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SrnrlVIBKANT ACADEIITY(India) Prirate Limited PHYSICSPractice Problems tRpTarget llT ADUANCE-ZO1 gDATE: 16/04/2013 TIME: 90 MIN. IRPDPP, NO..3

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Mechanics # 3Single Choice Questions1. Two identical rings each of mass m with their planes nutually perpendicular, radius R arewelded at their point of contact O. lf the system is free to rotate about an axis passing

through the point P perpendicular to the plane of the paper the moment of inertia of thesystem about this axis is equal to(A) 6.5 mR2 (B) 12 mR2 (C) 6 mR2 11.5 mR2A straight rod AB of mass M and length L is placed on a frictionless horizontal surface. A force havingconstant magnitude F and a fixed direction starts acting at the end A. The rod is initially perpendicular to theforce. The iniial acceleration of end B is(A) Zero (D) None

3. An uniform solid cylinder of mass 2 Kg and radius 1 m is imparted a velocity 10 m/s on a rough horizontalsurface as shown in figure.The coefficient of friction between the cylinder and surfece is 0.2. The workdone by friction force on the cylinder in its centre displacement by 9 m is [Assume cylinder is performingimpure rolling for this durationl(A)-34J-28 J34J

(D) None4. A rigid cube ABCDEFGH is in motion. At a certain moment, face ABCD is vertical, and the velocities ofverticesAand D are directed vertically downward and equal to v. At the same moment, the speed of point

H equals 2v. What point of the cube has the maximum speed at that moment ? What is that speed ?(A) Point Q 3v

4F(c) M

19#tt points between F & G .6v(B) All points between F & G 3v(D)Allpoints betwton B & C, r6v

5. A ball of mass m is located at the origin O of a frame of reference at a height h above the earth surface. Asthe ball starts falling from the point O without any initial velocity the frame of reference stiarts movinghorizontally with a constant velocity V. What is the increment in the angular momentum of the ball duringthe time of its free fall.(A) 2mVh p) more than mVh (D) less than mVh6. The frame shown in the figure is hinged at point O, so that it is free to rotate in a vertical plane about ahorizontal axis through O. The frame is made by welding two identical, slender and uniform rods of lengthL each. The rods are welded at 45o to each other, at the midpoint of the inclined rod. With the frame in itsequilibrium position, a sharp horizontal impulse is given to the vertical rod, at some point P, as shown. Findthe distiance OP if the hinge experiences no horizontal impulsive reactton.(A)

" =+ (B)(c) +3L4v-r-ry

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Mu ltiple Ghoice Questions7. A portion of a ring of radius R has been removed symmetrically aboutx-axis as shown in figure. Mass of the remaining portion is m. Centre ofthe ring is at origin O. Let lo and lo be the moment of inertias passingthrough points A and O and perpendicular to the plane of the ring. Then

(B) lo = lo

Comprehension # I (Q. No. 8 to 10)Two identical rods each of length L and Mass M are hinged at their endsas shown in figure. The horizontal separation between the hinges isInitially the rods are kept in the position shown and in this position thesystem is released from rest. Based on following diagram answerquestion32to 34.

(c)s/16

Intlo

L,l+-U2+1

8. lf just after the release of system angular accleration of \A is 4 radls2, then, the reaction force exertedby hinge an rod B is

(D) None of the above9. For the dat ofQ.64 the maximum value of L for which the rod's remain in contiact at least for some time just

afier the release of systeme is(A) 39/16 gt3210. Just after the release of system, the reaction force exerted by hinge on rod A is [Use data of Q.64]

8mI-343(c)

(D) None of these

(A) M ,F.rh.*.i| (c) M ,lr'.r(r-+l(D)Noneorthesev \ 313) \l \ 343)Subjective Questions11. In the figure shown, the mass of the wedge is 4 kg, while that of the spere (radius R) is unknown (say, m).

The inclined surface of the wedge is rough and makes an angle of 53o (sin53o = 0.8, cos53o = 0.6) with thehorizontal. The sphere can roll without sliding on the inclined surface of wedge. lt is released from rest atsome moment, with the wedge initially stationary.The wedge can move on the horizontal surface withoutfriction. Let acceleration of wedge = A, and angularacceleration of sphere = cr. lt is found that : 5A = Ro.(A) Draw the free body diagram of the sphere and the wedge.(B) Find the mass m of the sphere from the fact that 5A = Rs.(C) Find the value of A, using rotationaldynamics.

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A ring is placed on the top of an inclined plane and released from rest. The inclined plane makes an angleof 30o with the horizontal. The coefficient of friction between the ring and the incline is 0.2. Find the distancetraveled by the center of the ring by the time it completes one revolution, as it rolls down the incline.A thin uniform circular disc of radius R and mass m is hinged about its center point O, so that it is free torotate about a fixed horizontial axis through G. The plane of the disc is vertical. A small body A, of massm/2, is fixed at the rim of the disc, as shown. lnitially, the line OA makes an angle of 600 with the vertical.The disc is now released from rest.(A) Find the acceleration of pointA just after release.ts ) Fnd the m agnjh-rde ofhorZonbl-and vertbal reactbn 6rces : F;o. ?rtd Fu"n on the hinge, just after thedisc is released.

A thin wooden ring of mass M and radius R is placed on a horizontal rough floor. A slender metal rod oflength R mass M is fbced radially on the rim of the wooden ring, as shown. The rod is initially in the verticalposition shown' At this moment, the ring is set rolling to the right by imparting its center a linear velocity vo(and angular velocity votR). The ring rolls on the floor without stiding.(A)At the instant shown, find the normar reaction on the ring from the floor, if ve = n[tR(B) For what value of vo is the normal reaction between ring and he floor zero when the rod comes in thevertical position after a rotation of 180o ?

0,'LS-''f3) rlraba U" {-@

An uniform spherical shell of mass m and radius R starts frsrn rest with pure rolling on long inclined planeas shown in figure.The angular momentum of shell about point of contact after 1s of its starting isKmR.Determine the value of K. ,15.

A particle moves in a closed path under the action of a central field of force. The potential energy of theparticle is Kr2 where K is constant and r is the distance of the particle from the centre of the field. Find themass of the particle if the minimum distance of the pat the farthest point from the centre of the field is vr.Two particles of mass m., and m2 are moving rectilinshown in figure. The distance between the path followed by the two particles is L. Find the totral angularmomentum of the particle relative to the points 1 and 2.

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A thin uniform rod AB of mass m and length L is held in a vertical position above a frictionless horizontalfloor, with its lower end A touching the floor, Th stick is released with a negligible horizontal push so that itstarts falling'. Since the floor is frictionless, energy is conserved. Further, the absence of any horizontalfriction force on the rod ensures that its center of mass C follows a straight line path vertically down. Findthe speed of points A and C at the moment when the rod makes an angle of 30o with the floor'.In the figure shown, the uniform square plate, lying in a vertical plane, has mass m and side L. Initially,0 = 37o. The plate is released in this position. Find the normal reaction between the frictionless floor an thelower corner of the plate, just afier its release.

In the figure shown, each tiny ball has a mass = m. The balls are rigidly fixed to the massless rod of length.The bails + roC systeril is held horizontally and released from some height so that the center of mass fallsthrough a height H before the rod hits the rigid knife edge elastically, as shown. The dfstance of the ,knifeedge hit point' from the center of the rod is U4.(A) Find the ratio (r t) of the speeds of the two balts just after the coilision. X i I(B) Find the minimum value of H if the balls A ancl B, and the rod come back for a second collision with theknife edge in exactly the sanr position as the one shown.

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IU II1lba-"t1--iju'ff r I (_L-)

Answer of Q.l1 to 20(81 m=inn,., A=9m/s2 3. (A) A=fel ,,=+(c) Nv=#

taTv{H{->{)1

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14. (A) 3 mg (B) vo17. L,=1t,,(m,,v.,-m"v2) - Lmrvrlk ; Lz= Ilz(mzvz- mrvr)- Lm,v.,]k

VA=

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16. ^=ry18. rr-L@"" -Z! 26

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