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Quiz 3
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Q.1
A marble block of mass 2 kg lying on ice when given a velocity of 6m/s is stopped by friction in 10s. Then the coefficient of friction is [ AIEEE -2003]
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a) 0.02
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b) 0.03
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c)0.04
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d)0.06
Explanation
Retardation due to friction a=(0-6)/10=-0.6 m/s2 Frictional force f=ma=2(-0.6)=-1.2 N Coefficient of friction µ=f/N Normal force N=mgµ=f/mgµ=1.2 / (2×10)=0.06Answer: (d)
Q.2
A light spring balance hangs from the hook of the other spring balance and block of M kg hangs from the former one. Then the true statement about the scale reading is [ AIEEE 2003]
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a) Both the scale read M kg
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b) The scale of the lower one reads M kg and of upper one zero
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c)The reading of the scales can be anything but the sum of the reading will be M kg
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d)Both the scale read M/2 kg each
Explanation
The earth pulls the block by force M Kg. The block in turns exerts force M on the spring balance which therefore shows a reading of M kgThe spring is mass less. Therefore it exerts a force of M on the second spring balance which shows the reading M kg Answer:(a)
Q.3
A smooth block is released at rest on a 45° inclined and then slides distance 'd'. The time taken to slide is 'n' times as much to slide on rough incline than on a smooth incline. The coefficient of friction is [ AIEEE 2005]
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a)
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b)
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c)
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d)
Explanation
In absence of friction acceleration along inclined plane is gsinθ let time taken by the block be t1 to travel distance of 'd' ∴ d=½ g(t12) In presence of friction downward acceleration will be gsinθ - µgcosθ∴ d=½ g(t22) According to question t2=n t1 Here µ is kinetic friction as block moves over inclined plane given θ=45 Answer: (b)
Q.4
A block is kept on a frictionless inclined surface with angle of inclination 'α'. the incline is given an acceleration 'a' to keep the block stationary. Then a is equal to ..[ AIEEE 2005]
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a)g cosecα
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b) g /tanα
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c)gtanα
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d)g
Explanation
As shown in figure block will experience pseudo acceleration a' which is equal in magnitude of acceleration 'a' but opposite in directionComponent of a' parallel to inclined plane is a'cosα Given that block is stationary and a'=a thus a cosα=gsinαa=tanα Answer: (c)
Q.5
A player caught a cricket ball of mass 150g moving at a rate of 20m/s. If the catching process is completed in 0.1s, the force of the blow exerted by the ball on the hand of the player is equal to .. [ AIEEE 2006]
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a) 150 N
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b) 3 N
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c)30 N
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d)300 N
Explanation
Force=Change in momentum / time periodChange in momentum=m ( final velocity -initial velocity )=0.15(0-20)=-3Force on ball=-3 /0.1=-30NForce on hand=- force on ballForce on hand due to ball=30N Answer:(c)
Q.6
A particle is projected along a line of greatest slope up a rough plane inclined at an angle of 45° with the horizontal. If the coefficient of friction is 1/2, then the retardation is
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a) g/2
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b) g/2√2
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c) [g/√2][1+1/2]
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d) [g/√2] [ 1- 1/2]
Explanation
Retardation=[ µgcos45 + gsin45]Retardation=[g/√2][1+1/2] Answer: (c)
Q.7
A ball of mass 0.2kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2m while applying the force and the ball goes up to 2m height further, find the magnitude of the force( consider g=10 m/s2) .. [AIEEE 2006]
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a) 4 N
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b) 16 N
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c) 20 N
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d) 22 N
Explanation
Let the velocity of the ball when it leaves the hand is u then applying v2 -u2=2as for upward journey 0 -u2=2×10×2 u2=40 m/s Velocity due to motion of Hand final velocity=v2=40 m/s Applying again v2 -u2=2as 40=2a(0.2) a=100 m/s2 Thus resultant force F=m a=0.2×100=20N F=N - mg (N is the actual force applied by hand) 20=N - 0.2×10 N=20 + 2=22 N Answer: (d)
Q.8
A player caught a cricket ball of mass 150g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to [ AIEEE 2006]
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a)150 N
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b) 3 N
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c)30 N
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d)300 N
Explanation
F=m(v - u) / t F=0.15 (0 - 20 )/ 0.1=30 NAnswer: (c)
Q.9
The upper half of an inclined plane with inclination Φ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by [ AIEEE 2005]
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a) 2 cosΦ
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b) 2 sinΦ
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c)tanΦ
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d)2 tanΦ
Explanation
Let total length of slope be 2S For initial distance s, no friction, from equation of motion V2 = 2gsinΦ S For second half final velocity is zero and initial velocity u2 = 2gsinΦ S 0=2gsinΦ S+2aS a = - gsinΦ is the resultant acceleration. Of down word gsinΦ and upward friction µg cosΦ a = gsinΦ - µg cosΦ - gsinΦ = gsinΦ - µg cosΦ µ=2 tanΦ Alternate method Acceleration of block while sliding down upper half=g sinΦ Retardation of block while down lower half=- (gsinΦ - µg cosΦ) For a block to come rest at the bottom acceleration in first half=retardation in second half gsinΦ=-(gsinΦ - µg cosΦ) µ=2 tanΦAnswer: (d)
Q.10
A particle of mass 0.3 kg subject to a force F=-kx with k=15 N/m. What will be its initial acceleration if it is released from a point 20 cm away from origin [ AIEEE 2005]
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a) 15 m/s2
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b) 3 m/s2
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c)10 m/s2
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d)5 m/s2
Explanation
F=m ama=-kxa=-kx/mx=20 cm=0.2 m , k=15 N/ma=15×0.2 / 0.3=100m/s2 Answer:(c)
Q.11
A block of mass m is connected to another block of mass M by spring (mass less) of spring constant k. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unscratched. Then a constant force F starts acting on the block of mass M to pull it. Find the force of the block of mass m [ AIEEE 2007]
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a)
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b)
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c)
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d)
Explanation
Writing free body-diagram for m and M From diagram T=maAnd F-T=MaWhere T is due to spring∴ F - ma=Ma a(m + M)=F a=F / (M + m)Force acting on the block of mass m is F'=ma Answer: (d)
Q.12
A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If force P is applied at the free end of the rope, the force exerted by the rope on the block is [ AIEEE 2003]
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a)Pm / (M+m)
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b) Pm / (M-m)
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c)P
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d)PM/(M+m)
Explanation
Total mass pulled M'=(M+m) We get P=M'aa=P/M'∴ a=P / (m+M) Taking the block as a system, we get T=Ma∴ T=MP/(m+M)Answer: (d)
Q.13
A block rests on rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.If the frictional force on the block is 10N, the mass of the block ( in kg) is [ AIEEE 2004]
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a) 1.6
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b) 4.0
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c) 2.0
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d)2.5
Explanation
Body is at rest thus mg sinθ=f m×10×sin 30=10 5m=10 m=2.0 kgAnswer: (c)
Q.14
A mass of M kg is suspended by a weightless string. The horizontal force that is require to displace it until the string makes an angle of 45° with the initial vertical direction is [ AIEEE 2006]
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a)
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b)
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c)
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d)
Explanation
As shown in figure h=l -lcos45=l(1-1/√2) d=lcos√2=l/√2x=lsinθ=l/√2Work done by horizontal force=Change in Potential energy F×x=mghF×(l/√2)=mgl(1-1/√2)F=Mg(√2 + 1 ) Answer:(d)
Q.15
A bullet fired into fixed target loses half of its velocity after penetrating 3cm. How much further it will penetrate before coming to rest assume that it faces constant resistance to motion? [ AIEEE 2005]
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a) 2.0 cm
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b) 3.0 cm
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c) 1.0 cm
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d) 1.5 cm
Explanation
Case I From equation V2=u 2 - 2as (u/2)2=u2 - 2a(3) a =u2/8 Case II Initial velocity square=u2/4 and final velocity v=0 0=(u2 / 4) -2ax x=1 cm Answer: (c)
Q.16
A ship of mass 3×107 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, the speed of the ship is [ IIT 19980]
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a)1.5 m/s
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b) 60 m/s
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c)0.1 m/s
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d)5.0 m/s
Explanation
F=ma∴ 5×104=2×107 a a=(5/3) ×10-3 Now v2=u +2asu=0 given v2=2 ×(5/3)×10-3 ×3v=0.1m/s-1Answer: (c)
Q.17
A small block is shot into each of four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in [ IIT 2001]
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a)
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b)
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c)
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d)
Explanation
When body follows the path it presses the surface with force N. In given example it follows circular motion, centripetal force is required which is provided by (mg + N ) ∴ mg + N=mv2 / r where r is the radius of curvature at the top. If the surface is smooth then radius of curvature is maximum. In option "a" radius of curvature is minimum at top thus N is maximumAnswer: (a)
Q.18
A insect crawls up a hemispherical surface very slowly. The coefficient of friction between the insect and the surface is 1/if the line joining the center of the hemispherical surface is maximum possible value of α is given by [ IIT 2001]
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a)cot α=3
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b) tan α=3
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c)sec α
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d)cosec α=3
Explanation
For maximum value of α, mgsinα will also maximum.In case f is limiting friction. Two forces acting on the insect are mg and N. On resolving force mg in componentmgcosα=N f=mg sinα But µ=f/N µ=mgsinα/ mg cosα µ=tanα 1/3=tanα cot α=3 Answer:(a)
Q.19
The pulley and strings shown in the figure are smooth and of negligible mass. for the system to remain in equilibrium the angle θ should be [ IIT 2001]
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a) 0°
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b) 30°
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c) 45°
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d) 60°
Explanation
Due to symmetry tension in string passing over pulley is same. Let tension in strings T Force acting down word on √2 m is √2 mg system is in equilibrium On resolving tension 2Tcosθ=√2 mg And T=mg 2mgcosθ=√2 mg cosθ=1 /√2 θ=45° Answer: (c)
Q.20
A string of negligible mass going over a clamped pulley of mass m supports a block of M as shown in figure. The force on the pulley by the clamp is given by [ IIT 2001]
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a)√2 Mg
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b) √ mg
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c)
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d)
Explanation
At equilibrium T=Mg F.B.D of pulley F1=(m+M)gThe resultant force on pulley is Answer: (d)
Q.21
The mass of the lift is 100 kg which is hanging on the string. The tension in the string, when the lift is moving with constant velocity, is ( g=9.8 m/s2
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a)100 N
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b) 980 N
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c)1000 N
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d)none of the above
Explanation
Answer: (b)
Q.22
What is the maximum value of the force F such that the block shown in the arrangement, coefficient of friction between block and surface is 1/ (2√3) does not move? [ IIT 2003]
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a) 20N
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b) 10N
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c)12N
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d)15N
Explanation
F.B.D since the block is not moving forward for the maximum force F applied, therefore Fcos60°=f [ Note frictional force is limiting force] Now f=µN N=Fsin60 + mg Fcos60=µ[ Fsin60 + mg] Answer: (a)
Q.23
A block P of mass m is placed on a horizontal frictionless surface. A second block of same mass m is placed on it and is connected to a spring of constant k, the two blocks are pulled by distance A. Block Q oscillates without slipping. What is the maximum value of frictional force between the two blocks [ IIT 2004]
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a) kA/2
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b) kA
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c)µsmg
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d)zero
Explanation
Since both the block oscillates thus frictional force is greater then force due to springLet acceleration of blocks be aThus F=2ma ( since each block is of mass m) Force is exerted by spring thus F=kA ( k is spring constant, A is displacement) from above 2ma=kA Block P moves along with P due to friction between there surfaces thus f=maThus 2f=kAf=kA/2 Answer:(a)
Q.24
The string between blocks of mass m and 2m is mass less and inextensible. The system is suspended by a mass less spring as shown. If the string is cut find the magnitude of acceleration of mass 2m and m [ IIT 2006]
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a) g, g
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b) g, g/2
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c) b/2, g
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d) g/2, g/2
Explanation
Tension in string attached to m is mg Thus if string is cut mass m will fall with gravitational acceleration Tension in spring attached to mass 2m is T=3mg After cutting the string Tension in spring T'=2mg Thus change in tension=T - T'=2mg -mg=mg in upward direction Thus resultant acceleration a=mg/2m=g/2 Answer: (b)
Q.25
Two particle of mass m each are tied at the ends of light string of length 2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance of 'a' from the center P( as shown in figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result, the particle moves towards each other on the surface. The magnitude of acceleration, when the separation between them become 2x is [ IIT 2007]
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a)
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b)
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c)
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d)
Explanation
From diagram F=2Tsinθ --(1)The acceleration of mass m is due to the force Tcosθ Thus ma=Tcosθ --(2) From equation (1) and (2) we get Now tanθ=y/x but y=(a2 - x2) 1/2 Answer: (b)
Q.26
A particle moves in the X-Y plane under the influence of a force such that its linear momentum is Pt=A[icos(kt) - jsin(kt)], where A and k are constants. The angle between the force and the momentum is [ IIT 2007]
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a) 0°
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b) 30°
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c)45°
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d)90°
Explanation
We know that F=dP/dt Thus by differentiating momentum we get equation for force F=Ak[-isin(kt) - jcos(kt)]Now by taking dot product between F and P we get F⋅P=0 since F and P are nonzero ∴cosθ=00Thus θ=90°Answer: (d)
Q.27
A block of base 10cm × 10cm and height 15cm is kept on an inclined plane. The coefficient of friction between them is √The inclination θ of this inclined plane from the horizontal plane is gradually increased from θ°. Then [ IIT 2009]
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a)at θ=30°, the block will start sliding down the plane
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b) the block will remain at rest on the plane up to certain θ and then it will topple
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c)at θ=60°, the block will start sliding down the plane and continue to do so at higher angles
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d)at θ=60°, the block will start sliding down the plane and on further increasing θ, it will topples at certain angle
Explanation
For the block to slide, the angle of inclination should be equal to the angle of repose tanθ=µtanθ=√3θ=60°thus option' a' is wrong , and is on the verge of slipping For angle less than 60° For block to topple, the condition of the block will be as shown in figure For block to topple, the condition of the block will be as shown in figure Torque about C.M. due to N τN = N × (0.05) = mg cosθ(0.05) Torque about C.M. due to f τf = mg sinθ × (0.75) = mg sinθ(0.075) Note directions of τN and τf are opposite If τf > τN then block will topple for line mg sinθ(0.075)> mg cosθ(0.05) tanθ > 0.050/0.075; tanθ > 2/3 the block will remain at rest on the plane up to certain θ and then it will topple Answer:(b)
Q.28
An unloaded car moving with velocity u on a frictionless road can be stopped in a distance s. If the passenger add 40% to its weight and braking force remain same, the stopping distance at velocity u is now
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a) 1.4s
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b) √1.4 s
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c) (1.4) 2s
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d) 1 /1.4
Explanation
Let a be the acceleration of unloaded car and m be the mass Braking force F=ma Now new mass m'=1.4m Braking force is same F=m'a' Thus ma=m'a' ma=1.4m a' a'=a/1.4 v2=u2 -2as First case: 0=u2 -2as u2=2asSecond case:0=u2 -2a's' u2=2a's' Thus 2as=2a's' 2as=2 (a/1.4)s' ∴ s'=1.4s Answer: (a)
Q.29
A 2kg block is initially at rest. A 70N force is required to set the block in motion. After the motion, a force of 60N is applied to keep the block moving with constant speed. The coefficient of static friction is [ A.M.U. 1999]
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a)0.6
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b) 0.52
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c)0.44
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d)0.35
Explanation
Here static friction=force required to set body in motion f=70N But fractional force f=µN N=mg Thus f=µmg ∴ 70=µmg µ=70 /20 ×10=0.35Answer: (d)
Q.30
In an elevator moving vertically up with an acceleration 'g', the force exerted on the floor by a passenger of mass M is [ C.P.M.T. 1999]
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a) Mg
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b) (1/2)Mg
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c)zero
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d)2Mg
Explanation
Lift is moving vertically up with acceleration g thus pseudo acceleration is vertically downTotal acceleration a'=g+g=2g Total force=M(2g)Answer: (d)
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