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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 551. |
A good lubricant should be highlyA. viscousB. non-volatileC. both (1and2)D. transparent |
| Answer» Correct Answer - C | |
| 552. |
At a place where the acceleration due to gravity is `10"m sec"^(-2)` a force of 5 kg - wt acts on a body of mass 10 kg initially at rest. The velocity of the body after 4 second isA. `5 "m sec"^(-1)`B. `10"m sec"^(-1)`C. `20"m sec"^(-1)`D. `50"m sec"^(-1)` |
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Answer» Correct Answer - C |
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| 553. |
The coefficient of static friction between two surfaces depend onA. nature of surfaceB. the shape of the surface in contactC. the area of contactD. all of the above |
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Answer» Correct Answer - A Friction force is free from contact area. |
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| 554. |
Maximum value of static friction is .A. limiting frictionB. rolling frictionC. static frictionD. normal reaction |
| Answer» Correct Answer - A | |
| 555. |
The acceleration of a particle is found to be non zero when no force acts on the particle. This is possible if the measurement is made fromA. inertial frameB. non inertial frameC. bothD. some times intertial (or) some times non inertial |
| Answer» Correct Answer - B | |
| 556. |
Frictional force between two bodiesA. increases the motion between the bodiesB. destroys the relative motion between the bodiesC. sometimes helps and sometimes opposes the motionD. incerases the relative velocity between the bodies |
| Answer» Correct Answer - C | |
| 557. |
A ball of mass `0.5 kg` moving with a velocity of `2 m//s` strikes a wall normally and bounces back with the same speed . If the time of contact between the ball and the wall is 1 millisecond , the average force exerted by the wall on the ball isA. 2000 NB. 1000 NC. 5000 ND. 125 N |
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Answer» Correct Answer - A |
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| 558. |
The linear momentum `p` of a body moving in one dimension varies with time according to the equation `p=a+bt^(2)`, where a and b are positive constants. The net force acting on the body isA. Proportional to `t^(2)`B. A constantC. Proportional to `t`D. Inversely proportional to `t` |
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Answer» Correct Answer - C |
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| 559. |
A mass of `1kg` attached to one end of a string is first lifted up with an acceleration `4.9m//s^(2)` and then lowered with same acceleration What is the ratio of tension in string in two cases . |
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Answer» When mass is lifted up with acceleration `4.9m//s^(2)` `T_(1) =m(g +a) =1 (9.8 +4.9) =14.7N` When mass is lowered with same acceleration `T_(2) =m(g -a) =1(9.8 -4.9) =4.9N` `:. (T_(1))/(T_(2)) = (14.7)/(4.9) =3:1` . |
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| 560. |
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 toA. `0.3` NB. 30 NC. 300 ND. 3000 N |
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Answer» Correct Answer - B |
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| 561. |
A stretching force of `10N` is applied at one end of a spring balance and an equal force is applied at the other end at the same time The reading of the balance is .A. `5N`B. `10N`C. `20N`D. `0` |
| Answer» Correct Answer - B | |
| 562. |
The motion of a particle of mass m is given by `x=0` for `tlt0s`,`x(t)=Asin4pit` for `0lttlt((1)/(4))s(Agt0)`,and `x=0` for `tgt((1)/(4))s`.A. The force at `t = (1//8)s` on the particle is `m16pi^(2)A`.B. The particle is acted upon by an impulse of magnitude `m4 pi^(2) A` at `t =0s` and `t = (1//4)s`.C. The particle is not acted upon by any force.D. The particle is not acted upon by a constant |
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Answer» Correct Answer - A::B::D Given `x =0` for `t lt 0s` `x (t) =A sin 4pit, for 0 lt t lt (1)/(4)s` `x =0,` for `t gt (1)/(4)s` For `0 lt t lt (1)/(4) s` `v (t) = (dx)/(dt) =4pi A cos 4pi t` `a (t) = (dv(t))/(dt) = - 16pi^(2) a sin 4pit` `At = (1)/(8) s,a (t) = - 16pi^(2) A sin 4pi xx (1)/(8) = - 16pi^(2) A` `F =ma (t) = - 16pi^(2) A xx m = - 16 pi^(2) mA` impulse =Change in linear mo-mentum `I =F xx t = (-16 pi^(2) Am) xx (1)/(4)` ` =- 4pi^(2) Am` the impulse (Change in linear momentum) `at t =0` is same `as, t = (1)/(4)s` Clearly force depends upon A which is not constant Hence forces is also not constant . |
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| 563. |
The centripetal force required by a 1000 kg car that takes a turn of radius `50m` at a speed of 36kmph is .A. `1000N`B. `3500N`C. `1600N`D. `2000N` |
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Answer» Correct Answer - D `F = (mv^(2))/(r ) ` |
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| 564. |
A body under the action of a force `vec(F)=6hat(i)-8hat(j)N` acquires an acceleration of `5ms^(-2)` . The 10s mass of the body isA. 2kgB. 5kgC. 4kgD. 6kg |
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Answer» Correct Answer - A Given: `vec(F)=6hati-8hatjN` , `a=5ms^(-2)` . :. `Fsqrt((6)^(2)+(-8)^(2))=10N` `"Mass of the body"`, `m=(F)/(a)=(10N)/(5ms^(-2))=2kg` . |
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| 565. |
A man is standing at the center of frictionless pond of ice. How can he get himself to the shore ?A. By throwing his shirt in vertically upward directionB. By spitting horizontallyC. He will wait for the ice to melt in pondD. Unable to get at the shore |
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Answer» Correct Answer - B |
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| 566. |
A body of mass 5 kg is suspended by a spring balance on an inclined plane as shown in figure. The spring balance measure A. 50 NB. 25 NC. 500 ND. 10 N |
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Answer» Correct Answer - B |
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| 567. |
Three light rods from a right angled triangle. The tension in the rod PR, if a force of `300N` is applied vertically downward at `R` is. A. `400N`B. `200N`C. `300N`D. `500N` |
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Answer» Correct Answer - D Here `(PQ)/(T)=(PR)/(T)=(PQ)/(300)` `T_(PR)=500N` |
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| 568. |
Rocket engines lift a rocket from the earth surface because hot gas with high velocityA. Push against the earthB. Push against the airC. React against the rocket and push it upD. Heat up the air which lifts the rocket |
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Answer» Correct Answer - C |
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| 569. |
A man fires a bullet of mass 200 g at a speed of 5 `m //s` . The gun is of one kg mass. by what velocity the gun rebounds backwardsA. `0.1m//s`B. `10m//s`C. `1m//s`D. `0.01m//s` |
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Answer» Correct Answer - C |
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| 570. |
In the above question, the acceleration of the car will beA. `0.25m//sec^(2)`B. `2.5m//sec^(2)`C. `5.0m//sec^(2)`D. `0.25m//sec^(2)` |
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Answer» Correct Answer - D |
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| 571. |
The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration ‘ a ’ is `3 : 2`. The value of ‘ a ’ is ( g - Acceleration due to gravity of the earth)A. `(g)/(3)`B. `(g)/(2)`C. `g`D. `(4)/(3)g` |
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Answer» Correct Answer - A `(mg)/(m(g-a))=(3)/(2)` `2g=3(g-a)` `a=(g)/(3)` |
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| 572. |
An object is subjected to a force in the north-east direction. To balance this force, a second force should be applied in the directionA. North-EastB. SouthC. South-WestD. West |
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Answer» Correct Answer - C |
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| 573. |
When travelling freely a train is subjected to resistances which vary directly as the velocity and at 9= kph this is equal to 1 percent of the weight of the train. The brakes when applied create a further resistance equal to `l//16th` of the weight of the train. If the brakes are suddenly applied when the velocity is 90kph find the time and distance travelled before the train comes to rest. |
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Answer» Correct Answer - [3.7 sec, 467 m] |
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| 574. |
A body of mass `10kg` is moving with a constant velocity of `10m//s`. When a constant force acts for 4 seconds on it, it moves with a velocity `2m//sec` in the opposite direction. The acceleration produced in it isA. `3 m//s^(2)`B. `-3m//s^(2)`C. `0.3 m//s^(2)`D. `-0.3 m//s^(2)` |
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Answer» Correct Answer - B `a=(-2-10)/(4)=(-12)/(4)=-3m//s^(2)` |
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| 575. |
The velocity of a body at time `t = 0` is `10sqrt2m//s` in the north-east direction and it is moving with an acceleration of `2m//s` directed towards the south. The magnitude and direction of the velocity of the body after 5 sec will beA. `10m//s`, towards eastB. `10m//s`,towards northC. `10m//s`, towards southD. `10m//s`, towards north-east |
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Answer» Correct Answer - A |
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| 576. |
When 1 N force acts on 1 kg body that is able to move freely, the body receivesA. A speed of `1m//sec`B. An acceleration of `1m//sec^(2)`C. An acceleration of `980cm//sec^(2)`D. An acceleration of `1cm//sec^(2)` |
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Answer» Correct Answer - B |
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| 577. |
A body of mass `10kg` is moving with a constant velocity of `10m//s`. When a constant force acts for 4 seconds on it, it moves with a velocity `2m//sec` in the opposite direction. The acceleration produced in it isA. `3m//sec^(2)`B. `-3m//sec^(2)`C. `0.3m//sec^(2)`D. `-0.3m//sec^(2)` |
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Answer» Correct Answer - B |
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| 578. |
Find the acceleration of 3 kg mass when acceleration of 2 kg mass is `2ms^(-2)` as shown in figure-2.140 A. `3ms^(-2)`B. `2ms^(-2)`C. `0.5ms^(-2)`D. Zero |
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Answer» Correct Answer - B |
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| 579. |
In the last problem take M = m and `m_(0) = 2` m and calculate the acceleration of the wedge. |
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Answer» Correct Answer - `a_(0) = (48g)/(199)` |
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| 580. |
In the system shown, the initial acceleration of the wedge of mass 5M is (there is no friction anywhere) A. ZeroB. `2g//23`C. `3g//23`D. None |
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Answer» Correct Answer - B |
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| 581. |
The system shown in the fig. is in equilibrium. Pulleys A and B have mass M each and the block C has mass 2M. The strings are light. There is an insect (D) of mass M/2 sitting at the middle or the right string. Insect does not move. (a) Just by inspection, say if the tension in the string S1 is equal to, more than or less than 9/2 Mg. (b) Find tension in the string S2, and S1. (c) Find tension in S2 if the insect flies and sits at point E on the string. |
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Answer» Correct Answer - (a) More than 9/2 Mg (b) Tension in S2 = Mg/2 , Tension is S1 = 5 Mg (c) Tension in S2 = Mg/7 |
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| 582. |
The blocks B and C in the figure have mass m each. The strings AB and BC are light, having tensions `T_(1)` and `T_(2)` respectively. The system is in equilibrium with a constant horizontal force mg acting on C.A. `tantheta_(1)=(1)/(2)`B. `tantheta_(2)=1`C. `T_(1)=sqrt(5)mg`D. `T_(2)=sqrt(2)mg` |
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Answer» Correct Answer - A::B::C::D |
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| 583. |
A brick of mass 2kg begins to slide down on a plane inclined at an angle of `45^(@)` with the horizontal. The force of friction will beA. `19.6 sin 45^(@)`B. `9.8 sin 45^(@)`C. `19 .6 cos 45^(@)`D. `9.8 cos 45^(@)` |
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Answer» Correct Answer - A `f =mg sin theta` |
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| 584. |
A long helix made of thin wire is held vertical. The radius and pitch of the helix are R and `rho` respectively. A bead begins to slide down the helix. (a) Find the normal force applied by the wire on the bead when the speed of the bead is `upsilon`. (b) Eventually, the bead acquires a constant speed of `v_(0)`. Find the coefficient of friction between the wire and the bead. |
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Answer» Correct Answer - (a) `mg "cos"theta sqrt(1 + ((v^(2) "cos"theta)/(Rg))^(2)))` (b) `("tan"theta)/(sqrt(1 + ((v^(2) "cos"theta)/(Rg))^(2)))` where `"tan"theta = (rho)/(2piR)` |
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| 585. |
An astronaut accidentally gets separated out his small spaceship accelerating in interstellar space at a constant rate of `100ms^(-2)` . What is the acceleration of the astronaut the instant after he is outside the spaceship? (Assume that there are no nearby stars to exert gravitional force on him)A. zeroB. `10ms^(-2)`C. `50ms^(-2)`D. `100ms^(-2)` |
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Answer» Correct Answer - A Since there are no nearby stars to exert gravitational force on him and the small spaceship exerts negligible gravitational attraction on himm, the net force acting on the astronaut, once he is out of the spaceship, is zero. In accordance with the first law of motion the acceleration of the astronaut is zero. |
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| 586. |
If the above diagram initially there is no elongation in spring if the block is displaced towards right by `x_(0)`. Calculate the elongation of spring A. .A. `(3)/(7)x_(0)`B. `(x_(0))/(4)`C. `(x_(0))/(7)`D. `(x_(0))/(3)` |
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Answer» Correct Answer - A `(1)/(K_(eq))=(1)/(3K)+(1)/(4K)=(7)/(12K),T=(12K)/(2)x_(0)` `T =4K x_(1), (12)/(7) K//x_(0) =4K//x_(1),x_(1)=(3)/(7)x_(0)` . |
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| 587. |
The masses of `10kig` and `20kg` respectively are connected by a massless spring as shown in figure. A force of `200N` acts on the `20kg` mass. At the instant shown, the `10kg` mass has acceleration `12m//sec^(2)` . What is the acceleration of `20kg` mass? A. `12m//sec^(2)`B. `4m//sec^(2)`C. `10m//sec^(2)`D. Zero |
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Answer» Correct Answer - B As the mass of `10kg` has acceleration `12m//s^(2)` therefore it apply `120N` force on mass `20kg` in a backward direction. :. Net forward force on `20kg` mass `=200-120=80N` :. acceleration of the block is `=(4kx)/(m)` . |
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| 588. |
A heavy particle of mass `1kg` suspended from a massless string attached to a roof. A horizontal force F is applied to the particle such that in the equilibrium position the string makes and angle `30^(@)` with the vertical. The magnitude of the force F equals .A. 10 NB. `10 sqrt(3) N`C. 5 ND. `10//sqrt(3) N` |
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Answer» Correct Answer - D |
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| 589. |
The string between blocks of mass m and 2m is massless and inextensible. The system is suspended by a massless spring as shown. If the string is cut find the magnitudes of accelerations of mass 2m and m (immediately after cutting) A. `g,g`B. `g ,(g)/(2)`C. `(g)/(2),g`D. `(g)/(2),(g)/(2)` |
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Answer» Correct Answer - C For `m_(1), F =ma =m_(1) g,` for `m_(2),T -m_(2)g =m_(2) a^(1)` . |
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| 590. |
12N of force required to be applied on A to slip on B. Find the maximum horizontal force F to be applied on B so that A and B moves together. |
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Answer» Let `mu` be friction coefficient between A and B As 12N force on A is required for slipping so `mum_(A)g=12rArrmu=0.3` maximum force `(F_(B))` applied on B so that `A & B` move together. `F_(B)=(m_(A)+m_(B))a "where" a=mugrArrF_(B)=(m_(A)+m_(B))mug=(4+5)(0.3)(10)=27N` |
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| 591. |
Two blocks of masses `m_(1)` and `m_(2)` are connected through a massless inextensible string .A block of mass `m_(1)` is placed at the fixed rigid inclined surface while the block of mass `m_(2)` hanging at the other end of the string which is passing through a fixed massless frictionless pulley shown in figure The coefficient of static friction between the block and the inclined plane is `0.8` The system of masses `m_(1)` and `m_(2)` released from rest A. the tension in the string is 20 N after releasing the systemB. the contact force by the inclined surface on the block is along normal to the inclined surfaceC. the magnitude ofcontact force by the inclined surface on the block `m_(1)`, is `20sqrt(3) N`D. None of these |
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Answer» Correct Answer - A::B::C |
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| 592. |
The fricition coefficient between the table and block shown in fig. is 0.2 find the acceleration of the system (Take `g=10 m//s^(2)`) |
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Answer» For three blocks the equations of motion are- `a=(15g-5g-g)/(25)=((9g)/(25))` `15g-T_(1)=15arArrT_(1)=15g-(135g)/(25)=(375g-135g)/(25)=96N` `T_(2)-5g=5((9g)/(25))rArr T_(2)=5g+(9g)/(5)=(34g)/(5)=68N` |
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| 593. |
If the two blocks moves with a constant uniform speed then find coefficient of fricition between the surface of the block B and the table. The spring is massless and the pulley is fricitionless. |
| Answer» For block B `T=f=mum_(2)g &` For block `A T=m_(1)g` ltbr By solving above equation `mu=(m_(1))/(m_(2))` | |
| 594. |
A monkey of mass 40 kg climbs on a rope which can withstand a maximum tesion of 600 N In which of the following cases will the rope break ? The monkey (a) climbs up with an acceleration of `6 ms^(-2)` (b) climbs down with an acceleration of `4 ms^(-2)` (c) climbs up with a unifrom speed of `5 ms^(-1)` (d) falls down the rope nearly freely under gravity Ignore the mass of the rope . |
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Answer» (a) `T-40g=240 rArr T=632N` (b) `392-T=160 rArr=232N` (c) T=392N The rope will break in case (a) as `Tg600N`. |
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| 595. |
A suitcase is gently dropped on a conveyor belt moving at `3ms^(-1)` If the coefficient of friction between the belt and suitcase is `0.5` how far will the suitcase move on the belt before coming to rest ?A. `2.7m`B. `1.8m`C. `0.9m`D. `1.2m` |
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Answer» Correct Answer - C `a =mug, v =u+at, s_(1) = ut + (1)/(2) at^(2), s_(2) =vt` `S_("rel") =S_(1) -S_(2)` . |
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| 596. |
The resultant of two forces 3P and 2P is R. If the first force is doubled then resultant is also doubled.The angle between the two forces isA. `60^(@)`B. `120^(@)`C. `70^(@)`D. `180^(@)` |
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Answer» Correct Answer - B |
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| 597. |
A block of mass 2 kg is placed on the floor . The coefficient of static friction is `0.4` . If a force of `2.8` N is applied on the block parallel to floor , the force of friction between the block and floor is : (Taking g = `10 m//s^(2)`)A. 2.8NB. 8NC. 2.0ND. zero |
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Answer» Correct Answer - A `f_("max")=20xx0.4=8N` `therefore f=2.8N` |
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| 598. |
A `30 kg` mass is initially at rest on floor of a truck . The coefficient of static friction between the mass and the floor of truck in `0.3` and coefficient of kinetic friction is `0.2` . Initially the truck is travelling due east at constant speed.Find the magnitude and direction of the friction force acting on the mass , if : (take `g = 10 m//s^(2))` (a) the truck acceleration at `1.8 m//s^(2)` eastward (b) the truck acceleration at `3.8 m//s^(2)` westward |
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Answer» Correct Answer - [54 N, 59 N] |
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| 599. |
Given in figure are two blocks A and B of weight 20N and 100N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is: A. `100N`B. `80N`C. `120N`D. `150N` |
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Answer» Correct Answer - C `F.B.D` of block `A` and block `B` |
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| 600. |
A block of mass m placed on an incline just begins to slide when inclination of the incline is made `theta_(0) = 45^(@)`. With inclination equal to `theta = 30^(@)`, the block is placed on the incline. A horizontal force (F) parallel to the surface of the incline is applied to the block. The force F is gradually increased from zero. At what angle a to the force F will the block first begin to slide? |
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Answer» Correct Answer - `sin^(-1)((1)/(sqrt3))` |
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