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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.
| 251. |
The sum of moments of masses of all the particle in a system about the centre of mass is alwaysA. zeroB. maximumC. infiniteD. minimum |
| Answer» Correct Answer - A | |
| 252. |
A body performs rotational motion about the given axis, particles perform U.C.M. with same angular velocity. The particles at different positions haveA. same velocityB. constant velocityC. different velocityD. nothing can be said about velocity |
| Answer» Correct Answer - C | |
| 253. |
A body of radius `R` and mass `m` is rolling smoothly with speed `v` on a horizontal surface. It then rolls up a hill to a maximum height `h`. If `h = 3 v^2//4g`. What might the body be ?A. solid sphereB. hollow sphereC. discD. ring |
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Answer» Correct Answer - C |
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| 254. |
A cylinder of mass M and radius R is resting on a horizontal paltform (which is parallel to the x-y plane) with its exis fixed along the y-axis and free to rotate about its axis. The platform is given a motion in the x-direction given by `x =A cos (omega t).` There is no slipping between the cylinder and platform. The maximum torque acitng on the cylinder during its motion is .................. |
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Answer» Correct Answer - A::B::C Considering the motion of the platform `x=A cos omegat` `rArr (dx)/(dt) =- A omega sin omegat rArr(d^2x)/(dt^2) =- A omega^2 con omegat` The magnitude of the maximum acceleration of the platform is `:. |Max acceleration| = A omega^2` When platform moves a torque acts on the cylinder and the cylinder rotates about its exis. Acceleration of cylinder, `a_1 = (f)/(m)` Torque `pi = fR :. l alpha = fR` `alpha = (fR)/(I) = (fr)/(MR^2 /2)` or, `alpha = (2f)/(MR) or Ralpha = (2f)/(M)` `:. Equivalent linear acceleration (Ralpha = a_2)` `a_2 = (2f)/(M)` :. Total linear acceleration, `a_(max) = a_1 +a_2 = (f)/(M)+(2f)/(M) = (3f)/(M)` or,`Aomega^2 = (3f)/(M) or , f = (MAomega^2)/(3)` Thus, maximum torque, `tau_(max) = fxxR = (MAomega^2R)/(3) = (1)/(3) MARomega^2` |
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| 255. |
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 beada are at the centre of the rod and the system is rotating with an angular velocity `omega_0` about an axis perpenducular 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 ...... |
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Answer» Correct Answer - A Since no external force and hence no torque is applied the angular momentum remains constant `:. I_1 omega_1 = I_2 omega_2` `:. omega_2 = (I_1omega_1)/(I_2) = ((ML^2)/(12)xxomega_0)/((ML^2)/(12) +2mxx((L)/(2))^2) = (Momega_0)/(M+6m)` |
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| 256. |
Two balls A and B of angular velocities `omega_(A)` and `omega_(B)` collide with each other. Then, after collisionA. both have same angular velocitiesB. `omega_(A) gt Omega_(B)`C. `omega_(A) = omega_(B)` when balls are smoothD. `omega_(A) gt omega_(B)` when balls are smooth |
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Answer» c) a) This is possible, when balls are is smooth. b) `omega_(A) = omega_(B)` is not always possible. It depends upon conditions. But no conditions are mentioned . |
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| 257. |
If a circular concentric hole is made on a disc then about an axis passing through the centre of the disc and perpendicular to its planeA. moment of inertia decreasesB. moment of inertia increasesC. radius of gyration increasesD. radius of gyration decreases |
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Answer» Correct Answer - A::C |
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| 258. |
A rod of length l, density of material D and area of cross section is A . If it rotates about its axis perpendicular to the length and passing through its centre, then its kinetic energy of rotation will beA. `(Al^(3)omega^(2))/(12)`B. `(Al^(3)D. omega^(2))/(24)`C. `(Al^(3)D. omega^(2))/(6)`D. `(Al^(3)D. omega^(2))/(48)` |
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Answer» Correct Answer - B `KE = (1)/(2)I omega^(2)` `= (1)/(2)(ML^(2))/(12)xx omega^(2)` `= (1)/(2)xx(DAL^(3))/(12)omega^(2)` `= (Al^(3)D omega^(2))/(24)` |
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| 259. |
Two rod 1 and 2 are released from rest as shown in figure . Given, `l_(1)=4l,m_(1)=2m,l_(2)=2l "and "m_(2)=m.` There is no friction between the two rods . If `alpha` be the angular acceleration of rod 1 just after the rods are released . Then, What is the initial angular acceleration of rod 2 in terms of the given parameters in the question ? A. `[(2sqrt(3)g)/(2l)+2sqrt(3)alpha]`B. `[(3sqrt(3)g)/(l)-sqrt(3)alpha]`C. `[(6sqrt(3)g)/(8l)+5sqrt(3)alpha]`D. `[(3sqrt(3)g)/(8l)-(8)/(sqrt(3))alpha]` |
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Answer» Correct Answer - D |
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| 260. |
Two thin planks are moving on a four identical cylinders as shown. There is no slipping at any contact points. Calculate the ratio of angular speed of upper cylinder to lower cylinder |
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Answer» Correct Answer - 3 |
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| 261. |
In the given figure `F=10N, R=1m`, mass of the body is `2kg` and moment of inertia of the body about an axis passing through `O` and perpendicular to the plane of the body is `4kgm^(2)`. `O` is the centre of mass of the body. If the ground is smooth, what is the total kinetic energy of the body after `2s`?A. 25 JB. 50 JC. 75 JD. 100 J |
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Answer» Correct Answer - B |
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| 262. |
In the given figure `F=10N, R=1m`, mass of the body is `2kg` and moment of inertia of the body about an axis passing through `O` and perpendicular to the plane of the body is `4kgm^(2)`. `O` is the centre of mass of the body. If the ground is sufficiently rough to ensure rolling, what is the kinetic energy of the body now in the given time interval of `2s`?A. 10.33 JB. 25.67 JC. 16.67 JD. None of these |
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Answer» Correct Answer - C |
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| 263. |
A flywheel is in the form of a uniform circular disc of radius `1 m` and mass `2 kg`. The work which must be done on it to increase its frequency of rotation from `5` to `10 rev//s` is approximatelyA. `1.5xx10^(2)J`B. `3.0xx10^(2)J`C. `1.5xx10^(3)J`D. `3.0xx10^(3)J` |
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Answer» `W=K_(2)-K_(1)=(1)/(2)I(omega_(2)^(2)-omega_(1)^(2))` `=(1)/(2)((1)/(2)xx2xx1^(2))(10^(2)-5^(2))(4pi^(2))` `=150pi^(2)=1.5xx10^(2)J` |
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| 264. |
If a person standing on a rotating disc stretches out his hands, the angular speed willA. increasesB. decreasesC. become zeroD. remains constant |
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Answer» Correct Answer - D As no external torque is acting. |
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| 265. |
A ody is in pure rotation. The linear speed v of a particle, the distance r of the particle from the axis and the angular velocity `omega` of the body are related as `omega=v/r`. ThusA. `omega prop (1)/(r )`B. `omega` is independent of rC. `omega = 0`D. `omega prop r` |
| Answer» Correct Answer - B | |
| 266. |
A ody is in pure rotation. The linear speed v of a particle, the distance r of the particle from the axis and the angular velocity `omega` of the body are related as `omega=v/r`. ThusA. `omega prop 1//r`B. `omega prop r`C. `omega=0`D. `omega` is independent `r` |
| Answer» When a rigid body is in pure rotation about an axis, the angular speed of its all particles is same. | |
| 267. |
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» `x=(omega r)t` `y=(omegaxx2r)t` `y=2x` |
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| 268. |
A spherical body of radius `R` rolls on a horizontal surface with linear velociltly `v`. Let `L_(1)` and `L_(2)` be the magnitudes of angular momenta of the body about centre of mass and point of contact `P`. Then: A. `L_(2)=2L_(1)`, if radius of gyration about centroidal axis K=RB. `L_(2)=2L_(1)`, for all casesC. `L_(2)lt2L_(1)`, if radius of gyration about centroidal axis `KltR`D. `L_(2)lt2L_(1)`, if radius of gyration about centroidal axis `KgtR` |
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Answer» Correct Answer - A::C |
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| 269. |
A solid cylinder of mass M and radius R pure rolls on a rough surface as shown in the figure. Choose the correct alternative (s). A. The acceleration of the centre of mass is `F/(M)`B. The acceleration of the centre of mass is `(2)/(3)F/(M)`C. The friction force on the cylinder acts backwardD. The magnitude of the friction force is `F/(3)` |
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Answer» Correct Answer - B::C::D |
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| 270. |
A solid sphere and a solid cylinder of same mass are rolled down on two inclined planes of heights `h_(1)` and `h_(2)` respectively. If at the bottom of the plane the two objects have same linear velocities, then the ratio of `h_(1):h_(2)` isA. `2:3`B. `7:5`C. `14:15`D. `15:14` |
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Answer» Correct Answer - C |
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| 271. |
A solid sphere of radius R is rolled by a force F acting at the topo of the sphere as shown in the figure. There is no slipping and initially sphere is in the rest position, then (CM= centre of mass) A. Work done by force F when the centre of mass move a distance S is 2 FSB. speed of the CM when CM moves a distance S is `sqrt((20)/(7)(FS)/(M))`C. work done by the Force F when CM move a distance S is FSD. speed of the CM when CM moves a distance S is `sqrt((6)/(5)(FS)/(M))` |
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Answer» Correct Answer - A::B |
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| 272. |
A mass of 2 kg is rotating on a circular path of radius 0.8 m with angular velocity of `44 rad s^(-1)`. If the radius of the path becauses `1.0` m, what will be the value of angular velocity? |
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Answer» Correct Answer - `28.16 rad s^(-1)` |
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| 273. |
A uniform cubical block of mass M and side length L is lying on the edge of a rough table with `(1)/(4) th` of its edge overhanging. When a small block of mass m is placed on its top surface at the right edge (see fig.), the cube is on verge of toppling. The block of mass m is given a sharp horizontal impulse so that it acquires a velocity towards B. The small block moves on the top surface and falls on the other side. What is maximum coefficient of friction between the small block and the cube so that the cube does not rotate as the block moves over it. Assume that the friction between the cube and the table is large enough to prevent sliding of the cube on the table. |
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Answer» Correct Answer - `mu_("max") = (1)/(2)` |
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| 274. |
A uniform rectangular block is moving to the right on a rough horizontal floor (the block is retarding due to friction). The length of the block is L and its height is `h`. A small particle (A) of mass equal to that of the block is stuck at the upper left edge. Coefficient of friction between the block and the floor is `mu = (2)/(3)`. Find the value of h (in terms of L) if the normal reaction of the floor on the block effectively passes through the geometrical centre (C) of the block. |
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Answer» Correct Answer - `h = (L)/(2)` |
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| 275. |
A cylinder C rests on a horizontal surface. A small particle of mass m is held in equilibrium connected to an overhanging string as shown. The other end of the mass less string is being pulled horizontally by a force F as shown. Find F. |
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Answer» Correct Answer - `F = (mg)/(2)` |
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| 276. |
Rectangular block B, having height `h` and width `d` has been placed on another block A as shown in the figure. Both blocks have equal mass and there is no friction between A and the horizontal surface. A horizontal time dependent force `F = kt` is applied on the block A. At what time will block B topple? Assume that friction between the two blocks is large enough to prevent B from slipping. |
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Answer» Correct Answer - `t = (2d.g.m)/(kh)` |
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| 277. |
A cylinder rolls down on an inclined plane of inclination `37^(@)` from rest. Coefficient of friction between plane and cylinder is 0.5. Calculate the time ( in s) of travelling down the incline 8 m as shown in figure .`(g=10 m//s^(2))` |
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Answer» Correct Answer - 2 |
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| 278. |
The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is `1.1 Å`. Given, mass of carbon atom is 12 a.m.u. and mass of oxygen atom is 16 a.m.u., calculate the position of the centre of mass of the carbon monoxide molecule |
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Answer» Correct Answer - `0.645 Å` from C-atom |
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| 279. |
In a carbon monoxide molecule, the carbon and the oxygen atoms are separated by a distance `1.2xx10^-10m`. The distance of the centre of mass from the carbon atom isA. `0.64xx10^(-10) m`B. `0.56xx10^(-10) m`C. `0.51xx10^(-10) m`D. `0.48xx10^(-10) m` |
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Answer» Correct Answer - a |
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| 280. |
A sphere of radius R is supported by a rope attached to the wall. The rope makes an angle `theta = 45^(@)` with respect to the wall. The point where the rope is attached to the wall is at a distance of `(3R)/(2)` from the point where the sphere touches the wall. Find the minimum coefficient of friction `(mu)` between the wall and the sphere for this equilibrium to be possible. |
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Answer» Correct Answer - `mu_("min") = (1)/(2)` |
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| 281. |
A force `vecF=Ahati+Bhatj` is applied to a point whose radius vector relative to the origin is `vecr=ahati+bhatj`, where `a`, `b`, `A`,`B` are constants and `hati`, `hatj` are unit vectors along the `X` and `Y` axes. Find the torque `vectau` and the arm `l` of the force `vecF` relative to the point `O`. |
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Answer» `vectau=vecrxxvecF=|[hati,hatj,hatk],[a,b,0],[A,B,0]|` `vectau=(aB-bA)hatk` Lever arm `l= (|vectau|)/(|vecF)|=((aB-bA))/(sqrt(A^(2)+B^(2)))` `{{:(tau=rFsintheta implies r sintheta=(tau)/(F)),(r sin theta:"lever arm"):}}` |
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| 282. |
An equailaral 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 along AB and the other along AC as shown. Neglectig frictional effects, the equatities that are conserved as the beads slids 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 rotaitonD. total angular momentum and momento f inertia about the axis of rotation |
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Answer» Correct Answer - B The M.I. about the axis of rotation is not constant as the perpendicular distance of the bead with the axis of rotation increases. Also since no external torque is acting. `:. Tau_(ext) = (dL)/(dt) rArr L = constant rArr constant` Since, I increases, `omega` decreases. |
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| 283. |
One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. it is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is A. MR^2`B. `(1)/(4)MR^2`C. `(1)/(8)MR^2`D. `sqrt2MR^2` |
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Answer» Correct Answer - A The mass distribution of this sector is same about the axis of rotation as that of the complate disc about the axis. Therfore the formula remains the same as that of the disc. |
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| 284. |
A small sphere of mass 1 kg is rolling without slipping with linear speed `v=sqrt((200)/(7) m//s` It leaves the inclined plane at point C. Find the linear speed at point C.A. `sqrt((100)/(7)` m/sB. `sqrt((50)/(7)` m/sC. `sqrt((100)/(35)` m/sD. `sqrt((200)/(35)` m/s |
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Answer» Correct Answer - A |
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| 285. |
A uniform rod of mass m, length l is placed over a smooth horizontal surface along y -axis and is at rest as shown in figure. An impulsive force F is applied for a small time `Deltat` along x-direction at point A. The x-coordinte of end A of the rod when the rod becomes parallel to x-axis for the first is (in itially the coordinates opf centre of mass of the rod is (0,0). A. `(pil)/(12)`B. `l/2(1+(pi)/(12))`C. `l/2(1-(pi)/(6))`D. `l/2(1+(pi)/(6))` |
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Answer» Correct Answer - D |
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| 286. |
A uniform circular disc has a sector of angle `90^(@)` removed from it. Mass of the remaining disc is M. Write the moment of inertia of the remaining disc about the axis `x x` shown in figure (Radius is R) |
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Answer» Correct Answer - `(1)/(4) MR^(2)` |
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| 287. |
The following bodies have same mass which of them have largest M.I. ?A. Ring - about an axis perpendicular to its planeB. Disc - about an axis perpendicular to its planeC. Solid sphere - about an axis passing through its centreD. Bar magnet - about an axis through its centre and perpendicular to its plane |
| Answer» Correct Answer - A | |
| 288. |
the flywheel is so constructed that the entire mass of it is concentrated at its rim, becauseA. it increases the powerB. it increases the speedC. it increases the moment of inertiaD. it save the flywheel fan breakage |
| Answer» Correct Answer - C | |
| 289. |
Calculate the ratio of moment of inertia of a thin uniform disc about axis 1 and 2 marked in the figure. `O` is the centre of the disc. |
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Answer» Correct Answer - `(3)/(2)` |
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| 290. |
Moment of inertia of a body does not depend onA. mass of the bodyB. distribution of the mass in the bodyC. ais of rotation of the bodyD. angular velocity of the body |
| Answer» Correct Answer - D | |
| 291. |
Thye angular momentum and the moment of inertia are respectivelyA. vector and tensor quantitiesB. scalar and vector quantitiesC. vector and vector quantitiesD. scalar and scalar quantities |
| Answer» Correct Answer - A | |
| 292. |
The mass of a flywheel is concentrated at its rimA. to obtain a strong wheelB. to decrease the moment of inertiaC. to increase the moment of inertiaD. to obtain the stable equilibrium |
| Answer» Correct Answer - C | |
| 293. |
About which of the following axis, the moment of inertia of a thin circular disc is minimum ?A. Through centre perpendicular to the surfaceB. Tangential and perpendicular to the surfaceC. Through centre parallel to the surfaceD. Tangential and parallel to the surface |
| Answer» Correct Answer - C | |
| 294. |
Consider an equilateral prism as shown in the figure. The mass of the prism is M and length of each side of its cross section is `a`. Find the moment of inertia of such a prism about the central axis shown. |
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Answer» Correct Answer - `(Ma^(2))/(12)` |
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| 295. |
A thin carpet of mass `2m` is rolled over a hollow cylinder of mass `m`. The cylinder wall is thin and radius of the cylinder is R. The carpet rolled over it has outer radius 2R (see figure). This roll is placed on a rough horizontal surface and given gentle push so that the carpet begins to roll and unwind. Friction is large enough to prevent any slipping of the carpet on the floor. Also assume that the carpet does not slip on the surface of the cylinder. The entire carpet is laid out on the floor and the hollow cylinder rolls out with speed V. Find V. |
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Answer» Correct Answer - `V = sqrt(5gR)` |
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| 296. |
The moment of inertia of thin square plate `ABCD` of uniform thickness about an axis passing through the center `O` and perpendicular to the plane of the plate is (`i`) `I_(1)+I_(2)` (`ii`) `I_(2)+I_(4)` (`iii`) `I_(1)+I_(3)` (`iv`) `I_(1)+I_(2)+I_(3)+I_(4)` where `I_(1)`, `I_(2)`, `I_(3)` and `I_(4)` are repectively moments of inertia about axes `1`, `2`, `3` and `4` which are in the plane of the planeA. (`i`),(`ii`)B. (`i`), (`ii`), (`iii`)C. (`ii`), (`iii`)D. (`i`), (`iii`) |
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Answer» As explained in the previous problem `I_(1)=I_(2)=I_(3)=I_(4)` `I_(0)=I_(1)+I_(2)=I_(2)+I_(4)` (`bot^(ar)` axes theorem) `=I_(1)+I_(3)` |
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| 297. |
The radius of a wheel is R and its radius of gyration about its axis passing through its center and perpendicualr to its plane is K. If the wheel is roling without slipping. Then the ratio of tis rotational kinetic energy to its translational kinetic energy isA. `K^(2)/(R^(2)`B. `(R^(2)/K^(2))`C. `(R^(2)/(R^(2)+K^(2))`D. `(K^(2)/(R^(2)+K^(2))` |
| Answer» a) (Rotational KE)/(Translational KE) = ((1/2)lomega^(2))/(1/2Mv^(2)) = K^(2)/R^(2)` [`therefore` l = `MK^(2)`, v =`Romega] | |
| 298. |
A hollow cylinder and a solid cylinder having the same mass and diameter are released from rest simultaneously from the top of an inclined plane. They roll without slipping which will reach the bottom firstlyA. the solid cylinderB. the hollow cylinderC. both will reah the bottom togetherD. can not be predicted |
| Answer» Correct Answer - A | |
| 299. |
Three identical cylinders have mass M each and are placed as shown in the figure. The system is in equilibrium and there is no contact between B and C. Find the normal contact force between A and B. |
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Answer» Correct Answer - `(Mg)/(2)` |
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| 300. |
A 70 kg man standing on ice throws a 3 kg body horizontally at 8m//s. The friction coefficient between the ice and his feet is 0.02. The distance, the man slip isA. 0.3 mB. 2 mC. 1 mD. `infty` |
| Answer» Correct Answer - a | |