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Physics NEET MCQ
Centre Of Mass And Rotational Motion. Mcq
Quiz 3
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Q.1
Centrifugal force is inertial force when considered by
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a) an observer at the centre of circular motion
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b) an outside observer
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c)an observer who is moving with the particle which experiencing the force
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d)none of the above
Explanation
Answer: (c)
Q.2
A body is revolving with a constant speed along a circular path. If the direction of velocity is reversed, keeping speed unchanged, then
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a) Centripetal force suffer change in magnitude but not direction
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b) The centripetal force does not suffer any change in direction and magnitude
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c)centripetal force reverses in direction
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d)centripetal force is doubled
Explanation
Answer:(b)
Q.3
A disc of mass M and radius R is rolling with angular speed ω on a horizontal plane as shown in figure. The magnitude of angular momentum of the disc about origin is [ IIT 1999]
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a) ½ MR2ω
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b) MR2ω
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c) (3/2) MR2 ω
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d) 2 MR2ω
Explanation
The disc has two types of motion namely translational and rotational. therefore there are two types of angular momentum and the total angular momentum is the vector sum of these two In the case both the angular momentum have the same direction ( perpendicular to the plane of paper and away from the paper ) L=LT + LR Here LT=angular momentum due to translational motion LR=angular momentum due to rotational motion about C.M L=MV × R + Icmω Here Icm=M.I. about centre of mass C also v=ωR L=M( Rω)R + ½ MR2ω Icm=(3/2)MR2ω Answer: (c)
Q.4
A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity v. It collides elastically and head on with an identical sphere b at rest. Neglect friction everywhere. After collision, their angular speeds are ωA and ωB respectively. Then [ IIT 1999]
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a)ωA < ωB
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b) ωA=ωB
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c)ωA=ω
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d)ωB < ω
Explanation
As spheres are smooth there will be no friction ( no torque) and therefore there will be no transfer of angular momentum. Thus A, after collision will remain with its initial angular momentum ωA=ωAnswer: (c)
Q.5
A cubical block of side L rests on a rough horizontal surface with coefficient of friction µ. A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is [ IIT 2000]
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a) infinitesimal
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b) mg/4
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c)mg/2
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d)mg( 1 - µ)
Explanation
The applied force shifts the normal reaction to one corner as shown. At this situation, the cubical block starts topping about O. Taking torque about O F × L=mg × (L/2) F=mg/2 Answer: (c)
Q.6
A horizontal rod has a bead which can slide along its length and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration α. If the coefficient of friction between the rod and the bead is µ, and gravity is neglected, then the time after which the bead starts slipping is [ IIT 2000]
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a) √ ( µ/α)
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b) µ / √α
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c)1 / √(µα)
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d)infinitesimal
Explanation
When we are given an angular acceleration to the rod, the bead is also having an instantaneous tangential acceleration a= Lα and centrifugal acceleration ω2L this will happen when force is exerted on the bead by the rod. The bead has a tendency to move away from the centre. But due to the friction between the bead and rod, this does not happen to extent to which frictional force is capable of holding the beadThe frictional force here provides the necessary centripetal force. If instantaneous angular velocity is ω then Centripetal force=mLω2 Frictional force=µ(ma) mLω2=µ(ma) ∴ ω2=µα By applying ω=ω0 + αt We get ω=αt ( since ω=0 ) ∴ α2t2=µα ∴ t=√ ( µ / α) Answer:(a)
Q.7
A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved? [ IIT 2003]
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a) centre of the circle
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b) on the circumference of the circle
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c) inside the circle
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d) outside the circle
Explanation
The net force acting on particle undergoing uniform circular motion is centripetal force which always passes through the centre of the circle. The torque due to this force about the centre is zero Answer: (a)
Q.8
A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX' is .. [ IIT 2000]
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a)(ρL3)/ (8π2)
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b) (ρL3)/ (16π2)
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c)(5ρL3) / (16π2)
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d)(3ρL3) / (8π2)
Explanation
Moment of inertia about the diameter of the circular loop (ring)=½ MR2Using parallel axis theorem The moment of inertia of the loop about XX' axis is IXX'=MR2 / 2 + MR2 Where M=mass of the loop and R=radius of loop Here M=Lρ And R=L/2π ∴Answer: (d)
Q.9
A homogeneous disc of mass 2kg and radius 15cm is rotating about its axis ( which is fixed) with an angular velocity of 4 radian/sec with an angular velocity of 4 radian/sec. The linear momentum of the disc is ..[ CPMT 1989]
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a) 1.2 kg m/s
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b) 1.0 kg m/s
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c)0.6 kg m/s
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d)none of the above
Explanation
Disc does not have translatory motion so linear momentum is zero Answer:(d)
Q.10
An equilateral 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. Neglecting frictional effects, the quantities that are conserved as the beads slide down are [ IIT 2000]
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a) angular velocity and total energy ( kinetic and potential)
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b) total angular momentum and total energy
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c)angular velocity and moment of inertia about axis of rotation
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d)total angular momentum and moment of inertia about axis of rotation
Explanation
the M.I. about axis of rotation is not constant as the perpendicular distance of the bead with the axis of rotation increasesAlso since no external torque is acting ∴ τext=dL/dt L constant and Iω constant Since I increases, ω decreasesSince no force is acting mechanical energy is conserved Answer: (b)
Q.11
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 centre of the original to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is .. [ IIT 2001]
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a) ½ MR2
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b) ¼ MR2
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c)(1/8) MR2
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d)√2 MR2
Explanation
We cannot consider the quadrant as a single mass as the distance of different particles is different from the axis of rotation. So, we take the help of calculus. Let us consider a segment as shown in figure. All masses lying in this segment are at a distance r. from the axis and hence considered as small differential element d. Let the thickness of the segment be dr.The mass per unit area of quadrant Area of segment=¼ ( Circumference of circle )× drArea off segment=¼ (2πr) × dr=πrdr/2 ∴ Mass of segment dm=∴ Moment of Inertia of this segment about axis=dm×r2 Moment of inertia of the quadrant about axis=∫ dIALTERNATE SOLUTIONThe mass distribution of this sector is same about the axis of rotation as that of the complete disc about the axis. Therefore the formula remains the same as that of the disc Answer:(a)
Q.12
Consider a body, shown in figure, consists of two identical balls, each of mass M connected by a rigid rod. If an impels J=MV is imparted to the body at one of its ends, What would be its angular velocity [ IIT 2003]
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a) V/L
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b) 2V/L
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c) V/3L
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d) V/4L
Explanation
Change in angular momentum of the system=angular impulse given to the system of mass about centre of mass Final angular momentum - Initial angular momentum=MV × (L/2) --eq(1) Here centre of mass is at L/2 Let the system rotates with angular velocity ω Moment of inertia of the system about its axis of rotation [ centre of mass of the system] final Moment of inertia I=M ( L/2) 2 + M ( L/2) 2 I=ML2 / 2 --eq(2) from eq(1) Iω - 0=MV( L/2) ∴ ω=MVL/2I ---eq(3) On substituting value of I from eq(2)in eq(3) we get we get ω V/L Answer: (a)
Q.13
A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity ωA man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is K initially, its final kinetic energy will be [ IIT 2004]
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a)2K
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b) K/2
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c)K
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d)K/4
Explanation
The angular momentum is conserved, since τext is zero Here Ii=I0 and If=2I0 ωi=ω0 and ωf=ω (say)Then I0 ω0=2I0 ω∴ ω=ω0 / 2 Ki=½ I0 ω20=K --eq(1)Kf=½ (2I0) × (ω20 /4 ) --eq(2)Taking the ratio of eq(1) and eq(2)Kf=K/2Answer: (b)
Q.14
A disc is rolling without slipping with angular velocity ω. P and Q are two points equidistant from the centre C. The order of magnitude of velocity is [ IIT 2004]
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a) vQ > vC > vP
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b)vP > vC > vQ
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c)vP=vC , vQ=vC /2
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d)vP < vC > vQ
Explanation
In the given problem points will have two velocity linear velocity and Velocity tangential Now centre do not have any rotational motion thus velocity of C=vCAs shown in figureWhile Point P will have Two velocity linear VC and down direction v'=rω angle between two velocity is 90° Thus vP=[v2C + ω2r2]1/2 vQ=(v2C + ω2r2 + 2vCωrcosθ)1/2Answer: (b)
Q.15
From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is [ IIT 2005]
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a)4MR2
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b) (40/9)MR2
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c)10 MR2
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d)(37/9)MR2
Explanation
Let σ be the mass per unit area Total mass of disc=σ × πR2=9M The mass of circular disc cut =σ × π(R/3)2=σ × π(R)2 / 9 =M Let us consider the above system as complete disc of mass 9M and a negative mass M super imposed on itMoment of inertia of disc I1 of complete disc=(9/2)MR2 about axis passing through centre mass O and perpendicular to the plane of discM.I. of the cut out portion about an axis passing through O' and perpendicular to the plane of disc=½ × M × (R/3)2 ∴ M.I. I2 of the cutout portion about an axis passing through O and perpendicular to the plane of disc by using perpendicular axis theorem=Now I=I1 - I2 Answer:(a)
Q.16
A particle is confined to rotate in a circular path decreasing linear speed, then which of the following is correct [ IIT 2005]
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a) L angular momentum is conserved about the centre
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b) only direction of angular momentum L is conserved
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c) it spirals towards the centre
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d) its acceleration is towards the centre
Explanation
Since v is decreasing or changing, L is not conserved. Since it is given that a particle is confined to rotate in a circular path, it can not have spiral path. Since the particle has two acceleration ac and at therefore the net acceleration is not towards the centre. The direction of L remains same even when the speed decreases Answer: (b)
Q.17
A solid sphere of mass M and radius R having moment of inertia I about its diameter is recast into a solid disc of radius r and thickness t. The moment of inertia of the disc about an axis passing the edge and perpendicular to the plane remains I. Then R and r are related as [ IIT 2006]
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a)r=√(2/15) × R
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b) r=(2/√15) × R
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c)r=(2/15) R
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d)r=(√2 / 15)√R
Explanation
For solid sphere Given I=(2/5) MR2 --eq(1) For solid discAs per parallel axis theorem Moment of inertia about axis passing through edge of the disc I'=½Mr2 + Mr2=(3/2) Mr2 --eq(2) Given that Moment of inertia of sphere=moment of inertia of disc about axis passing through edge Thus (2/5)MR2=(3/2)Mr2∴ r=(2/√15)RAnswer: (b)
Q.18
A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up maximum height of 3v2 / 4g with respect to the initial position. The object is [ IIT 2007]
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a) ring
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b) solid sphere
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c)hollow sphere
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d)disc
Explanation
By the concept of energy conservationLiner kinetic energy + Rotational kinetic energy=potential energy ½ m v2 + ½ I ω2=mg ( 3v2/4g)For rolling v=ωR ∴ ½ m v2 + ½ I (v/R)2=mg ( 3v2/4g)on solving equation for we get I=½ m R2 This formula is for the momentum of inertia of discAnswer: (d)
Q.19
A body rolls down an inclined plane. If its kinetic energy of rotational motion is 40% of its kinetic energy of translational motion, then the body is a
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a)ring
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b) cylinder
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c)spherical shell
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d)solid sphere
Explanation
In case of solid sphere K.E=½ Iω2 but I=(2/5) m r2Substituting value of I in equation for energy we get K.E=(1/5) mv2 which is 40% of kinetic energy ½m v2Answer: (d)
Q.20
A bob of mass M is suspended by the massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach the point B. The angle θ t which the speed of the bob is half of that at A, satisfies.. [ IIT 2008]
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a)θ=π/4
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b) π/4 < θ < π/2
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c)π/2 < (3/4)π
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d)(3/4)π < θ < π
Explanation
This is the case of vertical motion when the body just completes the circle. Here v=√(5gL) --eq(1) Let MA=height of bob=hand MO=L - hApplying energy conservation Total energy at A=Total energy at P ½ m v2=½ m (v/2) 2 + mgh v2=(v/2)2 + 2gh (3/4)v2=2gh h=3v2/8g h=(3/8g) × (5gL)=15L/8 --eq(2) In ΔOPM ∴ θ = 5π/6 > 3π/4 Answer:(d)
Q.21
Look at the drawing given in figure which has been drawn with ink of uniform line-thickness. the mass of ink used to draw each of the two inner circles and each of the two line segment is m. The mass of the ink used to draw the outer circle is 6m. The coordinates of the centres of the different parts are : Outer (0, 0), left circle (-a, a), right inner circle (a, a), vertical line (0,0) and horizontal line (0, -a). The y-coordinate of the centre of mass of the ink in this drawing is[ IIT 2009]
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a) a/10
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b) a/8
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c) a/12
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d) a/3
Explanation
System is made up of five bodies, there circle and two straight lines. Mass distribution is uniform . Thus centre of mass of individual body will at there centre mass of big circle m1=6m, CM of circle=r1=(0, 0) mass of small circle on left m2 , CM of circle r2=(-a, +a) mass of small circle on right m3 , CM of circle r3=(+a, +a) mass of vertical line m4 , CM of vertical line r3=(0, 0) mass of horizontal line m5 , CM of horizontal line r3=(0, -a) The y-coordinate of centre of mass is Answer: (a)
Q.22
Moment of inertia of a circular wire of mass M and radius R about its diameter is [ AIEEE 2002]
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a)MR2 / 2
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b) MR2
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c)2MR2
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d)MR2/4
Explanation
M.I. of a circular wire about an axis nn' passing through the centre of the circle and perpendicular to the plane of the circle=MR2As shown in figure X-axis and Y-axis lie in the plane of the ring. Then by perpendicular axis theorem IX + IY=IZAs IX=IY by symmetry and IZ=MR2 we get 2IX=MR2 ∴ IX=½ MR2Answer: (a)
Q.23
Two identical particles move towards each other with velocity 2v and v respectively. The velocity of centre of mass is [ AIEEE 2002]
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a) v
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b) v/3
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c)v/2
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d)zero
Explanation
If we take 2v as positive velocity then second particle velocity will be -v The velocity of centre of mass of to particle system is given by Here m1=m2=m v1=2v and v2=-v on substituting above values we get vc=v/2Answer: (c)
Q.24
A solid sphere, a hollow sphere and a ring are released from top of an inclined plane ( frictionless, so that they slide down the plane). Then maximum acceleration down the plane is for [ AIEEE 2002]
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a)solid sphere
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b) hollow sphere
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c)ring
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d)all same
Explanation
The plane is being frictionless and therefore it is a case of sliding. Acceleration of all bodies is same gsinθ Answer:(d)
Q.25
The minimum velocity(in m/s) with which a car driver must traverse a flat curve of radius 150m and coefficient of friction 0.6 to avoid skidding is [ AIEEE 2002]
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a) 60
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b) 30
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c) 15
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d) 25
Explanation
For negotiating a circular curve on a leveled road, the maximum velocity of the car is vmax=√(µrg) Here µ=0.6, r=150m, g=9.8 ∴ vmax=√ (0.6 × 150 × 10)=30m/s Answer: (b)
Q.26
Four point masses, each of value m, are placed at the corners of square ABCD of side l. The moment of inertia of this system about an axis passing through A and parallel to BD. [ AIEEE 2006]
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a)2ml2
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b) √3 (ml2)
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c)3ml2
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d)ml2
Explanation
Moment of inertia about axis n'n due to point mass at B=m × perpendicular distance between point axis n'n=m(l/√2)2 Moment of inertia about axis n'n due to point mass at D=m(l/√2)2 Moment of inertia about axis n'n due to point mass at C=m(√2 l)2 Thus total moment of inertia about axis n'n=2×m( l/√2)2 + m(√2 l)2 Total momentum about n'n=3ml2 Answer: (c)
Q.27
Angular moment of the particle rotating with a central force is constant due to [ AIEEE 2007]
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a) constant torque
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b) constant force
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c)constant linear momentum
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d)zero torque
Explanation
We know that τ c=dLc / dt Where τc=torque about the centre of mass of the bodyand Lc=Angular momentum about the centre of mass of the body.Central forces act along the centre of mass . Therefore torque about centre of mass is zero Hence Angular momentum is constant Answer: (d)
Q.28
Consider a uniform square plate of side 'a' and mass 'm'. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of the axis perpendicular to its plane and passing through one of its corners is [ AIEEE 2008]
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a)(5/6) m a2
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b) (1/12)ma2
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c)(7/12)ma2
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d)(2/3)ma2
Explanation
Moment of inertia of rectangular plate is given by formula Inn'=(1/12) M ( l2 + b2) Since plate is square with side a, thus l=and b=a We get I=Ma2 ) /6Distance between nn’ and mm’ is a/√2 According to parallel axis theorem Imm' =Inn' + M(a/√2)2 Imm'=Ma2 / 6 + Ma2/ 2 Imm'=(2/3) Ma2 Answer:(d)
Q.29
Consider a two particle system with particles having masses m1 and mIf the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle is moved, so as to keep the centre of mass at the same position [ AIEEE 2006]
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a) (m2 / m1 ) d
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b) [m1 /(m1 + m2)] d
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c) (m1 / m2) d
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d) d
Explanation
Let the centre of mass be at origin position of m1 be - r1 position of mass m2 be r2 Then according to formula for centre of mass Finally after pushing the m1 towards centre by distance d. The centre of mass is at the origin. Let mass m2 be pushed by distance d' From equation (1) we get m1d=m2d' thus d'=(m1 / m2)d Answer: (c)
Q.30
The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is [ AIEEE 2005]
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a)(2/5)Mr2
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b) (1/4)Mr
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c)(1/2)Mr2
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d)Mr2
Explanation
The disc may be assumed as combination of two semicircular partsLet I be the moment of inertia of the uniform semicircular disc ∴ Moment of inertia of disc=2I=Mr2Thus half part moment of inertia I=Mr2 / 2Answer: (c)
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