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Physics NEET MCQ
Quiz 9
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Q.1
A cord is wound round the circumference of the wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about the centre is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through distance h, angular velocity of the wheel is [ AFMC 1998]
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a)
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b)
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c)
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d)
Explanation
Loss of potential energy=gain in kinetic energyLoss of potential energy=mghGain of kinetic energy=linear kinetic energy + rotational kinetic energyGain of kinetic energy=½m v2 + ½I ω2∴ mgh=½m v2 + ½I ω2on simplification we getAnswer: (c)
Q.2
A body of 2kg mass is rotating on a circular path of radius 0.8m with an angular velocity 44 rad/sec. If radius of the path becomes 1 m then value of angular velocity will be...[ AFMC 1998]
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a) 35.28 rad/sec
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b)14.08 rad/sec
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c)28.16 rad/sec
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d)24.08 rad/sec
Explanation
No external force acts hence according to law of conservation of angular momentumI1ω1=I2ω2but I=mr2 Thus mr12ω1=mr22ω2 r12ω1=r22ω2 (0.8)244=(1)ω2 ∴ ω2=28.16 Answer:(c)
Q.3
A stone tied to one end of string of 1m long is whirled in a horizontal circle with a constant speed. if the stone makes 10 revolutions in 20 sec, the magnitude of the acceleration of the stone is ...[ AFMC 1998]
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a) 988 cm/s²
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b) 490 cm/s²
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c) 986 cm/s²
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d) 996 cm/s²
Explanation
stone makes 10 revolution per 20 sec or 1/2 revolution per second ∴ f=1/2 Now ω=2Πf=π Centripetal acceleration=ω2r Centripetal acceleration=ππ × 100 Centripetal acceleration=986 cm/s2 Answer: (c)
Q.4
The moment of inertia of a regular circular disc of mass 0.4 kg and radius 100cm about the axis perpendicular to the plane of the disc and passing through its centre is [ AFMC 2000]
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a)0.002 kg m2
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b) 0.02kg m2
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c)2kg m2
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d)0.2 kg m2
Explanation
Moment of inertia of disc about its axis=½m r2m=0.4 kg, m=100cm=1 mI=½(0.4)(1)2I=0.2 kg/m2Answer: (d)
Q.5
The angular velocity of second's hand in a watch is ..[AFMC 2000]
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a) 0.82 rad/sec
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b) 0.105 rad/sec
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c)0.21 rad/sec
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d)0.052 rad/sec
Explanation
For a second hand T=60 secω=2π /T ω=2π /60ω=0.105 radian/secAnswer: (b)
Q.6
Two particles of equal mass revolving in circular paths of radii r1 and r2 respectively with the same angular velocity. The ratio of their centripetal force will be ....[ AFMC 2011]
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a) r₁ /r₂
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b) r₂ / r₁
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c)(r₂ / r₁) 1/2
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d)(r₂ / r₁)²
Explanation
Centripetal force=mω2rfor first particle centripetal force=mω2r1for second particle centripetal force=mω2r2taking ratio of above two equations we getRatio=r1 /r2 Answer:(a)
Q.7
A body of mass 5 kg is moving in a circle of radius 1m with an angular velocity of 2 rad/sec. The centripetal force acting on the body is ...[ AFMC 2002]
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a) 20 N
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b) 10 N
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c) 5N
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d) 15 N
Explanation
Centripetal force=mω2r m=5, ω=2, r=1 m on substituting values in above equation we get ω=20N Answer: (a)
Q.8
The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is IIts moment of inertia about an axis passing through one of its ends and perpendicular to its length is [ CBSE-PMT 2011]
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a)I0 + ML2 /2
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b) I0 + ML2 /4
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c)I0 + 2ML2
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d)I0 + ML2
Explanation
According to theorem of parallel axesI=Icm + Md2I=I0 + M(L / 2) 2I=I0 + M L2 / 4 Answer: (b)
Q.9
Moment of inertia of a uniform circular disc about a diameter is I. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be [ CBSE-PMT 1990]
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a)5I
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b) 3I
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c)6I
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d)4I
Explanation
M.I. of uniform circular disc about its diameter=IM.I. of disc about its axis=½ ( m r2)According to Theorem of perpendicular axis Iz=Ix + I yM.I. of disc about its axis Iz=I + I=2IThus 2I=½ ( m r2)∴ mr2=¼ (m r2)∴ mr2=4I According to Theorems of parallel axis M. I. of disc about the given axis=2I + mr2M. I. of disc about the given axis=2I + 4I=6I Answer: (c)
Q.10
A particle of mass m=5 is moving with a uniform speed v=3√2 in XOY plane along the line y=x +The magnitude of the angular momentum of the particle about the origin is [ CBSE-PMT 1991]
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a) 60 units
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b) 40√2 units
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c) zero
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d)7.5 units
Explanation
y=x + 4 line has been shown in figure . Slope of the line=1∴ tanθ=1 thus ∠ OQP=45°and ∠ OPQ=45°If we draw a perpendicular to this line=ORFrom ΔOPR OR=OPsin45 andOR=4 × (1 / √2)=2√2Angular momentum of this particle going along this line=OR×mvAngular momentum of this particle=2√2 × 5 × 3√2=60 unitAnswer: (a)
Q.11
particle of mass m is moving in a horizontal circle of radius R with uniform speed v. When it moves from one point to a diametrically opposite point, its [1989]
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a) kinetic energy changes by mv2/4
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b) momentum does not change
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c) momentum changes by 2 mv
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d)kinetic energy changes by mv2
Explanation
At A, momentum=mv Energy=½ (m v2) At B, momentum=-mv Energy=½ m (-v)2 Change in momentum=mv - ( - mv)=2mv Change in energy=0 Answer:(c)
Q.12
The speed of a homogenous solid sphere after rolling down an inclined plane of vertical height h from rest without sliding is [ CBSE-PMT 1992]
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a)
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b)
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c)
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d)
Explanation
Total energy=Kinetic energy linear + Kinetic energy rolling E=½ (m v2 ) + ½ I ω2 Now ω=v/r and for solid sphere moment of inertia I=(2/5) m r2On substituting values of ω and I in above equation we getE=½ m v 2 + (1/5) m v2E=(7/10)m v2Now Potential energy=Total kinetic energymgh=(7/10)m v2 Answer: (a)
Q.13
The angular speed of an engine wheel making 90 revolutions per minute is [ CBSE-PMT 1995]
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a)1.5 π rad/s
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b) 3 π rad/s
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c)4.5 π rad/s
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d)6 π rad/s
Explanation
Number of revolution n=90 per minuteone revolution=2 π ∴ angular displacement=90 × 2πAngular velocity=angular displacement / time Angular velocity=90 × 2π / 60Angular velocity=3 × π rad/ secAnswer: (b)
Q.14
Two racing cars of masses m1 and m2 , are moving in circles of radii r1 and r2, respectively. Their speeds are such that each makes a complete circle in the same time t. The ratio of the angular speeds of the first to the second car is [ CBSE - PMT 1999]
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a) 1 : 1
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b) m1 : m2
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c)r1 ,r2
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d) m1, m2 : r1r2
Explanation
As time taken by both car to complete one revolution is same ω ∝ (1/T) thus T is same in the both case. Hence ω will be same Answer: (a)
Q.15
O is the centre of an equilateral triangle ABC. F1, F2 and F3 are three forces acting along the sides AB, BC and AC as shown here. What should be the magnitude of F3, so that the total torque about O is zero? [ CBSE - PMT 1998]
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a)(F₁ + F₂) /2
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b) 2(F₁ + F₂)
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c)(F₁ + F₂)
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d)(F₁ - F₂)
Explanation
Perpendicular distance of each line of action of force from the centre is same, say r. Now taking momentum about OF1 × r + F2 × r - F3 × r = 0 ∴ F3=F1 + F2 Answer:(c)
Q.16
A disc is rotating with angular velocity ω. If a child sits on it, what is conserved ? [ CBSE-PMT 2002]
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a) Linear momentum
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b) Angular momentum
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c) Kinetic energy
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d) Moment of inertia
Explanation
Weight of child acts downwards hence external torque is zero. If external torque is zero, angular momentum remains conserved Answer: (b)
Q.17
A fly wheel rotating about a fixed axis has a kinetic energy of 360 joule when its angular speed is 30 radian/sec. The moment of inertia of the wheel about the axis of rotation is [ CBSE-PMT 1990]
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b) 0.15kgm2
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c)0.8 kg m2
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d)0.75 kg m2
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a)0.6kg/m²
Explanation
Rotational Kinetic Energy Er=½ I ω 2I=2 Er / ω 2On substituting the values we get I=0.8 kg m2Answer: (c)
Q.18
A solid cylinder and a hollow cylinder both of the same mass and same external diameter, are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?[ CBSE-PMT 2000]
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a) Both together
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b) Solid cylinder
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c) One with higher density
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d)Hollow cylinder
Explanation
Formula for acceleration, rolling without slipping is now K for hollow cylinder is=R K for Solid Cylinder=R / √ 2 from above since K is small for solid Cylinder acceleration of Solid Cylinder is more than hollow cylinderusing, s=ut + ½ a t2s=½ a t2⇒ t ∝ ( 1 / √a)'t' minimum means 'a' should be more. Therefore, solid cylinder will reach the bottom first.Answer: (b)
Q.19
A composite disc is to be made using equal masses of aluminum and iron so that it has as high a moment of inertia as possible. This is possible when [ CBSE-PMT 2002]
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a)the surfaces of the discs are made of iron with aluminum inside
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b) the whole of aluminum is kept in the core and the iron at the outer rim of the disc
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c) the whole of the iron is kept in the core and the aluminum at the outer rim of the disc
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d)the whole disc is made with thin alternate sheets of iron and aluminum
Explanation
Density of iron is more than density of aluminum, thus mass of small element of Iron will be more than mass of same size small element of AluminumMoment of Inertia is give by=∫ r2 dm. So if Iron is at outer ring it will increase M.I because of its distance from the centre and the higher mass of element. Thus Whole of aluminum is kept in the core and the iron outer rim of the disc. Answer:(b)
Q.20
A spherical ball rolls on a table without slipping. Then the fraction of its total energy associated with rotation is [ CBSE-PMT 1994]
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a) 2/5
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b) 2/7
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c) 3/5
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d) 3/7
Explanation
Rotational Kinetic energy Er=½ (MK2 ω 2) Total Energy E=½ M ω2[ K2 + R2] Now for spherical ball K2=(2/5) R2on substituting value of K2 in above equation we get ratio=2/7 Answer: (b)
Q.21
ABC is a triangular plate of uniform thickness. The sides are in the ratio shown in the figure. IAB, IBC. and ICA are the moments of inertia of the plate about AB, BC and CA as axes respectively. Which one of the following relations is correct?[ CBSE- PMT 1995]
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a)IAB > IBC
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b) IBC > IAB
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c)IAB + IBC=ICA
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d)ICA is maximum
Explanation
Intersection of medians is the centre of mass of the triangle. Since distance of centre of mass from the sides are related as xBC > xAB > xAC, therefore IBC > IAB > IACAnswer: (b)
Q.22
A couple produces [ CBSE-PMT 1997]
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a) no motion
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b) purely linear motion
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c)purely rotational motion
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d)linear and rotational
Explanation
A couple is formed of two equal and opposite forces at some separation, so net force is zero. hence a couple does not produce translatory motionAnswer: (c)
Q.23
A wheel of radius 1m rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is [ CBSE-PMT 2002]
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d)
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a)π
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b)2 π
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c)π √2
Explanation
Linear distance moved by wheel in half revolution=π r.Point P after half revolution reaches at Q which is vertically 2m above the ground∴ Displacement PQ is Answer:(d)
Q.24
The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge of the disc and normal to the disc is : [ CBSE-PMT 2005]
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a) MR2
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b) (1/2) MR2
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c) (3/2)MR2
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d) (7/2) MR2
Explanation
Moment of Inertia of a uniform circular disc of radius 'R' and mass 'M' about an axis passing through C.M. and normal to the disc is ICM=(1/2) M R2 Now Moment of Inertia about axis which is tangential to disc is From parallel axis theorem IT=ICM + MR2 IT=(1/2)MR2 + MR2=(3/2) MR2 Answer: (c)
Q.25
A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity ω. Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be [2003]
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a)
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b)
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c)
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d)
Explanation
Applying law of conservation of angular momentum I1ω=I2 ω1 Now after placing four object of mass m each on the ring Moment of Inertia I2=Mr2 + 4(m)r2=(M + 4m)r2 ∴ ω1 Answer: (c)
Q.26
A small mass attached to a string rotates on frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will [ CBSE-PMT 2011]
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a) remain constant
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b) increase by a factor of 2
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c) increase by a factor of 4
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d) decrease by a factor of 2
Explanation
K.E = L2 / 2I The angular momentum L remains conserved about the centre That is L = constant Now I= mr2 ∴ K.E. in first case = K.E'. in second case = But r' = r/2 ∴ K.E' = ⇒ K.E' = 4 K.E. ∴ K.E. is increased by a factor of 4 Answer: (c)
Q.27
A gramophone record is revolving with an angular velocity ω. A coin is placed at a distance r from the centre of the record. The static coefficient of friction is µ The coin will revolve with the record if: [ CBSE-PMT 2010]
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a) r = µgω 2
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b) r < ω2 / µg
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c) r ≤ µg / ω2
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d) r ≥ µg / ω2
Explanation
For the coin to revolve with the record, centrifugal force ≤ frictional force ⇒ mrω2 ≤ µ mg ∴ r ≤ µg / ω2 Answer: (c)
Q.28
A circular disk of moment of inertia It is rotating in a horizontal plane, its symmetry axis, with a constant angular speed ωi Another disk of moment of inertia Ib is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ωf. The energy lost by the initially rotating disk to friction is: [ CBSE-PMT 2010]
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a)
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b)
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c)
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d)
Explanation
Here Initial Angular momentum= It ω i After placing another disc Final angular momentum is= (It + Ib ) ω f By conservation of angular momentum we get It ω i = (It + Ib ) ω f ∴ ωf = Now loss in K.E ΔK = Initial K.E - Final K.E Now by substituting the value of ωf in above equation we get Answer:(d)
Q.29
A solid sphere, disc and solid cylinder all of the same mass and made of the same material are allowed to roll down (from rest) on the inclined plane, then [1993]
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a) solid sphere reaches the bottom first
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b) solid sphere reaches the bottom last
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c) disc will reach the bottom first
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d) all reach the bottom at the same time
Explanation
For solid sphere For disc and solid cylinder Since acceleration inversely proportional to ratio of K2 / R2 for sphere it is less thus Solid sphere will take minimum time to reach the bottom of the incline Answer: (a)
Q.30
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is [2004]
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a) 1 : √2
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b) 1:3
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c)2:1
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d)√5: √6
Explanation
Moment of Inertia around Axis passing through centre of mass Iy=MR2 /4 Now according parallel axis theorem momentum of Inertia around the tangential axis is I'y= K12 = 5/4 For circular ringMoment of Inertia around Axis passing through centre of mass Iy1=MR2 /2 Now according parallel axis theorem momentum of Inertia around the tangential axis is I'y1= K22 = 3/2 Answer: (d)
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