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Quiz 1
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Q.1
The time period of a second's pendulum in a satellite is
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a) zero
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b) 2
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c) infinity
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d) depends on mass of body
Explanation
Answer:(c)
Q.2
If radius of the earth were to shrink by one percent, its mass remaining the same, the acceleration due to gravity on the earth's surface would [ IIT 1981]
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a) decrease
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b) remain unchanged
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c)increase
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d)be zero
Explanation
Radius decreases without changing mass thus g ∝ 1/R2 Thus g increases Answer: (c)
Q.3
The escape velocity on the surface of earth is 11.2 km/s. What would be the escape velocity on the surface of another planet of the same mass but 1/4 times the radius of the earth? [ CBSE-PMT 2000]
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a)22.4 km/s
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b) 44.8 km/s
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c) 5.6 km/s
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d)11.2 km/s
Explanation
Escape velocity of EarthEscape velocity of planetgiven Me=Mp Radius Re /4=RPOn substitution we getTaking the ratio of vp to ve we get∴ vp=2 × ve vp=2× 11.2=22.4 km/s Answer:(a)
Q.4
Assuming the radius of the earth as R, the change in gravitational potential energy of a body of mass m. when it is taken from the earth's surface to a height 3R above its surface, is [ CBSE-PMT 2002]
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a) 3mgR
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b) (3/4) mgR
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c) mgR
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d) (3/2)mgR
Explanation
Gravitational potential energy (GPE) on the surface of earth E1=-GMm / R gravitational potential at 3R, E2=-GMm / 4R ∴ change in Gravitational potential Answer: (b)
Q.5
A roller coaster is designed such that riders experience "weight lessens" as they go round the lop of a hill whose radius of curvature is 20 m. The speed of the car at the top of the hill is between: [ CBSE-PMT 2008]
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a)14 m/s and 15 m/s
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b) 15 m/s and 16 m/s
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c)16 m/s and 17 m/s
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d)13 m/s and 14 m/s
Explanation
At the top mg is down ward force while centrifugal is up wardsmg≤ mv2 / 2v≥ √(gr)v ≥ √(9.8)20 v ≥ 14 m/sAnswer: (a)
Q.6
If the gravitational force between two objects were proportional to 1/R (and not as 1/R2) where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to [ CBSE-PMT 1989]
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a) 1/R2
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b) R0
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c) R
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d)1/R
Explanation
Now F ∝1/R ∴ F=k/R Gravitational force is centripetal force for Circular motion thus k/F=Mv2 / R Hence v ∝ R0Answer: (b)
Q.7
The earth ( mass=6 × 1024kg) revolves around the sun with an angular velocity 2 × 10-7 rad/sec in a circular orbit of radius=1.5 × 108km. The force exerted by sun on earth in newton is ...[ AFMC 1997, 1999]
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a)36 × 1021
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b) 18 × 1025
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c)29 × 1039
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d)zero
Explanation
Force exerted by sun on the earth is centripetal force.Centripetal force=mω2rm=6 × 1024kgω=2 × 10-7 rad/secr=1.5 × 1011kmsubstituting values in above equation we getCentripetal force=6 × 1024 × (2 × 10-7)2 × 1.5 × 1011=36 × 1021Answer: (a)
Q.8
The distance of two planets from the sun are 1013 and 1012 meters respectively. The ratio of time periods of these two planets is [ CBSE PMT 1988]
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a) 1/√10
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b) 100
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c) 10√10
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d) √10
Explanation
According to Kepler's law T2 ∝R3Taking the ratio of periodic time we getAnswer: (c)
Q.9
The largest and the shortest distance of the earth from the sun are r1and rIts distance from the sun when it is at perpendicular to the major-axis of the orbit drawn from the sun is [1988]
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a) (r1 + r2)/4
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b) (r1 + r2)/ (r1 - r2)
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c) 2(r1 × r2)/ (r1 + r2)
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d) (r1 + r2)/ 3
Explanation
Planets move around sun in elliptical path, Applying the properties of ellipse we have F and S are focus point and sun is at one focus point Verticle line is semi-minor axis Horizontal line is semi-major axis Point E represents position of Earth FE is the distance of earth from other focus F denoted by "p" SE is perpendicular distance of earth from sun denoted by "q" Answer:(c)
Q.10
If the gravitational force between two objects were proportional to 1/R (and not as 1/R2) where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to [1989]
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a) 1/R2
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b) R0
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c) R
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d) 1/R
Explanation
Given that Gravitational force is proportional to 1/R Hence Gravitational force=GMm/R Formula for centripetal force=mv2/R From above equation we get mv2/R=GMm/R ∴ v2=GM v ∝R0Answer: (b)
Q.11
A planet is moving in an elliptical orbit around the sun. If T, V, E, and L stands for kinetic energy, gravitational potential, total energy, and magnitude of angular moment bout the center of force respectively
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a) T is conserved
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b) V is always positive
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c) E is always negative
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d) L is conserved but direction of vector L changes continuously
Explanation
In a circular or elliptical orbital motion, torque is always acting parallel to displacement or velocity. So, angular momentum is conserved. In attractive field, potential energy is negative. Kinetic energy changes as velocity increase when distance is less. So, option (c) is correct.Answer: (c)
Q.12
For a satellite escape velocity is 11 km/s. If the satellite is launched at an angle of 60° with the vertical, then escape velocity will be [1989]
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a) 11 km/s
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b) 11 √3 km/s
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c) 11 /√3 km/s
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d) 33 km/s
Explanation
Formula for escape velocity ve=√(2gRe) is independent of angle of projection, so it will not change it will be 11 km/s Answer: (a)
Q.13
A satellite of mass am is orbiting around the earth in a circular orbit with a velocity v. What will be its total energy? [1991]
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a) (3/4) mv2
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b) (1/2) mv2
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c) mv2
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d)-(1/2)mv2
Explanation
Total energy of the satellite=kinetic energy + Potential energy=½mV2 - GMm/ h --re(1) here M is the mass of earth and h is the height of satellite from the center of the earthNow Centripetal force=Gravitational force mv2 / h=GMm/ h2multiplying both sides by h we getmv2=GMm/ hsubstituting value of GMm/ h in eq(1) we getTotal Energy=½mV2 - mv2Total energy=-(1/2)mv2negative sign indicates binding energy Answer:(d)
Q.14
The mean radius of earth is R, its angular speed on its own axis is ω and the acceleration due to gravity at earth's surface is g. What will be the radius of the orbit of a geostationary satellite ?[1992]
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a) (R2g/ω2)1/3
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b) (Rg/ω2)1/3
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c) (R2 ω2/g)1/3
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d) (R2 g/ ω)1/3
Explanation
For a geostationary satellite it's angular velocity should be same as earth's angular velocity ω Centripetal acceleration of geostationary satellite here 'r' is the radius of satellite v is the linear velocity of satellite Now v=ωr since angular velocity of geostationary satellite must be equal to angular velocity of earth. substituting the value of v in above equation we get Now g=GM/R2 GM=gR2 substituting value in above equation we get Answer: (a)
Q.15
satellite A of mass m is at a distance of r from the earth's center. Another satellite B of mass 2m is at a distance of 2r from the earth's center. Their time periods are in the ratio of
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a) 1:2
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b) 1:16
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c) 1 : 32
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d) 1 : 2√2
Explanation
According to Kepler's law T2 ∝R3 note periodic time is independent of massTaking the ratio of periodic time we getAnswer: (d)
Q.16
The escape velocity from earth is 11.2 km/s. If a body is to be projected in a direction making an angle 45° to the vertical, then the escape velocity is [1993]
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a) 11.2×2 km/s
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b) 11.2km/s
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c) 11.2 / √2 km/s
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d) 11.2 √2 km/s
Explanation
Escape velocity is independent of angle of projection Answer:(b)
Q.17
The distance of Neptune and Saturn from the sun is nearly 1013 and 1012 meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio [1994]
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a) 10
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b) 100
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c) 10√10
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d) 1000
Explanation
For solution refer Q2 Answer: (c)
Q.18
A satellite A of mass m is at a distance of r from the surface of the earth. Another satellite B of mass 2m is at a distance of 2r from the earth's center. Their time periods are in the ratio of..[ CBSE-PMT 1993]
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a)1:2
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b) 1:16
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c)1 : 32
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d)1 : 2√2
Explanation
Time period of satellite is independent of massT2 ∝ r3Answer: (d)
Q.19
A ball is dropped from a satellite revolving around the earth at a height of 120 km: The ball will -, [ CBSE-PMT 1996]
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a) continue to move with same speed along a straight line tangentially to the satellite at that time
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b) continue to move with the same speed along the original orbit of satellite
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c)fall down to earth gradually
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d)go far away in space
Explanation
The orbital speed of satellite is independent of mass of satellite, so the ball will behave as a satellite and will continue to move with the same speed in the original orbitAnswer: (b)
Q.20
Dependence of intensityof gravitational field(E) of earth with distance (r) from centre ofearth is correctly represented by
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a)
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b)
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c)
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d)
Explanation
Answer:(c)
Q.21
A seconds pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket [ CBSE-PMT 1991]
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a) comes down with uniform acceleration
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b) moves round the earth in a geostationary orbit
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c)moves up with a uniform velocity
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d)moves up with uniform acceleration
Explanation
Time period of pendulum is given by equation When rocket accelerates upwards g increases to (g + a) hence period decreases Answer:(d)
Q.22
A satellite in force free space sweeps stationary interplanetary dust at a rate dM/dt=αv where M is the mass and v is the velocity of the satellite and α is a constant. What is the deceleration of the satellite? [ CBSE-PMT 1994]
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a) -αv2
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b) -αv2/2M
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c) -αv2/M
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d) -2αv2/M
Explanation
For variable mass problems force= now dM/dt=αv substituting in above we get F=αv2 ∴ Retardation=-F / M=- αv2 / M Answer: (c)
Q.23
A body weighs 72 N on the surface of the earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface [CBSE-PMT 2000]
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a)32 N
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b) 28 N
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c)16N
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d)72N
Explanation
gravitational acceleration at surface of earth g=GM / R2 At height H=R/2, g'=Body weight at height R/2 is mg'Answer: (a)
Q.24
With what velocity should a particle be projected so that its height becomes equal to radius of earth? [ CBSE-PMT 2001]
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a)
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b)
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c)
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d)
Explanation
Let u be the velocity According to law of conservation of energyAnswer: (a)
Q.25
For a satellite escape velocity is 11 km/s. If the satellite is launched at an angle of 60° with the vertical, then escape velocity will be [ CBSE-PMT 1989]
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a)11 km/s
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b) 11 √3 km/s
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c)11/ √3 km/s
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d)33 km/s
Explanation
Escape velocity is independent of angle of projection, so it will not changeformula for escape velocity is Answer:(a)
Q.26
A planet is moving in an elliptical orbit around the sun. If T, V, E and L stand respectively for its kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the center of force, which of the following is correct ? [1990]
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a) T is conserved
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b) V is always positive
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c) E is always negative
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d) L is conserved but direction of vector L changes continuously
Explanation
In circular or elliptical orbital motion,torque is always acting parallel to displacement or velocity. So, angular momentum is conserved. In attractive field, potential energy is negative. Kinetic energy changes as velocity increases when distance is less. So, option 'c' is correct Answer: (c)
Q.27
Two spheres of masses m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravityThe gravitational force will now be [ CBSE-PMT 2003]
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a) F/9
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b) 3F
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c)F
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d) F/3
Explanation
Gravitational force is independent of the medium. Hence, it will remain independentAnswer: (c)
Q.28
The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satellite A is 3 V, then the speed of satellite B will be: [ CBSE-PMT 2010]
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a) 3V/4
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b) 6V
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c)12V
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d)3V/2
Explanation
Orbital velocity of a satellite in a circular orbit of radius 'R' inversely proportional ro square root of radius Thus If R1 and R2 Answer: (b)
Q.29
A particle of mass M is situated at the center of a spherical shell of same mass and radius 'a'. The gravitational potential at a point situated at a/2 distance from the center, will be: [ CBSE-PMT 2010]
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a)
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b)
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c)
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d)
Explanation
Potential at the point=Potential at the point due to the shell + Potential due to the particle Note potential inside the shell is uniform=-GM / aPotential at point at a/2 due to mass at center=-2GM/aThus Potential at the given point=-3GM / a Answer:(a)
Q.30
A satellite of mass m is orbiting around the earth in a circular orbit with a velocity v. What will be its total energy? [ CBSE-PMT 1991]
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a) (3/4) mv2
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b) (3/2) mv2
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c) mv2
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d) -(1/2)mv2
Explanation
total energy of the satellite=K.E. + P.E Now K.E=½ ( m v2) P.E=-GMm / R but v=( Gm/R)1/2 ∴ P.E=-mv Thus Total energy=½ ( m v2) Total energy=-½ ( m v2) Answer: (d)
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