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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.
| 351. |
A 50 kg man is standing at the centre of a 30 kg platform A. Length of the platform is 10 m and coefficient of friction between the platform and the horizontal ground is 0.2. Man is holding one end of a light rope which is connected to a 50 kg box B. The coefficient of friction between the box and the ground is 0.5. The man pulls the rope so as to slowly move the box and ensuring that he himself does not move relative to the ground. If the shoes of the man does not slip on the platform, calculate how much time it will take for the man to fall off the platform. Assume that rope remains horizontal, and coefficient of friction between shoes and the platform is 0.6. |
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Answer» Correct Answer - `t = sqrt((10)/(3))s ` |
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| 352. |
A solid block of mass m = 1 kg is resting on a horizontal platform as shown in figure. The z direction is vertically up. Coefficient of friction between the block and the platform is `mu = 0.2`. The platform is moved with a time dependent velocity givenby `vecV = (2thati + thatj+ 3thatk)m//s` . Calculate the magnitude of the force exerted by the block on the platform. Take `g = 10 m//s^(2)` |
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Answer» Correct Answer - `sqrt(174)N` |
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| 353. |
A cylinder of mass M and radius r is suspended at the corner of a room. Length of the thread is twice the radius of the cylinder. Find the tension in the thread and normal force applied by each wall on the cylinder assuming the walls to be smooth. |
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Answer» Correct Answer - `T = sqrt(2) Mg ; N = (Mg)/(sqrt2)` |
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| 354. |
Due to air drag the falling bodies usually acquire a constant speed when the drag force becomes equal to weight. Two bodies, of identical shape, experience air drag force proportional to square of their speed `(F_("drag") = kv^(2)`, k is a constant). The mass ratio of two bodies is `1 : 4`. Both are simultaneously released from a large height and very quickly acquire their terminal speeds. If the lighter body reaches the ground in 25 s, find the approximate time taken by the other body to reach the ground. |
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Answer» Correct Answer - 12.5 s |
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| 355. |
Consider the following statements about the blocks shown in the diagram that are being pushed by a constant force on a frictionless table ltbegt (a) All blocks move with the same acceleration (b) The net force on each block is the same Which of these statements are/is correctA. A onlyB. B onlyC. Both A and BD. Neither A nor B |
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Answer» Correct Answer - A |
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| 356. |
when a body is stationary:A. there is no force acting on itB. the force acting on its are not in contact with itC. the combination of force acting on it balance each otherD. the body is in vacuum |
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Answer» Correct Answer - C When a body is stationary its acceleration is zero. It means net force acting on the body is zero. i.e., `sumvec(F)=0` . Or we can say that all the force acting balance each other. |
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| 357. |
In fig., blocks A and B move with velocities `v_(1)` and `v_(2)` along horizontal direction. Find the ratio of `v_(1)//v_(2)` A. `(sin alpha)/(sin beta)`B. `(sin beta)/(sin alpha)`C. `(cosbeta)/(cosalpha)`D. `(cosalpha)/(cosbeta)` |
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Answer» Correct Answer - D `-T cos beta V_(2)= Tcos alpha V_(1)=0` `cos beta V_(2) = cos alpha V_(1), (V_(2))/(V_(1)) =(cos alpha)/(cos beta)` . |
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| 358. |
Figure shows the displacement of a particle going along the X-axis as a function of time. The force acting on the particle is zero in the region A. ABB. BCC. CDD. DE |
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Answer» Correct Answer - A::C |
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| 359. |
A 400 kg ice boat moves on runners on essentially frictionless ice.A steady wind blows,applying a constant force to the sail. At the end of 8.0 sec run,the acceleration is `0.5m/s^(2)`. (a) What was the acceleration at the beginning of the run? (b) What was the force due to the wind ? (c) What retarding force must be applied at the end of 4.0 sec to bring the ice boat to rest by the end of the next 4 sec ? (assume boat was at rest at time t = 0) |
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Answer» Correct Answer - `[0.5 m//s^(2), 200 N, 400 N]` |
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| 360. |
When we jump out of a boat standing in water it movesA. ForwardB. BackwardC. SidewaysD. None of the above |
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Answer» Correct Answer - B |
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| 361. |
The system shown in figure is in equilibrium. Surface PQ of wedge A, having mass M, is horizontal. Block B, having mass 2M, rests on wedge A and is supported by a vertical spring. The spring balance S is showing a reading of `sqrt2` Mg. There is no friction anywhere and the thread QS is parallel to the incline surface. The thread QS is cut. Find the acceleration of A and the normal contact force between A and B immediately after the thread is cut. |
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Answer» Correct Answer - `a = (g)/(sqrt2) ; N_(AB) = 0` |
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| 362. |
Statement-1 : If the net external force on the body is zero then its acceleration is zero. Statement-2 : Acceleration does not depend on forceA. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - C |
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| 363. |
In which of the following cases the net force acting on the body is not zero?A. A drop of rain falling down with a constant speed.B. A cork of mass 10g floting on the surface of water.C. A car moving with a contant speed of `20kmh^(-1)` on a rough road.D. A pebble of mass `0.05kg` is thrown vertically upwards. |
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Answer» Correct Answer - D As the rain drop is falling down with a constant speed, its acceleration, `a=0` . Hence, net force on the `drop=ma=0` . As the cork is floting on the surface of water, its weight is balanced by the upthurt. Hence, net force on the cork is zero. The force exerted by the engine is balance by the friction due to rough road. as the car is moving with constant velocity, its acceleration `a=0` . Hence, net force on the car, `F=ma=0` . Whenever a body is thrown vertically upward gravitational pull of earth gives it a unform acceleration, `a=g` in downward directiion. Hence, net force on the `pebble=mg:0.05xx10=0.5N` vertically downwards. |
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| 364. |
A main revolves a stone of mass `m` tied to the end of a string in a circle of radius R The net force at the lowest and beight point of the circle directed vertical downward are Here`T_(1),T_(2)` and `(v_(1),v_(2))` denote the tension in the string (and the speed of the stone) at the lowest and highest points, respectively.A. `mg-T_(1)` and `mg+T_(2)`B. `mg+T_(1)` and `mg-T_(2)`C. `mg+T_(1)-((mv_(1)^(2))/(R))` and `mg-T_(2)+((mv_(2)^(2))/(R))`D. `mg-T_(1)-((mv_(1)^(2))/(R))` and `mg+T_(2)+((mv_(2)^(2))/(R))` |
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Answer» Correct Answer - A At the lowest point, mg acts downwards and `T_(1)` upwards so that net force `=mg-T_(1)` . At the highest point, both mg and `T`, act downwards so that net force `=mg+T_(2)` . Hence, option (a) is correct. |
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| 365. |
A body of mass 50 grams is moving with a cosntant velocity 2 cm/s on a horizontal frictionless table. The force on the table is:-A. 39,200 dynes`160 dynessB. 80 dynesC. Zero dynesD. None of the above |
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Answer» Correct Answer - A `a=0, therefore F=0` |
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| 366. |
The statement 'acceleration is zero if and only if the net force is zero' is valid in .A. non-inertial framesB. inertial framesC. both in inertial frames and non-inertial framesD. neither inertial frames non-inertial frames |
| Answer» Correct Answer - B | |
| 367. |
In the figure given below, the position-time graph of a particle of mass `0.1kg` is shown. The impuslse at `t=2` sec is A. `0.2"kg m sec"^(-1)`B. `-0.2"kg m sec"^(-1)`C. `0.1"kg m sec"^(-1)`D. `-0.4:kg m sec"^(-1)` |
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Answer» Correct Answer - B |
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| 368. |
In the figure given below, the position time graph of a particle of mass 0.1kg is shown. The impulse at t=2 sec is A. `0.2kgms^(-1)`B. `0.2kgms^(-2)`C. `0.1 kgms^(-1)`D. `-0.4 kgms^(-1)` |
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Answer» Correct Answer - B `vecI=Deltavecp=vecp_(f)-vecp_(i)` `=0.-0.1=0.2ks ms^(-1)` |
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| 369. |
In the above question, the force acting on the object isA. 30 NB. `-30N`C. 3 ND. `-3 N` |
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Answer» Correct Answer - B |
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| 370. |
In the above question, the acceleration of mass m isA. `(F)/(m)`B. `(F-T)/(m)`C. `(F+T)/(m)`D. `(F)/(M)` |
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Answer» Correct Answer - B |
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| 371. |
In the above question, the impulse acting on the object isA. `120"newton"xxsec`B. `-120 "newtont sec"`C. `30 "newton"xxsec`D. `-30"newton"xxsec` |
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Answer» Correct Answer - B |
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| 372. |
Block B of mass m has been placed on block A of mass 3 m as shown. Block A rests on a smooth horizontal table. `F_(1)` is the maximum horizontal force that can be applied on the block A such that there is no slipping between the blocks. Similarly, `F_(2)` is the maximum horizontal force that can be applied on the block B so that the two blocks move together without slipping on each other. When `F_(1) "and" F_(2)` both are applied together as shown in figure. (a) Find the friction force acting between the blocks. (b) Acceleration of the two blocks. (c) If `F_(2)` is decreased a little, what will be direction of friction acting on B. |
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Answer» Correct Answer - (a) Zero (b) `(F_(1))/(3m) = (F_(2))/(m)` (c) To right |
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| 373. |
A nuclide at rest emits an alpha-particle. In this process:A. alpha-particle moves with large velocity and the nuclesus remains at restB. both alpha-particle and nucleus move with equal speed in opposite directionC. both move in opposite direction but nucleus with greater speedD. both move in opposite direction but alpha-particle with greater speed. |
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Answer» Correct Answer - D As `F_(ext)=0` , hence `vec(P)_(s)` =constant or `vec(P)_(alpha)+vec(P)_(N)` =constant Because nuclide was at rest brfore emission of alpha particle, hence acording to law of conservation of momentum `vec(P)_(alpha)+vec(P)_(N)=0` . or `m_(alpha)vec(v)_(alpha)+m_(N)vec(v)_(N)=0` or `vec(v)_(alpha)=-(m_(N))/(m_alpha)vec(v)_(N)` . Thus, alpha-particle and the nuncleus move in opposite directions. Further, as `m_(N)gtgtm_(oo)` hence `v_(alpha)gtgtv_(N)` . |
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| 374. |
A body at rest breaks into two pieces of equal masses. The parts will moveA. move in the same direction with equal speedsB. move in any direction with any speedC. move in opposite direction with equal speedsD. move in opposite direction with unequal speeds |
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Answer» Correct Answer - C As `F_(ext)=0` , hence `vec(P)_(1)+vec(P)_(12)` =constant Because the system was at rest initally, hence `vec(p)_(1)+vec(p)_(2)=0` or `vec(p)_(2)=-vec(p)_(1)` or `m_(2)vec(v)_(2)=-m_(1)vec(v)_(1)` Because `m_(1)=m_(2)` hence `vec(v)_(2)=-vec(v)_(1)` . |
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| 375. |
A body, under the action of a force `vec(F)=6hati -8hatj+10hatk` , acquires an acceleration of `1ms^(-2)` . The mass of this body must be.A. `10sqrt(2)kg`B. `2sqrt(10)kg`C. `10kg`D. `30kg` |
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Answer» Correct Answer - A Here, `F=sqrt(6^(2)+(-8)^(2)+10^(2))=10sqrt(2)` `m=(F)/(a)=(10sqrt(2))/(1)=10sqrt(2)kg` |
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| 376. |
Assertion: A man in a dosed cabin falling freely does not experience gravity. Reason: Inertial and gravitational mass have equivalence.A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false |
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Answer» Correct Answer - A Weight less state, both mases same thus forces cancle. |
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| 377. |
In the arrangement shown in the fig. a monkey of mass M keeps itself as well as block A at rest by firmly holding the rope. Rope is massless and the pulley is ideal. Height of the monkey and block A from the floor is h and 2h respectively [h = 2.5 m] (a) The monkey loosens its grip on the rope and slides down to the floor. At what height from the ground is block A at the instant the money hits the ground? (b) Another block of mass equal to that of A is stuck to the block A and the system is released. The monkey decides to keep itself at height h above the ground and it allows the rope to slide through its hand. With what speed will the block strike the ground? (c) In the situation described in (b), the monkey decides to prevent the block from striking the floor. The monkey remains at height h till the block crosses it. At the instant the block is crossing the monkey it begins climbing up the rope. Find the minimum acceleration of the monkey relative to the rope, so that the block is not able to hit the floor. Do you think that a monkey can climb with such an acceleration? (`g = 10ms^(-2)`) |
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Answer» Correct Answer - (a) The block is at height h = 2.5 (b) `V = 5sqrt(2) m//s` `25 m//s^(2) uparrow` |
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| 378. |
A body of mass `1.0` kg is falling with an acceleration of 10 `m//s^(2)` . Its apparent weight will be `(g=10m//sec^(2))`A. `1.0` kg wtB. `2.0` kg wtC. `0.5` kg wtD. Zero |
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Answer» Correct Answer - D |
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| 379. |
An ideal spring is in its natural length (L) with two objects A and B connected to its ends. A point P on the unstretched spring is at a distance `(2L)/(3)` from B. Now the objects A and B are moved by 4 cm to the left and 8 cm to the right respectively. Find the displacement of point P. |
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Answer» Correct Answer - Zero |
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| 380. |
In the arrangement shown in the fig. the pulley, the spring and the thread are ideal. The spring is stretched and the two blocks are in contact with a horizontal platform P. When the platform is gradually moved up by 2 cm the tension in the string becomes zero. If the platform is gradually moved down by 2 cm from its original position one of the blocks lose contact with the platform. Given M = 4 kg, m = 2 kg. (a) Find the force constant (k) of the spring (b) If the platform continues to move down after one of the blocks loses contact, will the other block also lose contact? Assume that that the platform moves very slowly. |
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Answer» Correct Answer - (a) K = 2.5 N/cm No |
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| 381. |
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. A constantB. Proportional to `t^(2)`C. Inversely proportional to `t`D. Proportional to `t` |
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Answer» Correct Answer - D |
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| 382. |
Assertion : Force is required to move a body uniformly along a circle. Reason : When the motion is uniform, acceleration is zero.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - B |
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| 383. |
An experimenter is inside a uniformly accelerated train. Train is moving horizontally with constant acceleration `a_(0)`. He places a wooden plank AB in horizontal position with end A pointing towards the engine of the train. A block is released at end A of the plank and it reaches end B in time `t_(1)`. The same plank is placed at an inclination of `45^(@)` to the horizontal. When the block is released at A it now climbs to B in time`t_(2)`. It was found that `(t_(2))/(t_(1))= 2^((5)/(4))`. What is the coefficient of friction between the block and the plank? |
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Answer» Correct Answer - `mu = ((3a_(0) - 4g)/(4a_(0) + 3g))` |
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| 384. |
Aeroplanes are streamlined to reduce .A. fluid frictionB. sliding frictionC. kinetic frictionD. limiting friction |
| Answer» Correct Answer - A | |
| 385. |
The quantity of motion of a body is best represented by .A. its massB. its speedC. its velocityD. its linear momentum |
| Answer» Correct Answer - D | |
| 386. |
Which of the following is the most significant law of motion given by Newton ? .A. First law of motionB. Second law of motionC. Third law of motionD. Zeroth law of motion |
| Answer» Correct Answer - B | |
| 387. |
A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to (a). the upper block, (b). the lower block. Take g=10 m/`s^2`.A. `2ms^(-2), 2ms^(-2)`B. `2ms^(-2), 1ms^(-2)`C. `3ms^(-2), 1ms^(-2)`D. `4ms^(-2), 1ms^(-2)` |
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Answer» Correct Answer - A `F` is greater than `mu_(s)` mg The blocks move with different acceleration `a_(u) =mu_(k) g, a_(L) = ((F -mu_(k)m_(u)g))/(m_(L))` . |
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| 388. |
Sand is dusted to the railway tracks during rainy season to .A. make it always wetB. increase frictionC. to reduce consumption of fuelD. make it always dry |
| Answer» Correct Answer - B | |
| 389. |
When forces `F_(1)` , `F_(2)` , `F_(3)` are acting on a particle of mass m such that `F_(2)` and `F_(3)` are mutually prependicular, then the particle remains stationary. If the force `F_(1)` is now rejmoved then the acceleration of the particle isA. `F_(1)//m`B. `F_(2)F_(3)//mF_(1)`C. `(F_(2)-F_(3))//m`D. `F_(2)//m` |
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Answer» Correct Answer - A For equilibrium of system, `F_(1)=sqrt(F_(2)^(2)+F_(3)^(2)` . As `theta=90^(@)` . In the absence of force `F_(1)` , Acceleration`=("Net force")/("Mass")` . `=sqrt(F_(2)^(2)+F_(3)^(2))/m=F_(1)/(m)` . |
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| 390. |
Block is attached to system of springs. Calculate equivalent spring constant. .A. `K`B. `2K`C. `3K`D. `4K` |
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Answer» Correct Answer - B `(1)/(K_(eq))=(1)/(4K)+(1)/(4K)=(1)/(2K),K_(eq)=2K` |
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| 391. |
Which of the following conclusion is correct regarding a stationary body?A. No force is acting on the body.B. Vector sum of force acting on the body is zero.C. The body is in vacuum.D. The force acting on the body, do not constitute a couple. |
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Answer» Correct Answer - B Net force on the body is zero. |
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| 392. |
A bullet is fired from a gun The force on a bullet is `F =600 -2 xx 10^(5) t` newton. The force reduces to zero just when the bullet leaves barrel Find the impulse imparted to the bullet . |
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Answer» `F =600 -2 xx 10^(5) t` F becomes zero as soon as the bullet leaves the barrel ` 0=600 -2 xx 10^(5) t rArr 600 =2 xx 10^(5) t` `t =3 xx 10^(3) s rArr` Impulse ` =underset(0)overset(t)int Fdt` `=underset(0)overset(t)int(600 -2 xx 10^(5)t) dt =[600t -2 xx 10^(5) (t^(2))/(2)]_(0)^(3xx10^(-3))` `=600 xx 3 xx 10^(-3) -10^(5) xx 9 xx 10^(-6) =0.9 Ns` . |
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| 393. |
Consider a car moving on a straight road with a speed of `100m//s`. The distance at which car can be stopped is `[mu_k=0.5]`A. `800 m`B. `1000 m`C. `100 m`D. `400 m` |
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Answer» Correct Answer - B Retardation due to friction `=mug` we know `v^(2) =u^(2) + 2as` `:. 0 = (100)^(2) -2 (mug)s` or `2mugs =100 xx 100` or `s = (100xx 100)/(2xx 0.5 xx 10) =1000m` . |
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| 394. |
In order to stop a car in shortest disatance on a horizontal road one shouldA. apply the brakes very hard so that the wheels stop rotating .B. apply the brakes hard enough to just prevent slippingC. pump the brakes (press and release)D. shut the engine off and not apply brakes . |
| Answer» Correct Answer - B | |
| 395. |
A body of mass `8kg` is moved by a force `F = (3x)N`, where x is the disatance covered Initial position is `x =2m` and final position is `x =10m` If initially the body is at rest find the final speed . |
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Answer» `F =ma rArr F =m (dv)/(dt) rArr 3 x = m (dv)/(dx) (dx)/(dt)` `3x = 8 (dv)/(dx) v rArr 3xdx = 8vdv` `3int_(2)^(10)xdx=int_(0)^(v)vdxrArr3[x^(2)/2]_(2)^(10) = [v^(2)/(2)]_(0)^(v)` `3[100-4] = 8 v^(2) rArr v^(2) = (3 xx 96)/(8) = 36 rArr v = 6 ms^(-1)` |
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| 396. |
The motion of a rocket is based on the principle of conservation ofA. MassB. Kinetic energyC. Linear momentumD. Angular momentum |
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Answer» Correct Answer - C |
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| 397. |
Assertion: A rocket works on the principle of conservation of linear momentum. Reason: Wheneven there is a change in momentum of one body, the same change occurs in the momentum of the second body of the same system but in the opposite directio.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - A |
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| 398. |
Assertion: A rocket works on the principle of conservation of linear momentum. Reason: Wheneven there is a change in momentum of one body, the same change occurs in the momentum of the second body of the same system but in the opposite directio.A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are ture but reason is not the correct explanation of assertion.C. If assertion is ture but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - A As the fuel in rocket undergoes combution, the gases so produced leave the body of the rocket with large velocity and give upthrust to the rocket. If we assume that the fuel is burnt at a constant rate, then the rate of change of momentum of the rocket will be constant. As more and more fuel gets burnt, the mass of the rocket goes on decreasing and it leads to increase of the velocity of rocket more and more rapidly. |
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| 399. |
Which of the following sets of concurrent force may be in equilibrium?A. `F_(1)=3N,F_(2)=5N,F_(3)=9N`B. `F_(1)=3N,F_(2)=5N,F_(3)=1N`C. `F_(1)=3N,F_(2)=5N,F_(3)=15N`D. `F_(1)=3N,F_(2)=5N,F_(3)=6N` |
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Answer» Correct Answer - D |
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| 400. |
A car of mass m starts from rest and acquires a velocity along east `upsilon = upsilonhati (upsilon gt 0)` in two seconds Assuming the car moves with unifrom acceleration the force exerted on the car is .A. `(mv)/(2)` eastward and is exerted by the car engine.B. `(mv)/(2)` eastward and is due to the friction on the tyres exerted by the road.C. more than `(mv)/(2)` eastward exerted due to the engine and overcomes the frictions of the road.D. `(mv)/(2)` exerted by the engine. |
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Answer» Correct Answer - B Given, mass of the car =m As car starts from rest `u =0` Velocity acquired along east `=vhati` Duration ` =t =2s` We know `v =u + at` `rArr vhati = 0 + a xx 2` `rArr a = (v)/(2) hati` Force `F = ma = (mv)/(2) hati` Hence, force acting on the car is `(mv)/(2)` towards east As external force on the system is only friction hence the force `(mv)/(v)` is by friction Hence, force by egine is internal force . |
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