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Quiz 4
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Q.1
A 7 kg object is subjected to two forces ( in Newton)F1=20i + 30j and F2=8i - 5j The magnitude of resulting acceleration in m/s2 will be [ C.P.M.T 1994]
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a) 5
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b) 4
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c)3
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d)2
Explanation
Resultant force F=F1 + F2 F=20i + 30j + 8i - 5j F=20i + 25j |F|=[202 + 252]1/2 |F|=37.5N a=F/m=(37.5)/7=5.3 m/s2 Answer:(a)
Q.2
A body of mass 10kg and velocity 10m/s collides with a stationary body of mass 5kg. After collision, both bodies stick to each other. ThE common velocity would be [ Pre-Medical/ Dental 1994]
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a) (3/20) m/s
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b) (20/3) m/s
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c) (10/3) m/s
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d) 3/10 m/s
Explanation
Applying the law of conservation of linear momentum m1 u1 + m2 u2 = (m1 + m2 )v 10 × 10 + 5×0 =(10+5)v 15v=100 v=(20/3) m/s Answer: (b)
Q.3
Two equal masses m1 and m2 moving along the same straight line with velocities 3 m/s and -5 m/s respectively, collide elastically. Their respective velocities after collision will be [ pre. Medical/Dental 1998]
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a)-3m/s , 5 m/s
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b) 4 m/s, 4 m/s
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c)-4 m/s, 4 m/s
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d)-5 m/s, 3m/s
Explanation
As masses are equal, their velocities on perfectly elastic collision, are just changed becomes -5m/s and 3 m/sAnswer: (d)
Q.4
For a body moving with constant speed in a horizontal circle, which of the following remains constant [M.P. P.M.T. 2000]
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a) velocity
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b) acceleration
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c)centripetal force
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d)kinetic energy
Explanation
K.E=½ m v2=constant While moving uniformly in a horizontal circle.Answer: (d)
Q.5
A 10 kg object collides with stationary 5kg object and after collision they stick to gather and move forward with velocity 4 m/s. What is the velocity with which the 10kg object hit the second one? [ kerala PET 2004]
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a) 4 m/s
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b) 6 m/s
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c) 10 m/s
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d) 12 m/s
Explanation
Here m1=10 kg , m2=5 kg v2=0, common velocity v=4 /m/s From the law of conservation of momentum 10v1 +5 ×0=4 ( 10+4) v1=6 m/s Answer: (b)
Q.6
Physical independence of force is a consequence of [ kerala PMT 2004]
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a)Third law of motion
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b) Second law of motion
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c)First law of motion
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d)All of these law
Explanation
Answer: (c)
Q.7
A cylinder rolls up an inclined plane, reaches some height and then rolls down ( without slipping throughout these motions) The directions of frictional force acting on the cylinder are [ IIT 2002]
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a) up the incline while ascending and down the incline while descending
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b) up the incline while ascending as well as descending
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c) down the incline while ascending and up to the incline while descending
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d)down the incline while ascending as well as descending
Explanation
Component of weight (mgsinθ) is always down the inclined plane, whether the cylinder is rolling up or it is rolling down. therefore, for no slipping,sense of angular acceleration must be the same in both the cases . therefore, force of friction (f) acts up the inclined plane in both the cases. Answer:(b)
Q.8
A bomb at rest explodes into three parts of same masse. The momentum of the two parts are -2pi and pj . The momentum of the third part will have a magnitude of
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a) p
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b) √(3p)
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c) p√5
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d) zero
Explanation
According to law of conservation of momentum Momentum before explosion=momentum after explosion 0=-2pi + pj + P P=2pi - pj |P|=P√5 Answer: (c)
Q.9
If a block moving up at θ=30° with a velocity 5m/s, stops after 0.5sec, then µ is [ C.P.M.T. 1995]
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a)0.5
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b) 1.25
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c)0.6
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d)none of the above
Explanation
Block is moving up Acceleration down the rough inclined plane is a=g(sinθ+ µcosθ) From equation v=u +at 0=5 -10(sin 30 + µcos30)×0.50=5 - 10×× 0.5[ 1/2 + (√3 /2) µ] 0=5 - [2.5 -2.5µ√3]∴ µ=1/ √3=0.57=0.6Answer: (c)
Q.10
A block of mass 3 kg in contact with a second block of mass 2kg rests on a horizontal frictionless surface. A horizontal force of 10N is applied to push the first block. The force with which the first block pushes the second block is [ J.E.E. Orissa 1998]
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a) 10N
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b) 6N
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c)4N
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d)zero
Explanation
Total mass M=2 +3=5kgTotal force=10N Thus acceleration of blocks a=F/M=10/5=2 m/s2force exerted by first on second f=m2 ×a f=2×2=4NAnswer: (c)
Q.11
Which of the following is true?
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a) friction can be reduced to zero
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b) frictional force can accelerate a body
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c)frictional force is proportional to the area of contact between the two surfaces
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d)kinetic friction is always greater than rolling friction
Explanation
Answer:(d)
Q.12
Which of the following is true?
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a) friction can be reduced to zero
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b) frictional force can accelerate a body
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c)frictional force is proportional to the area of contact between the two surfaces
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d)kinetic friction is always greater than rolling friction
Explanation
Answer:(d)
Q.13
A car of mass m is driven with an acceleration a along a straight level road against a constant external resistive force R. When the velocity of the car is V, the rate at which the engine of the car is doing work [ PMT MP 1998]
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a)RV
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b) maV
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c)(R+ma)V
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d)(ma+V)R
Explanation
let f be the resultant force on the car f=maThus f=f'-R here f' is the force exerted by engine f'=f+RPower P=f'V P=(f+R)V P=(ma+R)V Answer: (c)
Q.14
The mass of a lift is 500kg. What will be the tension in its cable when it is going up with an acceleration of 2.0m/s2.. [ MPPMT 2000]
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a) 5000 N
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b) 5600 N
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c)5900 N
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d)6200 N
Explanation
Let total force acting on the lift be f'let resultant force acting on lift upward be f=ma gravitational force acting on the lift downward W=mg Now f=f'- Wf'=f+ W f'=ma + mg=m(a+g) f'=500 ( 2 +9.8)f' =5900NAnswer: (c)
Q.15
A ship of mass 3×107kg initially at rest is pulled by a force of 5×104N through a distance of 3m. Assuming that the resistance due to water is negligible, what will be the speed of the ship? [ MPPMT 2000]
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a) 0.01 m/s
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b) 1.5 m/s
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c)5 m/s
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d)0.2 m/s
Explanation
acceleration a=F/m a=5×104 / 3×107 a=(5/3) × 10-3 v2 -u2=2asS=3m, u=0v2=2×(5/3)×10-3 ×3 v=10-2m/s Answer:(a)
Q.16
The time period of a simple pendulum of length 'l' is measured in an elevator descending with acceleration g/3 [ CPMT 2000]
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a)
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b)
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c)
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d)
Explanation
Lift is descending thus pseudo acceleration is upward gravitational acceleration downward Thus resultant acceleration=(g-g/3)=(2/3)g Formula for periodic time of pendulum Replacing value of g in above equation by (2/3) g we get Answer: (a)
Q.17
A particle of mass 10kg is moving in a straight line. If its displacement, x with time t is given by x=(t3 - 2t -10 )m, then the force acting on it at the end of 4 seconds is [cbse 2001]
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a) 24N
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b) 240N
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c)300N
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d)1200N
Explanation
By taking derivative two times with respect to time of above equation we get equation for acceleration a=6t Thus acceleration at t=4 sec is a=6×4=24 m/s2 force F=maF=10×24=240NAnswer: (b)
Q.18
A constant force acts on the body of mass 0.9 kg at rest for 10s. If the body moves a distance of 250m, the magnitude of the force is [ DCE 2001]
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a)3 N
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b) 3.5 N
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c)4 N
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d)4.5 N
Explanation
From s=ut + ½ at2s=250 m , u=0 and t=10s250=½ a(10) 2 a=5 m/s2 F=ma=0.9 ×5=4.5N Answer:(d)
Q.19
25N force is required to rise 75kg mass from a pulley. If rope is pulled 12 m, then the load is lifted to 3m, the efficiency of pulley system will be [ CBSE2001]
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a) 25%
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b) 33.3%
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c) 75%
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d) 90%
Explanation
Input energy=Force × displacement Input energy=25×12=300J Output energy=weight × height Output energy=75×3=225J efficiency η=Output / input efficiency η=225/300=0.75 % efficiency=η × 100=0.75×100=75% Answer: (c)
Q.20
A block of mass M is pulled along a horizontal surface by a rope of mass m by applying a force F at one end of the rope. The force which the rope exerts on the block is [ Haryana CEET 2000]
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a)
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b)
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c)
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d)
Explanation
Acceleration a=F /(M+m) force on the block=M×a=MF /(M+m)Answer: (d)
Q.21
A 1 kg particle strikes a wall with velocity 1m/s at an angle of 30° with the wall and reflects at the same angle. If it remains in contact with the wall for 0.1s, then the force is (UPCPMT 2001]
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a)0
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b) 10√3 N
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c)30√3 N
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d)40√3 N
Explanation
final momentum Pf=-mvsinθi-mvcosθj Initial momentum Pi=+mvsinθi-mvcosθj Change in momentum of particle ΔP=Pf - Pi ΔP=-2mvsinθ ΔP=-2(1)(1)sin60 ΔP=√3 Force=Change in momentum / time period of contact Force=√3 / 0.1=10√3Answer: (b)
Q.22
The coefficient of friction of a surface is 1 /√What should be the angle of inclination so that a body placed on the surface just begins to slide down [ PbCET 2002]
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a) 30°
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b) 45°
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c)60°
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d)90°
Explanation
we know that tanθ=µtanθ=1 /√3 θ=30°Answer: (a)
Q.23
A man weights 80kg. He stands on a weighing scale in the lift, which is moving upwards with uniform acceleration of 5m/sWhat would be the reading on the scale (g=10 m/s2 [ CBSE 2003]
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a) Zero
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b) 400N
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c)800 N
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d)1200 N
Explanation
Lift is going up thus pseudo acceleration on the man in downwards Resultant acceleration a'=g + aa'=10 +5=15 m/s2Force on scale=m×a' Force on scale=80×15=1200N Answer:(d)
Q.24
Two bodies of equal masses revolve in circular orbits of radius R1 and R2 with same period. Their centripetal forces are in the ratio [ Kerala PMT 2004]
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a)
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b)
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c)
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d)
Explanation
Centripetal force F=mv2 / r but v=2πr /T , T is periodic time since m and T is constant F∝ r Answer: (b)
Q.25
A body of mass m collides against a wall with the velocity v and rebounds with the same speed. Its change in momentum is [ kerala PMT 2004]
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a)2mv
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b) mv
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c)-mv
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d)zero
Explanation
Initial momentum=-mv, velocity is taken negative body is moving towards the wall Final momentum=mv , Velocity is taken as positive ball is going away from wall Change in momentum ΔP=final momentum - Initial momentum ΔP=mv -( -mv)=2mv Answer: (a)
Q.26
The recoil velocity of a 4.0kg rifle that shoots a 0.05kg bullet at a speed of 280 m/s is .. [ kerala PMT 2004]
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a) +3.5m/s
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b) - 3.5 m/s
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c)-√3.5 m/s
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d)√3.5 m/s
Explanation
According law of conservation of momentum Finial momentum of Rifle + Final momentum of bullet=Initial momentum of Rifle + Initial momentum of bullet vr M + vbm=0 + 0 vr × 4 + 280 × 0.05=-3.5 m/sAnswer: (b)
Q.27
In the motion of a rocket, physical quantity which is conserved is [ kerala PMT 2004]
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a) Angular Momentum
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b) Linear momentum
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c)Force
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d)work
Explanation
Answer:(b)
Q.28
An ice cart of mass 60kg rests on a horizontal patch with coefficient of static friction 1/assuming that there is no vertical acceleration, find the magnitude of the maximum horizontal force required to move the ice cart ( g=9.8 m/s-2)( keralaPET 2004)
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a) 100 N
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b) 110 N
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c)209 N
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d)196 N
Explanation
Maximum horizontal force required to move the cart=Force of friction (f) f=µN but N=mg ∴ f=µmg f=(1/3)× 60 × 9.8f=196 NAnswer: (d)
Q.29
A cyclist is moving in a circular track of radius 80m with velocity v=36 km/hr. He has to lean from the vertical approximately through an angle ( take g=10 m/s2)( kerala PET 2004)
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a) tan-1(4)
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b) tan-1(1/8)
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c)tan-1(1/4)
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d)tan-1(2)
Explanation
As shown in figure R is reactionary force centripetal force mv2 / r=Rsinθmg=Rcosθ Taking ratio we get v2 / rg=tanθ Here r=80m, v=36km/hr=10m/stanθ=10×10 / (80×10)=1/8 θ=tan-1(1/8) Answer:(b)
Q.30
A particle moves in a circular path with decreasing speed. Choose the correct statement [ IIT 2005]
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a) Angular momentum remains constant
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b) Acceleration is towards center
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c) Particle moves in a spiral path with decreasing radius
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d) The direction of angular momentum remains constant
Explanation
Angular momentum is an axial vector, so its direction is along axis, perpendicular to the plane of motion, which is not going to change because of change in speed. Therefore, the direction of angular momentum remains the same and its magnitude may vary. Answer: (d)
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