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Quiz 6
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Q.1
An object of mass 5 kg is acted upon by a force that varies with position of the object s shown. If the object starts out from rest at a point x=0, what is its speed at x=25m
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a) 5 m/s
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b) 10 m/s
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c)20 m/s
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d)28m/s
Explanation
Area covered under the graph F-x is work done Work done till 25 m=10×25=25 Joule∴ ½ mv2=250(5/2)v2=250 v=10 m/s Answer:(b)
Q.2
A 40 kg slab rests on the frictionless floor as shown in figure. a 10 kg block rests on the top of the slab. The static coefficient of friction between the block and slab is 0.6 while the kinetic friction is 0.The 10 kg block is acted upon by a horizontal force of 100N. If g=9.8 m/s2, the resulting acceleration of the slab will be
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a) 0.98 m/s2
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b) 1.47 m/s2
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c) 1.52 m/s2
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d) 6.1 m/s2
Explanation
Reaction of frictional force on upper block=force on the 40kg slab Let acceleration of 40 kg slab be 'a' Force on 40kg slab=40a Frictional force on upper block=µN But N=10g Frictional force on upper block=µ×10gThus 40a=µ×10g40a=0.4×10×9.8a=0.98 m/s2 Answer: (a)
Q.3
In Q281 if lift is is accelerating up with an acceleration 1 m/sec2 tension in string is
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a) 100 N
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b) 980 N
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c)1080 N
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d)880 N
Explanation
Answer: (c)
Q.4
Two blocks of masses 7 kg and 3 kg are connected by a spring of spring constant 1000 N/m and placed on a smooth horizontal surface. They are acted by horizontal surface. They are acted by horizontal forces F1=72N and F2=32N in opposite directions as shown in figure when acceleration of the blocks are equal and constant, the extension in cm of the spring is
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a)4cm
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b) 5cm
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c)6cm
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d)8cm
Explanation
Let common acceleration be a Tension in spring be T Thus 3a=72 -T and 7a=T -32 on solving above equations we get a=4 m/s2and T=60N Tension in spring=kx60=1000 × x x=0.06 m=6 cm Answer: (c)
Q.5
An impulse is applied to a moving object at an angle of 120° with respect to velocity vector. The angle between the impulse vector and change in momentum vector is [ BHU 1990]
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a) 120°
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b) 0°
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c)60°
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d)240°
Explanation
Change in momentum vector=Impulse vector Answer: (b)
Q.6
The block A and B, each of mass m are connected by mass less spring of natural length L and spring constant k. The block are initially resting on smooth horizontal floor with the spring at its natural length, as shown in figure. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B, and collides with A. Then [ IIT 1993]
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a)The K.E. of the A-B system, at maximum compression of the spring is zero
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b) The K.E. of the A-B system, t maximum compression of the spring is , mv2 / 4
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c)The K.E. of the A-B system, t maximum compression of the spring is , mv2 / 8
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d)The K.E. of the A-B system, t maximum compression of the spring is , mv2 / 2
Explanation
Applying law of conservation of momentummV=mV' + mV' V'=V/2 Therefore maximum Kinetic energy of system at compression mV'2/4 Answer:(b)
Q.7
The arm of nut cracker is 15cm long. The force needed to crack a nut without the help of the nut cracker is 22.5 kgwt. Where should a nut be placed in a nut cracker, so as to crack it with force of 2.25kg wt [ MPPMT 1989]
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a) 1 cm from fulcrum
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b) 1.5 cm from fulcrum
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c) 0.5 cm from fulcrum
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d) 2.0 cm from fulcrum
Explanation
Position of load , effort and fulcrum are shown in figure Load=22.5 kg wt Effort=2.25 kg wt 22.5 × x=2.25×15 10x=15 x=1.5 from fulcrumAnswer: (b)
Q.8
A 25 kg mass is accelerated from rest with a force of 100N. As it moves faster, friction and air resistance create an oppositely directed retarding force given by FR=A + Bv where A=25N and B=0.5 N-s/m. At what velocity does the acceleration is one half of the initial acceleration: [ AFMC 1994]
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a)25 m/s
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b) 50 m/s
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c)75 m/s
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d)100 m/s
Explanation
F net=Fapplied - Resistive forceF net=100 - (A+Bv)ma=100 - (A+Bv)Initial acceleration at v=0 15 ×a=100-[25+0.5×0] 15×a=75 a=5 m/s2 When acceleration is 2.5 m/s2 15×2.5=100 - [25+0.5v] 45=75 -v/2 v/2=37.5 v=75 m/sAnswer: (c)
Q.9
A flat plate moves normally with a speed v1 towards a horizontal jet of water of uniform area of cros-section. The jet discharges water at rate of volume V per second at a speed of vThe density of water is ρ. Assume that water splashes along the surface of the plate at right angles to the original motion. The magnitude of the force acting on the plate due to jet of water is [ IIT 1995]
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a) ρVv1
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b) ρV(v1 + v2)
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c)
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d)
Explanation
Force=change in momentumVρ=mass of water v1 - (-v2)=change in velocity F=ρV (v1 +v2)Answer: (b)
Q.10
A ball is dropped from height h, on a plane if the coefficient of restitution of the plane is e and if ball hits ground two times, the height up to which it reaches after two jumps will be [ Raj.PMT 1996]
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a) eh/2
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b) 2eh
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c)eh
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d)e4h
Explanation
Height attended after n jumpsh'=e2nh here n=2 ∴ h'=e4h Answer:(d)
Q.11
A bird is sitting in a large closed cage which is placed on a spring balance. It records a weight of 25N. The bird (mass=0.5kg) flies upward in the cage with an acceleration of 2m/sThe spring balance will now record a weight of ..[ MPPMT 1999]
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a) 24 N
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b) 25 N
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c) 26N
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d) 27 N
Explanation
Whether the bird sits or flies in the cage, the total weight will not change in any case. The only additional force on the base is ma=0.5×2=1N ∴ total wt=25 +1=26N Answer: (c)
Q.12
A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. The force F is allied at one end of the rope. What is the force, which the rope exerts on the block?
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a)F / (M-m)
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b) F/ M(m +M)
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c)FM/ (M+m)
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d)FM/ (M-m)
Explanation
Let a be the common acceleration of the system a=F/(m+M) Thus force on the block=MA Force on block=MF/(M+m) Answer: (c)
Q.13
Two bodies of mass m1 and m2 are connected by a light string passing over a smooth light fixed pully. the acceleration of the system is g/The ratio of their masses is
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a) 7:1
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b) 7:2
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c)4:3
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d)4:5
Explanation
From the formula for acceleration a=(m1 - m2)g / (m1 + m2)=g/7 ∴ 7m1 - 7m2=m1 + m26m1=8m2m1 / m2=4/3 Answer: (c)
Q.14
A block of mass 10kg rests on the floor of lift, which is accelerating upwards at 4m/s2 the reaction of the floor of the lift on the block is
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a)140 N
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b) 60N
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c)104 N
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d)96N
Explanation
Answer:(a)
Q.15
A bead is free to slide down a smooth wire tightly stretched between points A and B on a vertical circle. IF the bead starts from rest at A, the highest point on the circle.
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a) its velocity v on arriving at B is proportioanl to sinθ
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b) its velocity v on arriving at B is proportional to tanθ
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c) time to arrive at B is proportional to consθ
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d) time of arrival at B is independent of θ
Explanation
Distance AB=2Rcosθ --eq(1) Form equations of motion AB=½ gcosθt2 --eq(2) From above equations 2Rcosθ=½ gcosθt2 ∴t=√(4R/g ) time is independant of θ Answer: (d)
Q.16
With what minimum acceleration can be fireman slide down a rope whose breaking strength is two third of his weight?
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a)g/2
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b) 2g/3
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c)g/3
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d)3g/4
Explanation
Breaking strength=(2/3)mg Let fireman slides down with acceleration a then effective acceleration=g-a Force on rope=(g-a)m Thus (2/3)mg=(g-a)m 2g=3(g-a) a=g/3 Answer: (c)
Q.17
A rope of length L is pulled by a constant force F. What is the tension in the rope at a distance x from one end where the force is applied
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a) FL/x
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b)F(L-X) / L
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c)FL/(L-x)
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d)Fx/(L-x)
Explanation
LEt length of the sring be L and mass be M Mass per unit length=M/L Let P be a point at a distanxce x from B. Then Mass of PB=Mx/L Let a be the common acceleration a=F/M Mass of the remaining porion=(L-x) (M/L) Force on remaining part is tension at distance X from B=mass × acceleration Answer: (b)
Q.18
If force F=(3i + 4j - 10k) N produces an acceleration of 1m/s2 in a given mass, then the mass is
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a) 5√5 kg
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b) 10 kg
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c)7√2 kg
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d)125 kg
Explanation
Find magnitude of force then use formula F=ma to find mass Answer:(a)
Q.19
Forces of 30N, 40N and 50N acts along sides AB , BC and CA of an equilateral triangle ABC. The triangle is of mass 0.5 kg and kept in a verical plane as shown in figure, with the side AB vertical. The net vertical force acting on the triangle will be ( g=10m/s2)
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a) 125N
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b) 5N downwards
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c) 10N upwards
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d) 10N downwards
Explanation
By taking vertical components of all forces Net vertical force=50sin30 +40sin30-5 -30=10N upwards Answer: (c)
Q.20
A particle of mass 3kg slides down a smooth plane inclined at arcs in (1/3) to the horizontal. The acceleration of the particle is
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a)(1/3)g m/s2
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b) g m/s2
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c)1 m/s2
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d)0
Explanation
Answer: (a)
Q.21
Three forces of 30N, 40N and 50N are acting simultaneously on a mass of 10kg in the East, South and North East directions respectively. The acceleration of the mass will be
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a) 7.5 m/s2
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b) 6.55 m/s2
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c)20 m/s2
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d)12 m/s2
Explanation
Resultant force is verctor addition of all forces F=50cos45 i + 50sin45j + 30i -40j F=(50/√2)i + 50/√2j + 30i -40jAnswer: (b)
Q.22
A force F1 acting on a free mass m at rest produces in it acceleration of 1m/sAnother force F2 acting on the same mass at rest can produce in it a velocity of 10m/s after 5 s. The greatest acceleration of the mass m when both forces F1 and F2 acts together will be
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a) 2 m/s2
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b)4 m/s2
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c)3 m/s2
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d)4 m/s2
Explanation
F2 produces a velocity 10m/s after 5sec. IF acceleration produced by F2 is a then v=at, 10=5a, a=2 m/s2F1 produces acceleration of 1m/s2 Hence together can produce acceleration of 3m/s2 Answer:(c)
Q.23
A body of mass 1 kg is suspended from a spring balance graduated at g=10 m/sThe spring balance is fixed in a lift. Which is moving up with an acceleration of 5m/sWhat is the reading in the spring balance
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a) 0.5 kg
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b) 1.5 kg
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c) 1 kg
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d) 3.5 kg
Explanation
Resultant down ward acceleration=10+5=15 m/s2 Thus spring balance reads 1.5 kg Answer: (b)
Q.24
An inclined plane makes an angle 30° with the horizontal. A groove OA=5m cut in the plane makes an angle 30° with OX. A small smooth cylinder is free to slide down under the influence of gravity. The time taken by the cylinder to reach from A to O is ( g=m/s2)
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a)4 s
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b) 2 s
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c)2√2 s
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d)1 s
Explanation
Acceleration along CB=gsin30Acceleration along AO=gsin30cos60 From equation of motion s=½ at2 5=½ (gsin30cos60)t2t=2s Answer: (b)
Q.25
Two block of masses 4.4 kg and 4.8 kg are suspended by a string, which passes over a light frictionless pulley. Each block is 4m above the floor level initially. The time required for one of the block to reach the floor is
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a) 4 sec
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b) 0.4 sec
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c)8 sec
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d)4.3 sec
Explanation
Answer: (d)
Q.26
In the arrangement shown in the figure neglect the masses of the pully and string and also friction. The accelerations of block A and B are
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a)g, g/2
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b) g/2, g
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c)3g/2, 3g/4
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d)g, g
Explanation
since tension in the string is zero both blocks will have acceleration=g Answer:(d)
Q.27
Three equal weights A, B, C each of mass 2kg are hanging on a string passing over a fixed frictionless pulley as shown in figure. The tension in string connecting B and C is
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a) zero
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b) 13 N
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c) 3.3 N
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d) 19.6 N
Explanation
For mass C, resultant force on C=2g - T2 ∴ 2g - T2=2a --(i) For mass B, 2g+T2 - T1=2a --(ii) For mass A, T1 - 2g=2a --(iii) Adding (1), (2) and (3) a=2g/6=g/3 Tension T2=2g-2a=2g - 2g/3 T2=4g/ 3=(4×9.8)/3=39.2/3=13N Answer: (b)
Q.28
An elevator weighing 6000 kg is pulled upward by a cable with an acceleration of 5.0 m/s2 . The tension in cable is
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a)6000N
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b) 6000gN
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c)9000 N
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d)9000gN
Explanation
Acceleration of lift=Total force - gravitational force 6000 × 5.0=F - 6000×10 F=9000 N Thus tension in string=9000NAnswer: (c)
Q.29
A man thinks about four arrangement as shown to raise two small bricks each having mass m. Which of the arrangement would take minimum time ?
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a)
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b)
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c)
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d)
Explanation
Option(a) 2ma1=T2 - 2mg and 2T1=T2F=T1 from above equations Option(b) 2ma2=T2 - 2mg 2T1=T2F=2T1Option (c)2ma3=T-2mg T=F Option (d)ma3=T3 - mg T2=2T3 ∴ma3=T2 /2 - mg 2T3=T2 ma3=2T3 /2 - mg2T3=F ma3=F/2 - mg To raise brick in minium time acceleration should be maximum correct option is (a)Answer: (a)
Q.30
With what acceleration "a" should the box in figure descend so that a body of mass M placed in it exerts a force Mg/4 on the base of the box ?
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a)3g/4
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b)g/4
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c)g/2
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d)g/8
Explanation
If the box is accelerated downwards, from the frame outside the elevator, then resultant force on box M(g-a)=N given N=Mg/4 Thus M(g-a)=Mg/4a=g -g/4=3g/4 Answer:(a)
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