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Physics NEET MCQ
Centre Of Mass And Rotational Motion. Mcq
Quiz 6
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Q.1
railway tracks are banked on curves so that
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a)no frictional force may be produced between the track and wheels
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b) the train may not fall down inward
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c)necessary centripetal force may be obtained from the horizontal component of normal reaction due to the track
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d)none of the above
Explanation
Answer: (c)
Q.2
A motor cyclist rides on cylindrical wall of velodrome ( hollow metal cylinder kept with the axis vertical) of radius r rotating about its axis at the rate f rev/sec. the least coefficient of friction necessary so that he does not step down is
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a) g / (4π2f2r)
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b) (4π2f2r) / r
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c)(4π2f2) / gr
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d)gr/(4π2r2)
Explanation
From figure it is clear that N=mrω2 µN=mgµ mrω2=mg µ=g/(rω2) but ω=2πf ∴ µ=g / (4π2f2r)Answer: (a)
Q.3
If a particle of mass 'm' is moving in a horizontal circle of radius 'r' with a centripetal force ( -1/r2) the total energy is [ JIPMER 1998]
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a)-1/2r
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b) -1/r
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c)-2/r
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d)-4/r
Explanation
Given mv2 / r=+1/r2 mv2=1/r Kinetic energy ½ mv2=1/2r Potential energy=- ∫ F dr P.E=∫ F dr=∫ (1/r2)dr=-1/r Total energy=Kinetic energy + potential energy Total energy=1/2r - 1/r=-1/2r Answer:(a)
Q.4
A particle of mass m is executing uniform circular motion on a path of radius r. If p is the magnitude of its linear momentum, the radial force acting on the particle is : [ MPPMT 1994]
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a) pmr
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b) rm/p
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c) mp2 /r
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d) p2 / rm
Explanation
F=mv2 /r F=m2 v2 / rm F=p2 / mr Answer: (d)
Q.5
A body of mass m hangs at one end of string of length l, the other end of which is fixed. It is given a horizontal velocity so that the string would just become slack, when it makes an angle of 60° with the upward drawn vertical. The tension in the string at this position is [ ISM Dhanbad 1994]
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a) 4.5 mg
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b) mg
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c)3m
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d)√3 mg
Explanation
The string just becomes slack at B ∴ TB=0 from figure By Energy conservation at A and B Therefore tension at A=mg + 3.5mg=4.5mgAnswer: (a)
Q.6
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a is varying with time t, a=k2rt2 where k is a constant. The power delivered to the particle by the forces acting on it is [ IIT 1994]
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a) 2πmk2r2t
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b) mk2r2t
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c)(mk4r2t5) / 3
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d)zero
Explanation
Given a=k2rt2=v2 / r∴ v2=k2r2t2Kinetic energy K=½ m v2 By substituting value of v2 from above we get K=½ mk2r2t2According ro work energy theorem W=ΔK W=½ mk2r2t2 - 0W=½ mk2r2t2Power P=dW/dt, taking derivative of W with respect to time P=½ mk2r2 (2t)P=mk2r2 tAnswer: (b)
Q.7
The tension in the string revolving in a vertical circle with a mass m at the end which is .. [ EAMCET 1995]
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a) mv2 / r
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b) (mv2 / r) - mg
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c)(mv2 / r) + mg
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d)mg
Explanation
Centrifugal and gravitational force both are down ward direction at lower position Answer:(c)
Q.8
A pendulum is suspended from the ceiling of a car rotating in ac circular path with an acceleration of 0.49m/sThe angle which the string makes with the vertical is
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a) tan-1 ( 1/√3)
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b) tan-1√3
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c) tan-1 0.05
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d) tan-1 0.005
Explanation
from figure tanθ=a/g θ=tan-1 0.05 Answer: (c)
Q.9
The string of pendulum of length l is displaced through 90° from the vertical and released. then the minimum strength of the string in order to withstand the tension, as the pendulum passes through the mean position is [ MP 1986]
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a)mg
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b) 3mg
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c)5mg
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d)6mg
Explanation
To reach the point of suspension ( 90° from vertical) From law of conservation of energy ½ mv2=mgh or v2=2gl tension at the lowest point T=mg + mv2 / l T=mg + 2mg=3mgAnswer: (b)
Q.10
A stone of mass 1kg tied to a light inextensible string of length L=(10/3) m is whirling in a circular path of radius L in a vertical plane If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if g is taken to be 10 m/s2 the speed of the stone at the highest point of the circle is [ CPMT 1998]
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a) 20 m/s
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b) 10√3 m/s
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c)5√2 m/s
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d)10 m/s
Explanation
maximum tension will be at lower point A Minimum tension will be at highest point B By taking the ratio of max. tension to min. tension we get According to law of conservation of energy energy at A = energy at B we get V2A = 4gL- V2A On solving both equations we get V2B=3gL Velocity at highest point=[3 × 10 × (10/3)]1/2=10 m/sAnswer: (d)
Q.11
A heavy small sized sphere is suspended by a string of length l. the sphere rotates uniformly in horizontal circle with the string making angle θ with the vertical. Then the time period of conical pendulum is
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a)
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b)
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c)
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d)
Explanation
T cosθ=mg T sinθ=mrω2Taking ratio we get from figure r=lsinθ Answer:(c)
Q.12
A piece of stone is projected at angle θ with velocity u. If it executes nearly a circular motion at its maximum point for short time, the radius of the circle will be [CPMT]
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a) u2 / g
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b) (u2cos2θ) / g
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c) (u2sin2θ) / g
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d) (u2cos2θ) / 2g
Explanation
The velocity at maximum point be ucosθ ( only horizontal component ), the centripetal acceleration is given by a=v2 / r ∴ g=(u2cos2θ) / rr=(u2cos2θ) / g Answer: (b)
Q.13
A block of mass M is tied to one end of mass less rope. The other end of the rope is in the hands of a man of mass 2M as shown in figure. The block and man are resting on a rough wedge of mass M as shown in figure. The whole system is resting on smooth horizontal surface. The man pulls the rope. Pulley is mass less and frictionless. What is the displacement of the wedge when the block meets the pulley ( man does not leave his position during the pull)
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a)0.5 m
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b) 1 m
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c)zero
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d)23 m
Explanation
Since there no external force acting on the system and initially the centre of mass of the system was at rest, hence the centre of mass will not move Let d be displacement of wedgecentre of mass at initial position from diagram centre of mass at final position When block touches the pulley and displacement of wedge from diagram Since centre of mass should not move as no external force is acting (Xcm)i=(Xcm)f∴ d=(1/2) m=0.5 mAnswer: (a)
Q.14
Two meteorites in free space are 8 m apart. Their masses are 7kg and 9kg. Then their centre of mass which lies on the line joining the centre of two bodies is
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a) At 4 meters from each meteorites
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b) At 3.5 meter from meteorites with a mass of 9kg
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c)At 3.5 meter from the meteorites with mass 7 kg
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d)At 8 meters from one end and 16 meter from the other end
Explanation
let x be the distance of centre of mass from 7kg meteorite then 7x=9(8 - x) 7x=72 - 9x16x=72 x=72/16=4.5 from 7 kg or 8 - 4.5=3.5 m from 9 kgAnswer: (b)
Q.15
Two point masses m1 and m2 ( m >m2) attract each other with force which is inversely proportional to the square of the distance between them. The particles which are initially at rest, when released have centre of mass of the system moves
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a)moves towards m1
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b) moves towards m2
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c)move at right angles to the line joining m1 and m2
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d)does not move at all
Explanation
Since no external force acts on system centre of mass will not move Answer:(d)
Q.16
A heavy ball is thrown on rough horizontal surface in such a way that its slides with a speed Vo initially without rolling. It will roll without sliding when its speed falls to
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a) (2/7)Vo
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b) (3/7)Vo
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c) (5/7)Vo
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d) (7/5)Vo
Explanation
Retardation due to surface=a=F/m=µmg / m=µg ∴ Velocity after t sec, when sliding converted to rolling is V(t)=Vo - µgt --(i)Now, in case of rolling τ=µmg r ( here r is the radius of ball ) also τ=Iα Thus µmg r=Iα For sold sphere I=(2/5) mr2 Pure rolling starts when V(t)=rω(t) ∴ From equation (2) µgt=(2/5)V(t) Putting the value of µgt in equation (i) we get V(t)=Vo - (2/5)V(t) V(t) [ 1 + (2/5)]=Vo V(t)=(5/7)VoAnswer: (c)
Q.17
The centre of mass of a systems of two particles is
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a)on the line joining them and midway between them
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b) on the line joining them at a point whose distance from each particle is proportional to the square of the mass of that particle.
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c)on the line joining them at a point whose distance from each particle inversely proportional to the mass of that particle.
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d)On the line joining them at a point whose distance from each particle is proportional to the mass of that particle.
Explanation
Answer: (c)
Q.18
A solid sphere is rotating about a diameter due to increase in room temperature, its volume increases by 0.5%, If no external torque acts. The angular speed of the sphere will.
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a)increase by nearly 1/3 %
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b)decrease by nearly 1/3 %
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c)increase by nearly (1/2 %
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d)decrease nearly by (1/2) %
Explanation
Volume V ∝ r3 --eq(1) Thus dV ∝3r2 --eq(2) dividing eq(2) by eq(1) we get dV/V=3 dr/r given dV/V=0.5% Thus dr/r=(1/3) × 0.5=1/6%No external torque so Iω=constantBut for sphere I=(2/5) mr2 (2/5) mr2ω=constant ∴ r2ω ∝ constant Thus r2 ∝ ω-1 --eq(3) taking derivative 2rdr ∝ (-1)ω-2 --eq(4) dividing eq(4) by eq(3) we get 2dr/r=- dω/ ω dω/ ω=- (2) × (1/6)=-1/3% Answer: (b)
Q.19
A circular disc of radius R is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through an angle of 60? and released. Its angularvelocity when it reaches the equilibrium position will be__
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a)
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b)
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c)
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d)
Explanation
Cenre of mass of circular disc will be at it centre. at a distance R from axis of rotation at rim Potential energy of the disc at 60°=mgR(1-cos60)=mgR/2 At equilibrium point kinetic energy=½ I ω2 By using parallel axis theorem I=½ MR2 + MR2 I=(3/2) MR2 Now P.E=kinetic energy Answer:(b)
Q.20
The moment of inertia of a hollow sphere of mass M and inner and outer radii R and 2R about the axis passing through its centre and perpendicular to its plane is
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a)
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b)
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c)
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d)
Explanation
We can obtain hollow sphere as it solid sphere of radius R is removed from a solid sphereof 2R mass of hollow sphere M=M1 – M2If ρ is the density then Moment of inertia of hollow sphere Answer: (d)
Q.21
Two strings of length l = 0.5 m each are connected to a block of a mass m = 2 kg at one end and their ends are attached to the point A and B, 0.5 m apart on a vertical pole which rotates with a constant angular velocity ω = 7 rad/sec. Find ratio T1 / T2 . Tension in the upper sting is (T1) and tension in lower string is (T2) [ use g = 9.8 m/s2]
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a) 1
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b) 3
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c) 9
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d) 6
Explanation
From resolution of vectors as shown in figure radius r = 0.5cosθ T1cosθ + T2cosθ = mω2( 0.5cosθ) T1 + T2 = 0.5 × m × ω2 --eq(1) And T1sinθ = T2sinθ + mg since θ = 30 from geometry of figure T1 = T2 + 2mg -- eq(2) From substituting value of T1 from eq(2) in eq(1) we get T2 + 2mg + T2 = 0.5 × m × ω2 2T2 = 0.5 × m × ω2 - 2mg 2T2 = 0.5 × 2 × 72 - 2 × 2 × 9.8 2T2 = 49 - 39.2 = 9.8 T2 = 4.9 From equation (2) T1 = 4.9 + 2 × 2 × 9.8 = 44.1 T1 / T2 = 44.1/4.9= 9 Answer: (c)
Q.22
From a uniform circular disc of radius R, a circular disc of radius R/6 and having centre at a distance +R/2 from the centre of the disc is removed. Determine the centre of mass of remaining
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a) -R/70
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b) +R/70
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c) -R/7
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d) +R/7
Explanation
Let us consider the disc is made up of two parts small and remaining part Then centre of mass of the disc is at (0,0,0) Thus (centre of mass of small disc) × ( mass of the small disc) = - (centre of remaining part of disc) × ( mass of remaining part) --eq(1) Let ρ be the mass per unit area then mass of small disc = m Mass of remaining part = M Let centre of mass of remaining part be at x Substituting values in equation (1) we get Answer: (a)
Q.23
A wheel is rotating at 900 rpm about its axis. When power is cut off it comes to rest in 1 minute, the angular retardation in rad / sec is ___
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a) π/2
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b) π/4
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c) π/6
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d) π/ 8
Explanation
900 rpm = 900/60 = 15 rotation per second = 30π rad/sec Form the formula ω = ωo - αt 0 = 30π - α × 60 α = π/2 Answer:(a)
Q.24
A thin uniform rod AB of mass M and length L is hinged at one end A to the horizontal floor. Initially it stands vertically and becomes horizontal. It is allowed to fall freely on the floor in the vertical plane. The angular velocity of the rod when its end B strikes the floor is ____
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a) √(g/l)
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b) √(2g/l)
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c) √(3g/l)
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d) 2√(g/l)
Explanation
Loss in P.E = gain in Rotational K.E. centre of Mass of rod is at L/2 Potential energy of rod = MgL/2 Kinetic energy = ½ I ω2 Moment of inertia of rod for axis of rotation at the end = ML2 / 3 thus we get Answer: (c)
Q.25
A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter42 cm. is removed from +ve x edge of the plate. Find the position of centre of mass of theremaining portion with respect to centre of mass of whole plate.
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a)- 7 cm
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b)+ 9 cm
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c)- 9 cm
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d)+ 7 cm
Explanation
Let the centre of circular plate before cutting smaller plate be at (0,0) From figure it is clear that centre of mass of smaller circular plate is at 7 cm Let ρ be the surface mass density of the plate Mass of smaller plate=π(21)2ρ Mass of bigger pate=π[(28)2 - (21) 2]ρ Mass of bigger pate=π[(49)(7)]ρ From definition of cetre of mass Thus π(21)2ρ × 7=- π[(49)(7)]ρ × x 212=-72 × x x=-9cm Answer: (c)
Q.26
A particle performing uniform circular motion has angular momentum L., its angular frequency is doubled and its K.E. halved, then the new angular momentum is
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a) L/2
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b)L/4
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c)2L
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d)4L
Explanation
Kinetic energy E=½ Iω2 E=½ Iω × ω But Iω= L E=½ Lω L=2E/ω for given L'=2(E/2)/ 2ω L'=L/4Answer: (b)
Q.27
A solid sphere is rotating about a diameter at an angular velocity ω. if it cools so that its radius reduces to 1/n of its original value. Its angular velocity becomes_____
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a) ω/n
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b) ω/n2
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c)nω
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d)n2ω
Explanation
No external torque acts on it thus angular moment is conserved Use formula I1ω1=I2ω2 And I=(2/5) mr2 Answer:(d)
Q.28
Two blocks of masses 10 kg an 4 kg are connected by a spring of negligible mass and placedon a frictionless horizontal surface. An impulse gives velocity of 14 m/s to the heavier block inthe direction of the lighter block. The velocity of the centre of mass is
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a) 30 m/s
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b) 20 m/s
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c) 10 m/s
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d) 5 m/s
Explanation
From the formula for centre of mass Answer: (c)
Q.29
A circular disc of mass m and radius r is rolling on a smooth horizontal surface with a constant speed v. Its kinetic energy is _____
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a)(1/4) mv2
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b) (1/2) mv2
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c)(3/4) mv2
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d) mv2
Explanation
Since disc is rolling it will have two types of energy liner kinetic and rotational kinetic energy E=½ mv2 + ½ Iω2 since it is rolling without sliding ω=v/R and I=(1/2) MR2 on substituting values of I and ω we get option "c"Answer: (c)
Q.30
A body of mass m is tied to one end of spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimeter. If the angularvelocity is doubted, the elongation in the spring is 5 cm. The original length of spring is...
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a)16 cm
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b)15 cm
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c)14 cm
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d)13 cm
Explanation
Let original length be L and spring constant be k Centrifugal force=force of spring mω2r=k × x here x is increase in length of spring case I: Elongation x=1 mω2 ( L + 1 )=k --eq(1) case II : elongation x=5 cm and angular speed in doubled 4mω2 ( L + 5 )=5k --eq(2) on solving equation (1) and (2) we get L=15 cm Answer: (b)
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