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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.
| 51. |
A yo-yo is placed on a rough horizontal surface and a constant force `F`, which is less than its weight, pulls it vertically. Due to this A. friction force acts towards left, so it will move towards leftB. friction force acts towards right, so it will move towards rightC. it will move towards left, so friction force acts towards leftD. it will move towards right so friction force acts towards right |
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Answer» Correct Answer - A `F` will provide anticlockwise torque about the centre due to which the bottomost point will tend to move towards right, so friction will act towards left. So it will move towards left. |
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| 52. |
A yo-yo, arranged as shown, rests on a frictionless surface. When a force `F` is applied to the string, the yo-yo A. moves to the left and rotates counterclockwiseB. moves to the right and rotates counterclockwiseC. moves to the left and rotates clockwiseD. moves to the right and rotates clockwise |
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Answer» Correct Answer - B As the is no friction net force is `F` which is towards right. Torque due to this force will be in anticlockwise direction. |
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| 53. |
End `A` of the bar `AB` in figure rests on a frictionless horizontal surface and end `B` is hinged. A horizontal force `vecF` of magnitude `120 N` is exerted on end `A`. You can ignore the weight of the bar. What is the net force exerted by the bar on the hinge at `B`? A. `200N`B. `140N`C. `100N`D. none of these |
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Answer» Correct Answer - A The horizontal component of the force ederted on the bar by the hinge must balance the applied force `vecF`, nnd os has magnitude `120.0N` and is to the left. Taking torques bout point `A, (120.0N)(4.00m)+F_(v)(3.0m)` so the vertical componnet is `-160N, ` with the minus sign indicting a downward compnent, exerting torque in a directin opposite that of the horizontal component. the force exerted by the bar on the hinge is equal in magnitude and opposite in direction of the force exertd by the hinge on the bar. Net force `sqrt("["(160)^(2))+(140)^(2)"]"=200N` |
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| 54. |
Calculate the moments of inertia of the figures shown, each having mass `M`, radius `R` and having uniform mass distribution about an axis perpendicular to the plane and passing through the centre? |
| Answer» `(MR^(2))/2`. These cases are same as circular disc. The distributios with respect to axis of rotation is same pattern as circular disc. | |
| 55. |
Find the moment of inertia of a solid cylinder of mass M and radius R about a line parallel to the axis of the cylinder and on the surface of the cylinder. |
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Answer» The moment of inertia of the cylinder about is axis is `(MR^(2))/2`. Using parallel axis theorem. `I=I_(0)+MR^(2)=(MR^(2))/2+MR^(2)=3/2MR^(2)` Similarly the moment of inertia of a solid sphere about a tangent is `2/3 MR^(2)+MR^(2)=7/5MR^(2)` |
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| 56. |
In both the figures all other factors are same, except that in figure (i) `AB` is rough and `BC` is smooth while in figure (ii) `AB` is smooth and `BC` is rough. In figure (i), if a sphere is released from rest it starts rolling. Now consider the figure (ii), if same sphere is A released from top of the inclined plane, what will be the kinetic energy of the sphere on reaching the bottom: A. is same in both the casesB. is greater in case (i)C. is greater in case (ii)D. information insufficient |
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Answer» Correct Answer - D Let `J` be the linear impulse impacted to the ball. Applying impulse `=` change momentum we have `J=mv_(0)`……….i `J.h=Iomega_(0)=2/5mr^(2)omega_(0)`……….ii From eqn i and i we get `omega_(0)=5/2(v_(0)h)/r^(2)` |
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| 57. |
Two identical uniform solid spherical balls `A` and `B` of mass `m` each are placed on a the fixed wedge as shown in figure. Ball `B` is kept at rest and it is released just before two balls collides. Ball `A` rolls down without slipping on inclined plane and collide elastically with ball `B`. The kinetic energy of ball `A` just after the collision with ball `B` is: A. `(mgh)/7`B. `(mgh)/2`C. `(2mgh)/5`D. `(7mgh)/5` |
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Answer» Correct Answer - A Just before collision between two balls Potential energy lost by bal `A=` kinetic energy gained by ball `A`. `mgh/2=1/2I_(cm)omega^(2)+1/2mv_(cm)^(2)` `=1/2xx2/5 mR^(2)xx((v_(cm))/R)^(2)+1/2mv_(cm)^(2)=1/5mv_(cm)^(2)+1/2mv_(cm)^(2)` `implies5/7mgh=mv_(cm)^(2)implies(mgh)/7=1/5mv_(cm)^(2)` After collision only translattion kinetic energy is transferred to ball `B`. So just ater collision, rotational kinetic energy of ball `A=1/5mv_(cm)^(2)=(mgh)/7` |
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| 58. |
Two cylinders, one hollow (metal) and the other solid (wood) with the same mass identical dimensions are simulataneously 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. 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.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 - D `a=(gsintheta)/(1+I_(CM)/(mR^(2))), I_(CM)` of hollow cylinder is less so it will have more acceleration and will take less time to reach the bottom. |
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| 59. |
A body is rolling without slipping on a horizontal plane. The rotational energy of the body is `40% `of the total kinetic energy. Identify the body.A. ringB. Hollow cylinderC. solid cylinderD. hollow sphere |
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Answer» Correct Answer - D `(1/2lomega^(2))/(1/2Mv^(2))=40/60=2/3` or `((lv^(2))/R^(2))/(Mv^(2))=2/3` or `I=2/3MR^(2)` Clearly the body is hollow sphere |
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| 60. |
Two circular discs `A` and `B` of equal masses and thicknesses. But are made of metals with densities `d_A and d_B (d_A gt d_B)`. If their moments of inertia about an axis passing through the centre and normal to the circular faces be `I_A and I_B`, then.A. `I_(A)=I_(B)`B. `I_(A)gtI_(B)`C. `I_(A)ltI_(B)`D. `I_(A)geI_(B)` |
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Answer» Correct Answer - C Let `M` be the mass of each disc. Let `R_(A)` and `R_(B)` be the radii of discs `A` and `B`, respectively. Then `M=piR_(A)^(2)td_(A)=piR_(s)^(2) td_(B)` As `d_(A)=d_(B)`, so `R_(A)^(2)ltR_(s)^(2)`. Now `(I_(A))/(I_(B))=(R_(A)^(2))/(R_(B)^(2))lt1,` i.e. `I_(A)ltI_(B)` |
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| 61. |
`ABC` is a triangular plate of uniform `A` thickness. The sides are in the ratio shown in the figure. ` I_(AB), I_(BC), I_(CA)` are the moments of inertia of the plated about `AB, BC` and `CA` respectively. Which one of the following relation is correct? A. `I_(CA)` is maximumB. `I_(AB)gtI_(BC)`C. `I_(BC)gtI_(AB)`D. `I_(AB)+I_(BC)=I_(CA)` |
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Answer» Correct Answer - B Farther the mass from axis, greater will be the moment of inertia. |
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| 62. |
A trailer with loaded weight`F_(g)` is being pulled by a vehicle with a force `P`, as in figure. The trailer is loaded such that its centre of mass is located as shown. Neglect the force of rolling friction and let a represent the `x` component of the acceleration of the trailer. (a) Find the vertical component of `P` in terms of the given parameters. (b) If a` =2.00 ms^(-2)` and `h = 1.50 m`, what must be the value of `d` in order that `P = 0` (no vertical load on the vehicle)? |
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Answer» a. If the acceleration is a we have `P_(x)=ma` and `P_(y)+n-F_(g)=0` Taking the origin at the centre of gravity, the torque equation gives `P_(y)(L-d)+P_(x)h-nd=0` Solving these equations, we find `P_(Y)=(F_(g))/L(d-(ah)/g)` b. If `P_(y)=0,` then `d=(ah)/g=((2.00ms^(-2))(1.50m))/(10ms^(-2))=0.30m` |
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| 63. |
A solid iron sphere `A` rolls down an inclined plane. While an identical hollow sphere `B` of same mass sides down the plane in a frictionless manner. At the bottom of the inclined plane, the total kinetic energy of sphere `A` is.A. less than that of `B`B. equal to that of `B`C. more than that of `B`D. Sometimes more and sometimes less |
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Answer» Correct Answer - B In both the cases the loss of gravitational potential energy and the resulting gain total kinetic energy is same. |
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| 64. |
A solid iron sphere A rolls down an inclined plane, while another hollow sphere B with the same mass and same radius also rolls down the inclided plane. If `V_(A)` and `V_(B)` are their velocities a the bottom of the inclined plane. ThenA. `V_(A)gtV_(B)`B. `V_(A)=V_(B)`C. `V_(A)ltV_(B)`D. `V_(A)gt = ltV_(B)` |
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Answer» Correct Answer - A `a =(g sin theta)/(1+I/(MR^(2)))` `a_(A) gt a_(B)` So, `V_(A) gt V_(B)` |
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| 65. |
An isosceles triangular piece is cut a square plate of side `l`. The piece is one-fourth of the square and mass of the remaining plate is `M`. The moment of inertia of the plate about an axis passing through `O` and perpendicular to its plane is A. `(Ml^(2))/6`B. `(Ml^(2))/12`C. `(Ml^(2))/24`D. `(Ml^(2))/3` |
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Answer» Correct Answer - A Mass of each of the four parts `=M/3` Mass of the plane including the cut piece `=(4M)/3` Moment of inertia of the whole plate (including the cut piece ) about the said axis `=((4M)/3)(l^(2))/6` Now moment of inertia of the remaining portioin should be `3/4` of the above `=Ml^(2)//6` |
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| 66. |
A force `vec(F)=4hati-10 hatj` acts on a body at a point having position vector `-5hati-3hatj` relative to origin of co-ordinates on the axis of rotation. The torque acting on the body isA. `38 hatk`B. `-25 hatk`C. `62 hatk`D. none of these |
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Answer» Correct Answer - C `tau=vec(r)xxvec(F)` `=50hatk+12 hatk=62 hatk` |
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| 67. |
Two particles of mass 1 kg and 3 kg move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is 2m/s, their centre of mass has a velocity of 0.5 m/s. When the relative velocity of approach becomes 3m/s, the velocity of the centre of mass is 0.75 m/s. |
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Answer» Since no external force is acting on the two particle system `:.a_(CM)=0` `implies V_(CM)=`constant |
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| 68. |
A sphere has to purely roll upwards. At an instant when the velocity of sphere is `v,` frictional force acting on it is A. downwards and `mumgcostheta`B. downwards and `(2mgsintheta)/7`C. upward and `mumgcostheta`D. upwards and `(2mgsintheta)/7` |
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Answer» Correct Answer - D In case of rolling in the inclined plane, friction is static and acts in the upward direction and is given by `f=(mgsintheta)/(1+(R^(2))/(k^(2)))`………..i Form sphere `k^(2)=2/5R^(2)` ……….ii Form eqn i andii `f=2/7 mgsintheta` (upwards) |
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| 69. |
A particle of mass `m` is released from rest at point `A` in Fig., falling parallel to the (vertical) `y`-axis. Find the angular momentum of the particle at any time `t` with respect to the same origin `O`. |
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Answer» `vecL=vecrxxvecp` The magnitude of `vecL` is given by `L=rp sintheta(hatk)` in this example velocity of the particle at any time `v="gt",rsintheta=x_(0)` and `p=mv=m"gt"` `:.vecL=mgx_(0)t(hatk)`……..i The right hand rule shows that `vecL` is parallel to torque, i.e. direcrected perpedicular to paper in wards. |
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| 70. |
In the figure shown, a ball without sliding on a horizontal, surface. It ascends a curved track up to height `h` returns. The value of h is `h_(1)` for sufficiently rough curved track to avoid sliding and is `h_(2)` for smooth curved track then A. `h_(1)=h_(2)`B. `h_(1)lt h_(2)`C. `h_(1)gth_(2)`D. `h_(2)=2h_(1)` |
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Answer» Correct Answer - C If the track is smooth (case A), only transitional kinetic energy changes to the gravitational potential energy. But, if the track is rough (Case B), both translational and rotational kinetic energy changes to potential energy. Therefore, potential energy (=mgh) will be more in case B than in case A hence `h_(1) gt h_(2)` hence (3) |
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| 71. |
In the figure shown, a ball without sliding on a horizontal, surface. It ascends a curved track up to height `h` returns. The value of h is `h_(1)` for sufficiently rough curved track to avoid sliding and is `h_(2)` for smooth curved track then A. `h_(1)=h_(2)`B. `h_(1)lth_(2)`C. `h_(1)gth_(2)`D. `h_(2)=2h_(1)` |
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Answer» Correct Answer - C If the track is smooth (case `A`) only translational kintetic energy changes to gravitational potential energy. But,if the track is rough (case `B`) both translational and rotational kinetic energies change to potential energy. Therefore potential energy `(=mgh)` will be more in case `B` than in case `A`. Hence `h_(1)lth_(2)` |
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| 72. |
A smooth tube of certain mass is rotated in a gravity-free Space and released. The two balls shown in Fig move towards the ends of the tube. For the whole system, which of the following quantities is not conserved. A. Angular momentumB. Linear momentumC. Kinetic energyD. Angular speed |
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Answer» Correct Answer - D As `Sigmatau=0` , Angular momentum, linear momentum remains conserved. As the two balls will move radially out, I changes. In order to keep to the angular momentum (`L=I omega`) conserved, angular speed `(omega)` should change. Hence.(D) |
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| 73. |
A smooth tube of certain mass is rotated in a gravity-free Space and released. The two balls shown in Fig move towards the ends of the tube. For the whole system, which of the following quantities is not conserved. A. Angular mometumB. linear momentumC. kinetic energyD. angular speed |
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Answer» Correct Answer - D As `Sigmatau=0` angular momentum and linear moimentum remain conserved. As the two balls will move radially out `I` changes. In order to keep the agular momentum `L=Iomega` conserved, angular speed `omega` should change. |
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| 74. |
A disc of circumference `s` is at rest at a point `A` on a horizontal surface when a constant horizontal force begins to act on its centre. Between `A` and `B` there is sufficient friction toprevent slipping, and the surface is smooth to the right of `B.AB = s`. The disc moves from `A` to `B` in time `T`. To the right of `B`, .A. the angular acceleration of the disc will disappear linear acceleration will remain unchangedB. . linear acceleration of the disc will increaseC. the disc will make one rotation in time `T//2`D. the disc will cover a distance greater than s in further time `T` |
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Answer» Correct Answer - B::C::D Let `P=`external force `F=` force of friction between `A` and `B` `a_(1)=` acceleration between `A` and `B a_(2) =` acceleration beyond `B` `P-F=ma_(1)` `P=ma_(2` `a_(2)gta_(1)` Let `alpha=`angualar acceleration between `A` and `B` For one rotation `theta=2pi=1/2alphaT^(2)` `T=sqrt((4pi)/alpha)= ` time of travel from `A` to `B` angular velocity at `B=omega_(0)=alphaT` For one rotation to the right of `B` `theta=2pi=omega_(B)t` `t=(2pi)/(aT)=(1/2alphaT^(2))/(alphaT)=T/2` |
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| 75. |
A thin rod of length 4l, mass 4 m is bent at the point as shown in the figure. What is the moment of inertia of the rod about the axis passing through O and perpendicular to the plane of the paper?A. `(Ml^(2))/3`B. `(10Ml^(2))/3`C. `(Ml^(2))/12`D. `(Ml^(2))/24` |
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Answer» Correct Answer - B Total `MI=I_(1)+I_(2)+I_(3)+I_(4)` `=2I_(1)+2I_(2)` `=2(I_(1)+I_(2))[I_(1)=I_(3),I_(1)=I_(4)]` now `I_(2)=I_(3)=(Ml^(2))/3` using parallel axis theorem, we have `I=I_(CM)+Mx^(2)` and `x=sqrt(l^(2)+(l/2)^(2))` `I_(1)=I_(4)=(Ml^(2))/12+M[sqrt(l^(2)+(l/2)^(2))]` putting all values we get `I=(10 Ml^(2))/3` |
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| 76. |
A uniform thin rod is bent in the form of closed loop `ABCDEFA` as shown in the figure. The ratio of moment of inertia of the loop about `x`-axis to that about `y`-axis is A. `gt1`B. `lt1`C. `=1`D. `1//2` |
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Answer» Correct Answer - B Moment of inertia of semicirular portions about `x` and `y` axes are same. But moment of inertia of straight portions about `x`-axis is zero. `:. I_(x)ltI_(y)` or `(I_(x))/(I_(y))lt1` |
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| 77. |
A particle, moving horizontally, collides perpendicularly at one end of a rod having equal mass and placed on a smooth horizontal surface. A. Particle comes to rest if collision is perfectly elastic and centre of m rod starts to move with the same velocity.B. Particle continues to move along the same direction, whatever is the value of `e`.C. Particle may get rebound back.D. Velocity of mid-point of the rod will be less than `v//2` if the particle gets stuck. |
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Answer» Correct Answer - B::D Whenever two paticles having equal mass collide head on elastically, their velocities get interchanged. Therefore, if the particle collides at midpoint of the rod then velocities would get interchanged. In that case option a would be correct. But now the particle strikes at an end of the rod, hence particle head on collision does not take place. Therefore, option a is incorrect. If the particle gets rebound back to its original path, then its final mometum will become negative. Since mass of the particle and rod is equal, therefore law of conservation of momenum will be satisfied only when velocity of centre of mass of the rod is greater than orginal velocity of te particle. Hence, kinetic energy of the system (just after collision) will become greater than that (just before the collision), which is impossible. Hence, the particle cannot rebound back or it will colntinue to move along the same direction. Hence, option b is correct but option c is incorrect. if the paticle gets stick to the rod, centre of mass o the system will lie at a distance `l//4` from theend at which the particle sticks. According to the law of conservation of momentum cenre of mass will start ot move wit velocity `v//2`. But the particle will simultaneously start to rotate about tecentre of mass m an anticlockwise direction. Angular velocity `omega` of that rotational motion can be calculated by applying law of mass `m` an anticlockwise direction. Angular velocity `omega` of that rotational motion can be calculated by applying law of conservation of angular momentum. Hence the resultant velocity of the midpoint of the rod will be equal to `[(v_(0))//2)-(lomega//4)]` which is obviously less than `v_(0)//2`. Therefore option d correct. |
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| 78. |
A sphere `A` moving with speed `u` and rotating With angular velocity `omega` makes a head-on elastic collision with an identical stationary sphere `B`. There is no friction between the surfaces of `A` and `B`. Choose the conrrect alternative(s). Disregard gravity.A. `A` will stop moving but continue to rotate with an angular velocity `omega`B. `A` will come to rest and stop rotatingC. `B` will move with speed `u` without rotatingD. `B` will move with speed `u` and rotate with an angular velocity `omega` |
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Answer» Correct Answer - A::C::D There will be exchange of linear velocities only. However the two spheres cannot exert torques on each other, as their surfaces are firctionless and the angular velocities of the spheres do not change. |
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| 79. |
A equilaterial triangle `ABC` formed from a uniform wire has two small identical beads initially located at `A`. The triangle is set rotating about the vertical axis `AO`. Then the beads are released from rest simultaneously and allowed to slide down. one long. `AB` and the other along `AC` as shown. Neglecting frictional effects, the quantities that are conserved as the beads slide down, are. .A. angular velocity and total energy (kinetic and potential)B. total angular momentum and total energyC. angular velocity and moment of inertia about the axis of rotationD. total angular momentum and moment of inertia about the axis of rotation |
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Answer» Correct Answer - B The `MI` about the axis of rotation is not rotation is not constant as the perpendicular distance of the bead with the axis of rotation increases. Also since no external torque is acting, therefore, `tau_(ext)=(dL)/(dt)` `implies L=` constt. `Iomega=`constt. Since `l` increases `omega` decreases. |
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| 80. |
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 exis 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 `omega` the motion at any instant can be taken as a combination of (i) 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 system 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^@` 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 `omega` about the z-axis. Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct?A. It is `sqrt(2)omega` for both the casesB. It is `omega` for case a and `omega/(sqrt(2))` for case bC. It is `omega` force case a and `sqrt(2)omega` for case bD. It is `omega` for both the case |
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Answer» Correct Answer - D Angular speed about the instantaneous axis passing through centre of mass is `omega` for both the cases. |
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| 81. |
A solid sphere rolls down two different inclined planes of the same height but of different inclinationsA. the speeds will be same but time of descent will be differentB. . in both cases, the speeds and time of descent will be sameC. in both cases, the speeds and time of descent will be sameD. speeds and time of descent both will be different |
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Answer» Correct Answer - A As ` sqrt((2gh)/(1+1/(MR^(2))))` Hence velocity is independent of te inclination of the plane and depend any on height `h` through which body descends. But because `t+1/(sintheta)sqrt((2h)/g(1+I/(MR^(2)))` depends on the inclination also, hence greater the inclination lesser will be the time of desced. Hence, in present cae the speeds will be same (became `h` is same) but time is decreased will be difference (because of different inclination) |
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| 82. |
A hole of radius R/2 is cut from a thin circular plate of raduis R as shown in the figure. If the mass of the remaining plate is M, then find moment of inertia of the plate about an axis through O perpendicular to plane A. `13/25 MR^(2)`B. `13/24 MR^(2)`C. `11/24 MR^(2)`D. `11/25 MR^(2)` |
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Answer» Correct Answer - B Let the mass of disc without hole =m The mass of cut out hole of radius `R/2` is `m/4` `:. m-m/4=M` or `m=4/3M` Moment of inertia of given body about axis passing through O `=MI` of complete disc -MI of cut out hole . `=1/2((4M)/3)R^(2)-[1/2(M/3)(R/2)^(2)+M/3(R/2)^(2)]=13/24 MR^(2)` |
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| 83. |
If a spherical ball rolls on a table without slipping, the fraction of its total energy associated with rotation isA. `3/5`B. `2/7`C. `3/5`D. `3/7` |
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Answer» Correct Answer - B `f_(r)=(1/2Iomega^(2))/(1/2mv^(2)+1/2Iomega^(2)), where I=2/5mr^(2)` `omega=v/r` solve to get `f_(r)=2/7` |
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| 84. |
A solid sphere of mass `M` and radius `R` is placed on a rough horizontal surface. It is stuck by a horizontal cue stick at a height `h` above the surface. The value of `h` so that the sphere performs pure rolling motion immediately after it has been stuck is A. `(2R)/5`B. `(5R)/2`C. `(7R)/5`D. `(9R)/5` |
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Answer» Correct Answer - C Let `v` be the velocity of the centre of mass of the sphere and `omega` be the angular velocity of the body about an axis passing through the centre of mass. `J=Mv` `J(h-R)=2/5MR^(2)xxomega` From the above two equations `=v(h-R)=2/5r^(2)omega` From the condition of pure rolling `v=Romega` `h-R=(2R)/5impliesh=(7R)/5` |
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| 85. |
Two points of a rod move with velocity `3 v` and `v` perpendicular to the rod and in the same direction. Separated by a distance `r`. Then the angular velocity of the rod is :A. `(3v)/r`B. `(4r)/r`C. `(5v)/r`D. `(2v)/r` |
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Answer» Correct Answer - D `omega_(rod)=(v_(rel))_(_|_)/r` `(V_(rel))_(_|_)` is the velocity of one point w.r.t other perpendicular to rod. `omega_(rod)=(3v-v)/r` and `r` being the distance between the points `omega_(rod)=(2v)/r` |
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| 86. |
A smooth uniform rod of length L and mass M has two identical beads of negligible size each of mass m which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with an angular velocity `omega_0` about an axis perpendicular to the rod and passing through the midpoint of the rod. There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is ...... A. `(Momega_(0))/(M+tm)`B. `(Momega_(0))/m`C. `(Momega_(0))/(M+12m)`D. `omega_(0)` |
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Answer» Correct Answer - A From conservation of angular momentum `(ML^(2))/12 omega_(0)=[(ML^(2))/12+2m(L/2)^(2)]omega` `omega(M/(M+6m))omega_(0)` |
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| 87. |
A man of mass `100 kg` stands at the rim of a turtable of radius `2 m` and moment of inertia `4000 kgm^(2)` mounted on a vertical frictionless shaft at its centre. The whole system is initially at rest. The man now walks along the outer edge of the turntable with a velocity of `1m//s` relative to the earth a. With what angular velocity and in what direction does the turntable rotate? b. Through what angle will it have rotated when the man reaches his initial position on the turntable? c. Through what angle will it have rotated when the man reaches his initial position relative to the earth?A. The table rotates through `2pi//11` radians anticlockwiseB. The table rotates through `4pi//11` radians clockwiseC. The table rotates through `4pi//11` radians anticlock-wiseD. The table rotates through `27pi//11 radians clockwise |
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Answer» Correct Answer - D If the man completes one revolution relative to the table, then `theta_(mt)=2pi` `implies 2pi=theta_(m)-theta_(t)` `2pi=theta_(m)t-omega_(t)t` (where `t` is the time taken) `t=(2pi)/(omega_(m)-omega_(t))=(2pi)/(0.5-(-0.05))=(2pi)/0.55s` Angular displacement of the table is ` theta_(t)=omega_(r)t=-0.05xx((2pi)/0055)` `=-((2pi)/11)` radians The table rotates through `2pi//11` radians clockwise. |
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| 88. |
A conical pendulum consists of a mass `M` suspended from a strong sling of length `l`. The mass executes a circle of radius `R` in a horizontal plane with speed `v`. At time `t`, the mass is at position `Rhati` and `vhatj` velocity. At time `t` the angular momentum vector of mass `M` about the point from which the string passes on the ceiling is A. `MvRhatk`B. `Mvlhatk`C. `Mvl[sqrt((l^2-R^2)/l)hati+R/lhatk]`D. `-Mvl[sqrt((l^(2)-R^(2))/l)hati+R/lhatk]` |
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Answer» Correct Answer - C From the figure in questions position vector if mass `vecr=Rhati-sqrt(l^(2)+R^(3)hatk)` Linear moment `vecp=Mvhatj` `vecL=vecvxxvecp` substituting the values, we get result |
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| 89. |
Two heavy particles having masses `m_(1)` and `m_(2)` are situated in a plane perendicular to line `AB` at a distance or `r_(1)` and `r_(2)` respectively. a. What is the moment of inertia of the system about axis `AB`? b. What is the moment of inertia of the system about an axis passing through `m_(1)` and perpendicular to the line joining `m_(1)` and `m_(2)`? c. What is the moment of inertia of the system about an axis passing through `m_(1)` and `m_(2)`? |
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Answer» a. Moment of inertia of the particle on the left `I_(1)=m_(v)r_(1)^(2)`,. Moment of inertia of the particle on the right is `I_(2)=m_(2)r_(1)^(2)`. Moment of inertia of the system `AB` is `I=I_(1)+I_(2)=m_(1)r_(2)^(2)+m_(2)r_(2)^(3)` b. Moment of inertia of the particle on the left is `I_(1)=0`. Moment of inertia of the particle on the right is `I_(2)=m_(2)(r_(1)+r_(2))^(2)`. moment of inertia of the system about the given axis is `I=I_(1)+I_(2)=0+m_(2)(r_(1)+r_(2))^(2)` c. Moment of inertia of the particle on theleft is `I_(1)=0` Moment of inertia of the particle on the right is `I_(2)=0` Moment of inertia of the system about the given axis is `I=I_(1)+I_(2)=0+0` |
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| 90. |
Find out the moment of inertia of the circular arcs shown, each having mass `M`, radius `R` and having uniform mass distribution about an axis passing through the centre and perpendicular to the plane ? |
| Answer» `MR^(2)` (infat `MI` of any part of mass `M` of a ring of radius `R` about an axis passing through the geometrical centre and perpendicular to the plane of the ring is `MR^(2)`). | |
| 91. |
In Fig, the bar is uniform and weighing `500 N`. How large must `W` be if `T_(1)` and `T_(2)` are to be equal? A. `500N`B. `300N`C. `750N`D. `1500N` |
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Answer» Correct Answer - D Taking torque about the attachmet point for `W` get `-T_(1)(0.4L)+T_(2)(0.3L)+500(0.2L)=0` `T=1000N`, were `T_(1)=T_(2)=T` `SigmaF_(y)=0gt2T-W-500=0` `implies W=1500N` |
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| 92. |
A mass `m` is moving at speed `v` perpendicular to a rod of length `d` and mass `M= 6m` pivots around a frictionless axle running through its centre and stickes to the end of the rod. The moment of inertia od the rod about its centre is `Md^(2)//12`. Then the angular speed of the system right after the collision us -A. `(2v)/(3d)`B. `(2v)/d`C. `v/d`D. `(3v)/(2d)` |
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Answer» Correct Answer - A By conservation of angular momentum about pivot `L=Iomega` `(mvd)/2=[(Md^(2))/12+m(d/2)^(2)]omega` `=((md^(2))/2+(md^(2))/4)omega` `=3/4md^(2)omegaimplies 2/3v/d=omega` |
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| 93. |
A uniform solid. cylinder `A` of mass can freely rotate about a horizontal axis fixed to a mount of mass `m_(2)`. A constant horizontal force `F` is applied to the end `K` of a light thread tightly wound on the cylinder. The friction between the mount and the supporting horizontal plane is assumed to be absent. Find the acceleration of the point `K`. |
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Answer» The acceleration whole system `a_(1)=F/(m_(1)+m_(2))` The acceleration of the point `K` w.r.t the axis of the cylinder `a_(2)=alphaR` where `alpha` is given by `FR=Ialpha` or `alpha=(FR)/(m_(1)R^(2)//2)=(2F)/(m_(1)R)` `implies a_(2)=(2F)/m_(1)` The acceleration of the point `K` w.r.t ground `=a_(1)+a_(2)=F/(m_(1)+m_(2))+(2F)/m_(1)=F((3m_(1)+2m_(2)))/(m_(1)(m_(1)+m_(2)))` |
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| 94. |
Figure shows a small wheel fixed coaxially on a bigger one of double the radius. The system rotates about the comon axis. The strings supporting A and B do not slip on the wheels. If x and y be thedistances travelled by A and B in the same time interval, then A. `x=2y`B. `x=y`C. `y=2x`D. none of these |
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Answer» Correct Answer - C Double torque, double angular acceleration, double, linear acceleration double distance |
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| 95. |
A ring of mass `3kg` is rolling Without Slipping with linear velocity `1 ms^(-1)` on a smooth horizontal surface. A rod of same mass is fitted along its one diameter. Find total kinetic energy of the system (in `J`). |
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Answer» Correct Answer - 5 `K=K_(R)+K_(T)=1/2[mR^(2)+((2R)^(2))/12][V/R]^(2)+1/2(2m)V^(2)` `=5/3Mv^(2)=5/3(3)(1)^(2)=5J` |
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| 96. |
A thin ring of mass 2kg and radius 0.5 m is rolling without on a horizontal plane with velocity 1m/s. A small ball of mass 0.1kg, moving with velocity 20 m/s in the opposite direction hits the ring at a height of 0.75m and goes vertically up with velocity 10m/s. Immediately after the collision A. the ring has pure rotation about its stationaryB. the ring comes to a complete stopC. friction between the ring and the ground is to the leftD. there is no friction between the ring and the ground |
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Answer» Correct Answer - C Friction force on the ring |
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| 97. |
A block of base `10cmxx10cm` and height 15cm is kept on an inclined plane. The coefficient of friction between them is `sqrt3`. The inclination `theta` of this inclined plane from the horizontal plane is gradually increased from `0^@` ThenA. at `theta=30^@` the block will start sliding down the planeB. the block will remain at rest on the plane up to certain `theta` and ten it will toppleC. at `theta=60^@` the block will start slidng down the plane and continue to do so at higher anglesD. at `theta=60^@` the block will start sliding down the plane and on further increasing `theta`, it will topple at certain `theta` |
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Answer» Correct Answer - B For sliding `tanthetagesqrt(3)(=1.732)` For toppling `tanthetage2/3(=0.67)` |
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| 98. |
A uniform flag pole of length `L `and mass `M` in pivoted on the ground with a frictionless hinge. The flag pole makes an angle `theta` with the horizontal. The moment of inertia of the flag pole about one end is `(1//3) ML^(2)`. If starts falling from the position shown in figure, the linear acceleration of the free end of the flag pole-labeled `P` - would be. .A. `(2/3)gcostheta`B. `(2/3)g`C. `g`D. `(3/2)gcostheta` |
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Answer» Correct Answer - D `(MgL)/2costheta=(ML^(2))/3alpha` `alpha=3/2(gcostheta)/L` `a=alphaL=3/2gcostheta` |
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| 99. |
Calculate the moment of inertia of each particle in Fig. about the indicated axis of rotation. |
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Answer» a. `I=m(l/2)^(2)+m(l/2)^(2)=(ml^(2))/2` b. `I=4ml^(2)` c. `I=ml^(2)+m(l/2)^(2)=5/4ml^(2)` `d. I=2m(2l)^(2)=8ml^(2)` |
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| 100. |
A uniform rod of mass `m` and length `l` is placed over a smooth horizontal surface along the `y`-axis and is at rest as shown in Fig. An impulsive force `F` is applied for a small time `/_ ` along `x`-direction at point `A`. The `x`-coordinate of end `A` of the rod when the rod becomes parallel to `x`-axis for the first time is [initially, the coordinate of centre of mass of the rod is `(0, 0)`] A. `(pil)/12`B. `l/2(1+pi/12)`C. `1/2(1-pi/6)`D. `l/2(1+pi/6)` |
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Answer» Correct Answer - D As torque `=`change in momentum `F/_ =mv`(linear momentum)………i and `(Fl/2)/_ =(ml^(2))/12 omega` (angular momentum)..ii Dividing eqn i and ii we get `2=(12v)/(omegal)implies omega=(6v)/l` using `S=ut` Displacement of `CM` is `pi/2=omegat=((6v)/l)t` and `=vt` Dividing we get `(2x)/pi=l/6impliesx=(pil)/12` coordinates of A will be `[(pil)/12+l/2, 0]` |
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