by axles perpendicular to their planes. The large cylinder is initially rotating clockwise with angular velocity `omega_(0)`. The small cylinder is moved to the right until it touches the large cylinder and is caused to rotate by the frictional force between the two. Eventually slipping ceases and the two cylinders rotate at constant rates in opposite directions. During this A. angular momentum of system is conservedB. kinetic energy is conservedC. neither the angular momentum nor the kinetic energy is conservedD. both the angular momentum and kinetic energy are conserved |
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Answer» Correct Answer - C Angular momentum of system can not remain consrved as some external unbalanced torque is present due to forces at axels. Kinetic energy is not conserved, because slipping is there and work is done against friction. |
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| 113. |
Two discs `A` and `B` are mounted coaxially ona vertical axle. The discs have moments of inertia `l` and `2l` respectively about the common axis. Disc A is imparted an initial angular velocity `2 omega` using the centre potential energy of a spring compressed by a distance `x_(1)`. Disc `B` is imparted angular velocity `omega` by a spring having the same spring constant and compressed by a distance `x_(2)`. Both the disc rotate in the clockwise direction. The loss of kinetic energy the above process is -A. `(Iomega^(3))/2`B. `(Iomega^(2))/3`C. `(Iomega^(2))/4`D. `(Iomega^(2))/6` |
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Answer» Correct Answer - B `KE_(i)-KE_(f)` `implies 1/2I(2omega)^(2)+1/22Iomega^(2)-1/23Iomega_(1)^(2)=1/3omega^(2)` |
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| 114. |
Two discs `A` and `B` are mounted coaxially ona vertical axle. The discs have moments of inertia `l` and `2l` respectively about the common axis. Disc A is imparted an initial angular velocity `2 omega` using the centre potential energy of a spring compressed by a distance `x_(1)`. Disc `B` is imparted angular velocity `omega` by a spring having the same spring constant and compressed by a distance `x_(2)`. Both the disc rotate in the clockwise direction. The rotation `x_(1)//(x_(2)` is.A. `2`B. `1/2`C. `sqrt(2)`D. `1/(sqrt(2))` |
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Answer» Correct Answer - C `1/2I(2omega)^(2)=1/2kx_(1)^(2)` and `1/2I(omega)^(2)=1/2kx_(2)^(2)` dividing then we get `(x_(1))/(x_(2))=sqrt(2)` |
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| 115. |
Two discs A and B are mounted coaxiallay on a vertical axle. The discs have moments of inertia I and 2 I 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. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period isA. `(2Iomega)/(3t)`B. `(9Iomega)/(2t)`C. `(9Iomega)/(4t)`D. `(3Iomega)/(2t)` |
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Answer» Correct Answer - A Let final angular velocity be `omega_(1)`. Applying conservation of angular momentum. `(I+2I)omega_(1)=I(2omega)+2Iomega` `implies omega_(1)=(4omega)/3` now angular impulse `=` change in angular momentum `taut=2I(omega_(1)-omega)` `implies taut=2Iomega/3impliest=(2Iomega)/(3t)` |
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| 116. |
A ring, a disc a sphere and spherical shells are simutaneously released to roll down from the top of an inclined plane of height h. the four bodies will reach the bottom in the following orderA. sphere,disc,shell and ringB. ring,spherical,shell,disc and sphereC. sphere,spherical shell, disc and ring.D. sphere,spherical shell, ring and disc |
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Answer» Correct Answer - A `a=(g sin theta)/(1+I/(MR^(2)))` So, sphere, disc, shell and ring |
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| 117. |
In this passage a brief idea is given of the motion of the rolling bodies on an inclined plane. We will consider three cases: objects are released on an incline plane Case-A: which is smooth. Case-B: where friction is insufficient to provide pure rolling. Case-C: where friction is sufficient to provide pure rolling. Force diagram for three cases are as follows: (where symbols have their usual meanings) Two children `A` and `B` use bicycles, having wheels of ring type and disc type respectively. During a race, bicycles are given the same velocity from the bottom of the inclined bridge to ascend the bridge without pedalling, then (assuming pure rolling).A. both bicycles with reach up to same heightB. bicycle of child `A` will reach a greater heightC. bicycle of child `B` will reach a greater heightD. bicycle of child `B` will reach a greater height |
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Answer» Correct Answer - B For ring `a=(gsintheta)/2` (for pure rolling) is less than of disc |
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| 118. |
In this passage a brief idea is given of the motion of the rolling bodies on an inclined plane. We will consider three cases: objects are released on an incline plane Case-A: which is smooth. Case-B: where friction is insufficient to provide pure rolling. Case-C: where friction is sufficient to provide pure rolling. Force diagram for three cases are as follows: (where symbols have their usual meanings) If the four objects given in the above question are of same mass, same radius having the same friction coefficient and are released from the same height, then at the bottom the object which will have least kinetic energy for case `B` will be the:A. hollow sphereB. solid sphereC. ringD. disc |
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Answer» Correct Answer - C As given in the equation of case b `muNR=Mk^(2)alpha` and `N=Mgcostheta` As: `theta, M, R , mu` are same for all `alpha` will be least for the object for which `k` and hence `I` is maximum. Therefore `alpha` for ring (`k=R`) and hence `omega` for ring at the bottom is minimum. also, `Mgsintheta-muN=Ma` Since `M, mu, theta, N` are same for all objects, they same linear acceleration and hene same linear velocity and hence same `1/2Mv_(cm)^(2)` `:. KE=(1/2Mv_(cm)^(2)+1/2I_(cm)omega^(2))` is least for the ring, |
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| 119. |
In this passage a brief idea is given of the motion of the rolling bodies on an inclined plane. We will consider three cases: objects are released on an incline plane Case-A: which is smooth. Case-B: where friction is insufficient to provide pure rolling. Case-C: where friction is sufficient to provide pure rolling. Force diagram for three cases are as follows: (where symbols have their usual meanings) We have four objects: a solid sphere, a hollow sphere, a ring and a disc, all of same radius. When these are released on an inclined plane, it may happen that all of them do not perform pure rolling. But from the information of pure rolling, if one object can be confirmed to be purely rolling then it can be said that rest all will perform pure rolling. This object whose pure rolling confirms pure rolling of all other objects isA. hollow sphereB. solid sphereC. ringD. disc |
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Answer» Correct Answer - C From passage for case `c` `mu_("min")=(tantheta((k^(2))/(R^(2))))/((1+(k^(2))/(R^(2))))rarr("pure rolling")` Putting the values of `k` for different objects given in the table (in passege) we get `mu_("min"("ring"))=(tantheta)/2` `mu_("min"("Disc"))=(tantheta)/3, mu_("min"("solid sphere"))=2/3tantheta` `mu_("min"("Hollow sphere")=2/7tantheta` `implies mu_("min"("Ring")), ` is greater than either of `mu_("min"("Ring")), mu_("min"("Solid Sphere")), mu_("min"("Hollow sphere"))` Therefore, the pure rolling of ring will confirm pure rolling of all other bodies. |
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| 120. |
In this passage a brief idea is given of the motion of the rolling bodies on an inclined plane. We will consider three cases: objects are released on an incline plane Case-A: which is smooth. Case-B: where friction is insufficient to provide pure rolling. Case-C: where friction is sufficient to provide pure rolling. Force diagram for three cases are as follows: (where symbols have their usual meanings) Three solid uniform spheres are released on an inclined plane as shown in the figure. The distance between the spheres remains constant during motion in: A. all three massesB. case `A` and `B`C. only case `C`D. depends on the mass of the spheres |
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Answer» Correct Answer - A Since acceleration is same for all the three spheres, they cover equal distances in equal intervals of time in all the case `a. b` and `c` hence `a`. |
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| 121. |
The steel balls `A` and `B` have a mass of `500 g` each and al rotating about the vertical axis with an angular velocity of `4 rad//s` at a distance of `15 CM` from the axis. Collar `C` is now forced down until the balls are at a distance `5 CM` from the axis. How Much work must be done move the collar down? |
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Answer» In process angular momentum remains constant `I_(1)omega_(1)=I_(2)omega_(2)` Where `I_(1)=2m(0.15)^(2), omega_(1)=4rad//s` `I_(2)=2m(0.05)^(2)` Substituting these values in above equation, we get `omega_(2)=(I_(1)omega_(1))/I_(2)=36rad//s` Work done on the collar `W=1/2I_(2)omega_(2)^(2)-1/2I_(1)omega_(1)^(2)` or `W=1/2(0.05)^(2)xx36^(2)-1/2 2m(0.015)^(2)xx4^(2)` where `m=500/1000=0.05kg=1.44J` |
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| 122. |
A string is wrapped around a cylinder of mass `M` and radius `R`. The string is pulled vertically upwards to prevent the centre of mass from falling as the cylinder unwinds the string. The tension in the string isA. `(Mg)/6`B. `(Mg)/3`C. `(Mg)/2`D. `(2Mg)/3` |
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Answer» Correct Answer - B Equation of motion `Mg-T=Ma` ………….i Taking torque about the axis passing through the centre of the spool and perpendicular to it `TR=Ialpha=1/2MR^(2)(a/2)` `T=1/2Ma` …………ii From eqn i and ii `Ma=Mg-1/2ma` `a=(2g)/3` `:. T=(Mg)/3` |
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| 123. |
A spool (consider it as a double disc system joined by a short tube at their centre) is placed on a horizontal surface as shown Fig. A light string wound several times over the short connecting tube leaves it tangentially and passes over a light pulley. A weight of mass `m` is attached to the end of the string. The radius of the connecting tube is r and mass of the spool is `M` and radius is `R`. Find the acceleration of the falling mass `m`. Neglect the mass of the connecting tube and slipping of the spool. |
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Answer» Since there is no slipping, the velocity of the point of contact of the string with the pulley is the vector sum of the velocity of the point of contact due to rotation plus the velocity of the centre of mass of the spool. That is `v_(B)=v_(CM)=omegar, omega` is in clockwise direction. But `v_(CM)=omegaR` `v_(B)=v_(CM)-v_(CM)r/R=v_(CM) ((R-r)/R)` This also the velocity v of the rolling mass `v_(B)=v_(CM) ((R-r)/R)=v((R-r)/R)` Cosidering the conservation of energy when `m` lowers by `h`, we get `mgh=1/2 mv^(2)+(1/2Mv_(CM)^(2)+1/2Iomega^(2))` `=1/2mv^(2)+(1/2Mv_(CM)^(2)+1/2xx1/2MR^(2)v_(CM)^(2)/R^(2))` `=1/2 mv^(2)+3/4Mv_(CM)^(2)` `=1/2 mv^(2)+3/4Mv^2(R/(R-r))` `v^(2)=(4mgh)/(2m+3M[R/(R-r)]^(2))` Comparing with `v^(2)=2ah` `a=(v^(2))/(2h)=(2mg)/(2m+3M[R//(R-r)]^(2))` |
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| 124. |
A spool is pulled horizontally on rough surface by two equal and opposite forces as shown in the figure. Which of the following statements are correct?A. The centre of mass moves towards left.B. The centre of mass moves towards right.C. The centre of mass remains stationary.D. The net torque about the centre of mass of the spool is zero. |
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Answer» Correct Answer - B Friction force will act towards right. |
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| 125. |
Figure shows a spool with thread wound on it placed on a smooth plane inclined at angle from horizontal. The spool has mass m, edge radius `R`, and is wound up to a radius r. its moment of inertia about its own axis is `I`. The free end of the thread is attached as shown in the figure. So that the thread is parallel to the inclined plane. `T` is the tension in the thread. Which of the following is correct? A. The linear acceleration of the spool axis down the slope mg is `(mgsintheta-T)/m`B. Angular acceleration is `(Tr)/(2I)`C. The linear acceleration of the spool axis down the plane is `(Tr^(2))/I`D. The acceleration of the spool axis down the slope is `(gsintheta)/(1+I/(mr^(2)))` |
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Answer» Correct Answer - A::C::D Linear acceleration `=(mgsintheta=T)/m` Angular acceleration `=("Torque" (Tr))/("Momentum of inertia" (I))` Linear acceleration `=r[(Tr)/I]=(Tr^(2))/I` Since the two linear acceleration must match, `:. (Tr^(2))/I=(mgsintheta-T)/m` or `(Tmr^(2))/I+T=mgsintheta` or `T[1+(mr^(2))/I]=mgsintheta` or `T=(mgsintheta)/(1+(mr^(2))/I)` |
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| 126. |
Inner and outer radii of a spool are `r` and `R I`, respectively. A thread is wound over its inner surface and spool is placed over a rough horizontal surface. Thread is pulled by a force `F` as shown in Fig. In case of pure rolling, which of the following statements are false? A. Thread unwinds, spool rotates anticlockwise and friction acts leftwards.B. Thread winds, spool rotates clockwise and friction acts leftwards.C. Thread winds, spool moves to the right and friction acts rightwards.D. Thread winds, spool moves to the right and friction does not come into existence. |
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Answer» Correct Answer - A::C::D Since, the spool rolls over the horizontal surface, instantaneous axis of rotation passes through the point of contact of spool with the horizontal surface. About the instantaneous axis if rotation, moment produced by `F` is clockwise. Therefore, the spool rotates clockwise. In that case, acceleration will be rightward and thread will wind. If rotational motion of the spool is considered about its own, then the resultant moment on it must be clockwise. But moment produced by the force is anticlockwise and its magnitude is equal to `F_(1)//r`. Hence, moment produced by the friction (about its own axis) must be clockwise and its magnitude must be greater than `Fr`. It is possible only when friction acts leftwards. Therefore, Option (b) is correct |
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| 127. |
A ball of mass `m` and radius `r` rolls inside a hemispherical shell of radius `R`. It is released from rest from point `A` as shown in figure. The angular velocity of centre of the ball in position `B` about the centre of the shell is. .A. `2sqrt(g/(5(R-r)))`B. `2sqrt(g/(7(R-r)))`C. `sqrt((2g)/(5(R-r)))`D. `sqrt((5g)/(2(R-r)))` |
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Answer» Correct Answer - B `KE` of ball in position `B=mg(R-r)` Here `m=`mass of ball. Since it rolls without slipping the ratio rotational to translational kinetic energy will be `2/5` `(K_(R))/(K_(T))=2/5` `K_(T)=2/7mg(R-r), 1/2mv^(2)=2/7mg(R-r)` `v=2/(sqrt(7))sqrt(g(R-r)), omega=v/(R-r) =2 sqrt(v/(7(R-r)))` |
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| 128. |
A uniform circular disc of mass `M` and radius `R` rolls without slipping on a horizontal surface. If the velocity of its centre is `v_0`, then the total angular momentum of the disc about a fixed point `P` at a height `(3R)//2` above the centre `C`. A. increases continuously as the disc moves awayB. decreases continuously as the disc moves awayC. is equal to `2MRv_(0)`D. is equal to `MRv_(0)` |
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Answer» Correct Answer - D `vecL=I_(CM)vecomega=mvecrxxvecv_(CM)` `=(MR^(2))/2omega(-hatk)+M3/2 R(-hatj)xxv_(0)hati` `=MRv_(0)hatk` |
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| 129. |
A small ball of radius r rolls without sliding in a big hemispherical bowl. of radius R. what would be the ratio of the translational and rotaional kinetic energies at the bottom of the bowl.A. `2:1`B. `3:2`C. `4:3`D. `5:2` |
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Answer» Correct Answer - D `(K.E._(T))/(K.E._(R))=(1/2 Mv^(2))/(1/2Iomega^(2))=(1/2Mv^(2))/(1/2 2/5 MR^(2)(v/R)^(2))=5:2` |
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| 130. |
In Fig a sphere of radius `2 m` rolls on a plank. The accelerations of the sphere and the plank are indicated. The value of `alpha` is A. `2rad//s^(2)`B. `4rad//s^(2)`C. `3rad//s^(2)`D. `1rad//s^(2)` |
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Answer» Correct Answer - C As sphere rolls the lowest of the point of the sphere should have the same acceleration as the plank hence `a_(1)=alphaR-a_(2)` `2=2alpha-4implies alpha=3rad//s^(2)` |
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| 131. |
A force `F` acts tangentially at the highest point of a disc of mass `m` kept on a rough horizontal plane. If the disc rolls without slipping, the acceleration of centre of the disc is: A. `(2F)/(3m)`B. `(10F)/(7m)`C. zeroD. `(4F)/(3m)` |
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Answer» Correct Answer - D `F+f=Ma` ………..i `FR=fR=(MR^(2))/2 a/R`……ii `F-f=(Ma)/w`………..iii `2F=(2Ma)/wimpliesa=(4F)/(3M)` |
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| 132. |
A uniform circular disc of radius `r` is placed on a rough horizontal surface and given a linear velocity `v_(0)` and angular velocity `omega_(0)` as shown. The disc comes to rest after moving some distance to the right. It follows that A. r/2B. rC. 3r/2D. 2 |
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Answer» Correct Answer - A `mV_(0)R-(mR^(2))/2. omega_(0)=0 rArr (V_(0))/(omega_(0))=R/2` |
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| 133. |
A cylinder of mass at and radius `R` rolls on a stationary plank of mass `M`. The lower surface of the plank is smooth and the upper surface is sufficiently rough with a coefficient of friction `mu`. A man is to hold the plank stationary with respect to the ground, as shown Fig. The force exerted by the man to keep the plank stationary is equal to…… |
| Answer» Zero. Since lowest point of the sphere is stationary wit respect to the plank, therefore friction force acting on te plank is zero. Thus, no force is required to keep the plank statiorary. | |
| 134. |
A solid homogeneous sphere is moving on a rough horizontal surface, partly rolling and partly sliding. During the king of motion of the sphere.A. Total kinetic energy is conservedB. Angular momentum of the sphere about the point of contact is conservedC. Only the rotational kinetic energy about the centre of mass is conservedD. Angular momentum about the centre of mass is conserved |
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Answer» Correct Answer - B Since all the forces pass through point of contact, so angular momentum remains conserved. |
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| 135. |
Figure shows a uniform smooth solid cylinder `A` of radius `4 m` rolling without slipping on the `8 kg` plank which, in turn, is supported by a fixed smooth surface. Block `B` is known to accelerate down with `6 m//s^(2)` and block `C` moves down with acceleration `2 m//s^(2)`. What is the angular acceleration of the cylinder?A. `4/5rads^(-2)`B. `6/5 rads^(-2)`C. `2rads^(-2)`D. `1 rads^(-2)` |
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Answer» Correct Answer - D Acceleration of the bottom of the cylinder = acceleration of the plank `=2m//s^(2)` towards the right Acceleration of the top of te cylinder = acceleration of `B=6m//s^(-2)` towards the left Let linear acceleratiion of the cylinder be a towards the left and angular acceleration of it `alpha` in an anticlockwise sense. writing constraint, we get `a+Ralpha=6` and `Ralpha-a=2` `impliesalpha=1rads^(-2),a=2m//s^(2)` |
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| 136. |
A cylinder executes pure rolling without slipping with a constant velocity on a plank, whose upper surface is rough enough, but lower surface is smooth. The plank is kept at rest on a smooth horizontal surface by the application of an external force `F`. Choose the correct alternative: A. The direction of `F` is towards right.B. The direction of `F` towards left.C. The value of `F` is zeroD. The direction of `F` depends on the ratio of the relative masses of disc and plank. |
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Answer» Correct Answer - C Since cylinder rolls with a constant velocity so there is no friction between the upper surface of the plank and the cylinder. The plank is at rest, unjder the influence of the weight of the cylinder. So no external force is required to keep the plank in equilibrium. |
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| 137. |
A plank `P` is placed on a solid cylinder `S`, which rolls on a horizontal surface. The two are of equal mass. There is no slipping at any of the surfaces in contact. The ratio of kinetic energy of `P` to the kinetic energy of `S` is: A. `1:1`B. `2:1`C. `8:3`D. `11:8` |
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Answer» Correct Answer - C `KE` of `P=1/2m(2v)^(2)=2mv^(2)` `KE` of `S=1/2mv^(2)+1/2Iomega^(2)=3/4mv^(2)implies (KE "of" P)/(KE "of" S)=8/3` |
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| 138. |
Figure shows a uniform smooth solid cylinder `A` of radius `4 m` rolling without slipping on the `8 kg` plank which, in turn, is supported by a fixed smooth surface. Block `B` is known to accelerate down with `6 m//s^(2)` and block `C` moves down with acceleration `2 m//s^(2)`. What is the ratio of the mass of the cylinder to the mass of block `B`?A. `1`B. `2`C. `3`D. `4` |
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Answer» Correct Answer - B For block `B,m_(B)g-T=m_(B)6`…………i For cylinder `T=m_(cyl)2`………..ii Solving eqn i and ii `m_(cyl),m_(B)=2:1` |
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| 139. |
Figure shows a uniform smooth solid cylinder `A` of radius `4 m` rolling without slipping on the `8 kg` plank which, in turn, is supported by a fixed smooth surface. Block `B` is known to accelerate down with `6 m//s^(2)` and block `C` moves down with acceleration `2 m//s^(2)`. If unwrapped length of the thread between the cylinder and block `B` is `20 m` at the beginning, when the system was released from rest, what would it be `2 s` laterA. `28m`B. `30m`C. `22m`D. `32.5m` |
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Answer» Correct Answer - A The acceleration of the top part of the cylinder (thread) with respect to `CM` of the cylinder is `a_(AO)=alphaR=1xx4=4m//s^(2)` Length of the thread `=20+1/2(Ralpha)t^(2)=28m` |
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| 140. |
A `198-cm` tall girl lies on a light (massless) board which is supported by two scales one under the top of her heal and one beneath the bottom of her feet. The two scales read respectively `36` and `30 kg`. What distance is the centre of gravity of this girl from the bottom of her feet? . A. `99cm`B. `90cm`C. `108cm`D. `82cm` |
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Answer» Correct Answer - C Let the required distance is `x`, then balancing the torque about centre of gravity `(30g)x=(36g)(198-x)x=108cm` |
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| 141. |
A sphere is released on a smooth inclined plane from the top. When it moves down, its angular momentum isA. conserved about every pointB. conserved about the point of contact onlyC. conserved about the centre of the sphere onlyD. conserved about any point on a line parallel to the inclined plane and passing through the centre of the ball |
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Answer» Correct Answer - D As the inclined plane is smooth, the sphere an never, roll, rather it will just slip down. Hence the angular momentum remains conserved about any point on a line parallel to the inclined plane and passes through the centre of the ball. |
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| 142. |
Angular momentum of the particle rotating with a central force is constant due toA. constant forceB. constant linear momentumC. zero torqueD. constant torque |
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Answer» Correct Answer - C Central force is directed towards a point, therefore torque of the central force is zero. |
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| 143. |
Three spools `A, B` and `C` each having moment of inertia `I=MR2//4` are placed on rough ground and equal force `F` is applied at positions as shown in the figures (a), (b) and (c). Then A. frictional force on spool `A` acts in forward directionB. frictional force on spool `A` acts in backward directionC. frictional force on spool `B` acts in forward directionD. frictional force on spool `B` and `C` acts in backward direction |
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Answer» Correct Answer - A::D For figure `A` and `B` If `k^(2)gtRx` friction will act in backward direction otherwise in forward direction. For given spool `K^(2)=(R^(2))/4` Figure a `Rx=R.R/3=(R^(2))/3` friction forward Fig b `R=R.R/4=(R^(2))/5` friction backward. In figure c as force is below the diametric plane, hence friction backward. |
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| 144. |
We have two spheres, one of which is hollow and the other solid. They have identical masses and moment of intertia about their respective diameters. The ratio of their radius is given by.A. `5:7`B. `3:5`C. `sqrt(3):sqrt(5)`D. `sqrt(3):sqrt(7)` |
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Answer» Correct Answer - C `2/3MR_(h)^(2)=2/5MR_(S)^(2)` or `(R_(h)^(2))/(R_(S)^(2))=3/5` or `(R_(h))/(R_(s))=sqrt(3/5)` |
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| 145. |
Two rigid bodies `A` and `B` rotate with angular momenta `L_(A)` and `L_(B)` respectively. The moments of inertia of `A` and `B` about the axes of rotation are `I_(A)` and `I_(B)` respectively. If `I_(A)=I_(B)//4` and `L_(A)=5L_(B)`, then the ratio of rotational kinetic energy `K_(A)` of `A` to the rotational kinetic energy `K_(B)` of `B` is given byA. `(K_(A))/(K_(B))=25/4`B. `(K_(A))/(K_(B))=5/4`C. `(K_(A))/(K_(B))=1/4`D. `(K_(A))/(K_(B))=100` |
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Answer» Correct Answer - D `K_(A)=(L_(A)^(2))/(2l_(A))=(25L_(B)^(2))/(2l_(B))` `K+B=(L_(B)^(2))/(2l_(B))` Now, `(K_(A))/(K_(B))=25xx4=100` |
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| 146. |
A solid sphere and a solid cylinder having the same mass and radius, rolls down the same incline. The ratio of their acceleration will beA. `15:14`B. `14:15`C. `5:3`D. `3:5` |
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Answer» Correct Answer - A `a=(g sin theta)/(1+I/(MR^(2)))rArr (a_(1))/(a_(2))=(1+1/2)/(1+2/5)=15:14` |
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| 147. |
A solid sphere is given an angular velocity `omega` and kept on a rough fixed incline plane. The choose the correct statement. A. If `mu=tantheta` then sphere will be in linear equilibrium for some time and after that pure rolling down the plane will startB. If `mu=tantheta` then sphere will move up the plane and frictional force acting all the time will be `2 mg sin`C. If `mu=(tantheta)/2` there will never be pure rolling (consider inclined plane to be long enough).D. If incline plane is not fixed and it is on smooth horizontal surface then linear momentum of the system (wedge and sphere) can be conserved in horizontal direction. |
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Answer» Correct Answer - A::D Due to torque of friction about `CM omega` eventually decreases to zero, initially there is no translation. Friction is sufficient for pure rolling therefore after sometime pure rolling begin. There is no external force in `x` direction therefore momentum is conserved along `x` direction. |
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| 148. |
A body of mass `m` slides down an smooth incline and reaches the bottom with a velocity, Now smooth incline surface is made rough and the same mass was in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been:A. `sqrt(2v)`B. `v`C. `(sqrt(2/5))v`D. `v//sqrt(2)` |
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Answer» Correct Answer - D For a sliding body of mass `m` `v_("body")=sqrt(2gh)=v`………..i For a rolling of same mass `m` `v_("ring")=sqrt((2gh)/g)` `=sqrt((2gh)/(1+I/(mR^(2))))=sqrt((2gh)/(1+(mR^(2))/(mR^(2))))=sqrt((2gh)/2)=sqrt(gh)=v/(sqrt(2))` |
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| 149. |
A ball rolls down an inclined plane and acquires a velocity `v_(r)` when it reaches the bottom of the plane. If the same ball slides without friction and acquires rolling from the same height down an equally inclined smooth plane and acquires a velocity `v_(s)` (then which of the following statements are not correct?A. `v_(r)ltv_(s)` because a work is done by the rolling ball against the frictional force.B. `v_(r)gtv_(s)`, because the angular velocity acquired makes the rolling ball to travel fasterC. `v_(r)=v_(s)` because kinetic energy of the two balls is same at the bottom of the planes.D. `v_(r)gtv_(s)` because the rolling ball acquires rotational as well as translational kinetic energy. |
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Answer» Correct Answer - A::B::C::D When the ball moves along a smooth plane, the accelerating force on it is `mgsintheta`. Hence its acceleration is equal to `g sintheta`. When the ball rolls down the rough inclined plane, the `mg sintheta` acts down the plane but a frictioncomes into existence which acts up the plane. That friction prodces angular acceleration on the ball and the net accelerating force on the ball becomes equal to `(mgsintheta-"friction")`. Therefore, it has a smaller acceleration. Therefore its velocity at the bottom of the plane is less than that of the ball moving down a smooth plane or `v_(s)gtv_(r)`. In fact, at the bottom of the planes both the balls have the same `KE` because loss of potential energy of both the balls is the same. But the ball rolling down a rough plane has translational as well as rotational kinetic energy at the botom of the plane while ball sliding down a smooth plane has translational `KE` only. Therefore, translation `KE` of the ball will be less than the translational `KE` of the ball sliding down the smooth plane. Hence, option d is correct. |
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| 150. |
A small stone of mass `m` is attached to a light string which passes through a hollow tube. The tube is held by one hand and the free end of the string by the other hand. The mass is set into revolution in a horizontal circle of radius `r_(1)` with a speed `v_(1)` . The string is pulled down shortening the radius of the circular path to `r_(2)`. Which of the following is not correct? (`omega_(1)` and `T` have usual meanings)A. `(omega_(2))/(omega_(1))=(r_(1)^(2))/(r_(2)^(2)`B. `(E_(k1))/(E_(k2))=(r_(2)^(2))/(r_(1)^(2))`C. `(E_(k1))/(E_(k2))=(r_(2)^(2))/(r_(1)^(2))`D. `(T_(2))/(T_(1))=(r_(1)^(3))/(r_(2)^(3))` |
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Answer» Correct Answer - D Using conservation of angular momentum `mr_(2)^(2)omega_(2)=mr_(1)^(2)omega_(1)` or `(omega_(2))/(omega_(1))=(r_(1)^(2))/(r_(2)^(2)` Again `E_(k_(1))/E_(k_(2))=(1/2mv_(1)^(2))/(1/2mv_(2)^(2))=(r_(1)^(2)omega_(1)^(2))/(r_(2)^(2)omega_(2)^(2))=(r_(1)^(2))/(r_(2)^(2))xx[(r_(2)^(2))/(r_(1)^(2))]^(2)=(r_(2)^(2))/(r_(1)^(2)` Again `(T_(2))/(T_(1))=(mr_(2)omega_(2)^(2))/(mr_(1)omega_(1)^(2))=(r_(2))/(r_(1)=[(r_(1)^(2))/(r_(2)^(2))]=(r_(1)^(3))/(r_(2)^(3))` |
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