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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.
| 251. |
A monkey of 25 kg is holding a vertical rope. The rope does not break if a body of mass 30 kg is suspended with the rope exceeds 30 kg. What will be the maximum acceleration with which the monkey can climb up along the rope ? ( Take ,`=10 ms^(-2)`)A. `2 ms^(-2)`B. `2.5 ms^(-2)`C. `3 ms^(-2)`D. `4 ms^(-2)` |
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Answer» Correct Answer - A (a) The maximum weight which can be suspended with the rope without breaking it =30 kg-wt. `=30xx10=300N` `therefore 300N=mg+ma` `implie ma=300- mg` `=300-25xx10=50N` `implies a=(50)/(m)=(50)/(25)=2 ms^(-2)` |
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| 252. |
A system is shown in the figure. A man standing on the block is pulling the rope. Velocity of the point of string in contact with the hand of the man is `2 m//s` downwards. The velocity of the block will be [assume that the block does not rotate] A. `3 m//s`B. `2 m//s`C. `1//2 m//s`D. `1 m//s` |
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Answer» Correct Answer - B |
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| 253. |
Table -2 gives certain situatios involving two blocks of mass 2 kg and 4 kg the 4 kg block lies on a amooth horizontal table there is sufficiet friction between both the block and there is no relatve motion between both the block in all situations. Horizontal forces on one or both blocks in all situations . Horizontal forces act on one o rboth blocks as shown . table -1 given certain statement realate to figures given in table -2 Match the statement s in Table -1 with the figure in table -2. |
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Answer» Correct Answer - (A) Q,(B ) S,(C ) PS,(D )QR |
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| 254. |
In the figure a block of mass `10kg` is in equilibrium. Identify the string in which the tension is zero A. BB. CC. AD. None of the above |
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Answer» Correct Answer - D `T_(A) = 10 g = 100 N` :. `T_(B) cos 30^(@) = T_(A)` `T_(B) = (200)/(sqrt 3) N` `T_(B) sin 30^(@) = T_(C)` `T_(C) = (100)/(sqrt 3) N` |
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| 255. |
The fast moving vehicles are given streamline shape . Why ? |
| Answer» This reduces the force of friction due to air . | |
| 256. |
Is frictions independent of actual area of contact ? |
| Answer» Yes , force of friction does not depend upon actual area of contact . | |
| 257. |
Conservation of momentum in a collision between particles can be understood from .A. conservation of energyB. Newton s first law onlyC. Newton s second law onlyD. both Newton s second and third law |
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Answer» Correct Answer - d Conservation of momentum in a collision between particles can be understood from both, Newton s`2nd` law and `3rd` law . |
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| 258. |
Anoop pushes eight identical blocks on the horizontal frictionless surface with horizontal force F. the that block-1 (where F is applied) exerts on block 2 has magnitudes `F_(21)` and the force that block-7exerts on the block -8 is `F_(87)`. Find `(F_(21))/(F_(87))`A. 2B. 8C. 5D. 7 |
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Answer» Correct Answer - d |
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| 259. |
A block slides down an incline of `30^(@)` with the horizontal starting from rest it covers `8m` in the first two seconds Find the coefficient of kinetic friction between the two . |
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Answer» Correct Answer - `0.11` Here, `theta = 30^(@), u = 0, s = 8 m, t = 2s` From `s = y t+ (1)/(2)a t^(2)` `8 = 0 + (1)/(2) xx a (2)^(2)` or `a = 4 ms^(-2)` As `a = g (sin theta - mu cos theta)` `:. 4 = 9.8 (sin 30^(@) - mu cos 30^(@))` `= 9.8 ((1)/(2) - mu sqrt3/(2))` `mu = (0.9)/(4.9 sqrt 3)=0.11` . |
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| 260. |
Can coefficient of friction exceed unity ? |
| Answer» Coefficient of friction is normally lass than one but when the surfaces in contact are very irregular having sharp minute projections and cavities the coefficient of friction between them may exceed unity . | |
| 261. |
Why are ball bearings used in machinery ? |
| Answer» By using ball bearings between the moving parts of a machinery the sliding friction gets converted into rolling friction As rolling friction is much less than sliding friction power dissipation is reduced . | |
| 262. |
A cyclist goes round a circular track of `440` metres length in `20` seconds Find the angle that his cycle makes with the verticle. |
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Answer» Correct Answer - `35^(@) 12` . Let r be radius of the track `2pi r = 440 m` `r = (440)/(2pi) m, upsilon = (2pi r)/(t) = (440)/(20) = 22ms^(-1)` As `tan theta = (upsilon^(2))/(rg)` `:. Tan theta =(22xx22xx2pi)/(440 xx 9.8) = 0.7055` `theta = 35^(@) 12` . |
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| 263. |
The acceleration of a train travelling at ` 40 m//s ` as it goes round a curve of 160 m in radius is ? |
| Answer» ` a = (upsilon^(2))/(r) = (40 xx 40)/(160) = 10 m//s^(2) ` . | |
| 264. |
The upper portion of an inclined plane of inclination a is smooth and the lower portion is rough. A particle slides down from rest the top and just comes to rest at the foot . If the ratio of smooth length to rough length is `m:n`, find the coefficient of friction .A. `[(m+n)/(n)]tan alpha`B. `((m+n)/(n))cot alpha`C. `((m-n)/(n))cot alpha`D. `((m+n)/(m))tan alpha` |
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Answer» Correct Answer - a |
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| 265. |
A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined at an angle `theta` to the vertical. The minimum value of coefficient of friction of friction between the block and the surface for which the block will remain in equilibrium is A. `(P+Q sin theta)/(mg+Q cos theta)`B. `(P cos theta+O)/(mg-Q sin theta)`C. `(P+Q cos theta)/(mg+Q sin theta)`D. `(Psin theta-Q)/(mg-Q cos theta)` |
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Answer» Correct Answer - a |
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| 266. |
In the three figure shown, find acceleration of block and force of friction on it in each case. |
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Answer» Figure (a) `N = mg = 40N` `mu N = 24N` Since `F lt mu_(s)N` Block will remain stationary and `f = F = 20N` Figure (b) `N = mg = 20 N` `mu_(S) N = 12N` and `mu_(K) N = 8 N` Since, `F gt mu_(S) N`, block will side and kinetic friction `(= 8N)` will act. `a = (F - f)/(m) = (20 - 8)/(2) = 6 m//s^(2)` |
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| 267. |
Pseudo force with respect to a frame moving with constant velocity is zero. Is this statement true or false? |
| Answer» Constant velocity means acceleration of frame is zero | |
| 268. |
Two identical blocks of weigth w are placed one on top of the other as shown in figure. The upper block is tied to the wall. The coefficient of static friction between B and ground is `mu` and friction between A and B is absent. When `F=mu w`force is applied on the lower block as shown. The tension in the string will be A. `mu w`B. `(mu w)/(2)`C. `0`D. `2 mu w` |
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Answer» Correct Answer - C |
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| 269. |
A block of unknown mass is at rest on a rough, horizontal surface. A horizontal force `F` is applied to the block. The graph in the figure shows the acceleration of the block with respect to the applied force. The mass of the block is A. `1.0 kg`B. `0.5 kg `C. `2.0 kg`D. `1.5 kg` |
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Answer» Correct Answer - C |
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| 270. |
Forces of action and reaction never cancel each other as they areA. always equalB. always oppositeC. acting on same bodyD. acting on different bodies |
| Answer» Correct Answer - d | |
| 271. |
A lift a mass `1200 kg` is raised from rest by a cable with a tension `1350 kg f` . After same time the tension drops to `1000 kg f` and the lift comes to rest at a height of `25 m` above its intial point `(1 kg - f = 9.8 N)` What is the height at which the tension changes ?A. `10.8 m`B. `12.5 m`C. `14.3 m`D. `16 m` |
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Answer» Correct Answer - C Acceleration `a_(1) = (1350 xx 9.8 - 1200 xx 9.8)/(1200)` `= 1.225 m//s^(2)` Retardation, `a_(2) = (1200g - 1000g)/(1200)` `= 1.63 m//s^(2)` `h_(1) + h_(2) = 25` …(i) `nu = sqrt(2 a_(1) h_(1))` or `sqrt(2 a_(2) h_(2))` or `2 a_(1) h_(1) = 2 a_(2) h_(2)` `:. (h_(1))/ ( h_(2)) = (a_(2))/(a_(1))` ...(ii) `= (1.63)/(1.225) = 1.33` Solving these equations,we get `h_(1) = 14.3 m` |
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| 272. |
A lift a mass `1200 kg` is raised from rest by a cable with a tension `1350 kg f` . After same time the tension drops to `1000 kg f` and the lift comes to rest at a height of `25 m` above its intial point `(1 kg - f = 9.8 N)` What is the greatest speed of lift?A. `9.8 ms^(-1)`B. `7.5 ms^(-1)`C. `5.92 ms^(-1)`D. None of these |
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Answer» Correct Answer - C `nu = sqrt(2 a_(1) h_(1)) = sqrt(2 xx 1.225 xx 14.3)` `= 5.92 m//s` |
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| 273. |
Why has a horse to pull a cart harder during the first few steps of his motion ? |
| Answer» During the first few steps of his motion the horse has to pull a cart harder because the horse has to work against the limiting friction whereas once the motion starts the horse has to work against the dynamic friction which is less than the limiting friction . | |
| 274. |
An aeroplane requires for take off a speed of `80 km//h` the run on the ground being 100 m The mass of aeroplane is `10^(4)` kg and ground is 0.2 What is the maximum force required by the engine of the plane for take off ? |
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Answer» Here ,` upsilon = (80 xx 1000)/(60 xx 60) = (200)/(9) m//s ` ` s = 100 m , u = 0 , mu = 0.2 , m = 10^(4) kg F = ? ` From ` upsilon^(2) - u^(2) = 2 as ` ` ((200)/(9))^(2) - 0 = 2 xx a xx 100 ` ` a = (200 xx 200 )/(9 xx 9 ) xx (1)/(200) = (200)/(81) m//s^(2) ` Force required to produce acceleration `F_(1) = ma = 10^(4) xx (200)/(81) = 1.96 xx 10^(4) N` Force required to overcome friction `F_(2) = mu R = mu ng = 0.2 xx 10^(4) xx 9.8 = 1.96 xx 10^(4) N` `:.` maximum force required by the engine to take off `= F_(1) + F_(2)` `=2.47 xx 10^(4)+1.96 xx 10^(4) = 4.43 xx 10^(4) N`. |
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| 275. |
Why are wheels of an automobile made circular ? |
| Answer» Circular wheels roll on the road . Therefore motion of the vehicle is opposed by rolling friction which is much smaller than the sliding friction . | |
| 276. |
Automobile tyres are generally provided with irregular projections over their surfaces, why ? |
| Answer» Irreular projections over the surface of automobile tyres increase the force of friction between the tyres and the road . | |
| 277. |
A point particle of mass m, moves long the uniformly rough track PQR as shown in figure. The coefficient of friction, between the particle and the rough track equals `mu`. The particle is released, from rest from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The value of the coefficient of friction `mu` and the distance x `(=QR)`, are, respectively close to: A. (a) 0.29 and 3.5mB. (b) 0.29 and 6.5mC. (c) 0.2 and 6.5mD. (d) 0.2 and 3.5m |
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Answer» Correct Answer - A Loss in P.E.= Work done against friction from `prarrQ` +work done against friction form `QrarrR` `mgh=mu(mg cos theta)PQ+mu mg(QR)` `h=mu cos thetaxxPQ+mu(QR)` `2=muxxsqrt3/2xx(2)/(sin 30^@)+mux` `2=2sqrt3mu+mux` …(i) `[sin 30^@=(2)/(PQ)]` Also work done `PrarrQ=` work done `QrarrR` `:.` `2sqrt3mu=mux` `:.` `x~~=3.5m` From (i) `2=2sqrt3mu+2sqrt3mu=4sqrt3mu` `mu=(2)/(4sqrt3)=(1)/(2xx1.732)=0.29` |
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| 278. |
A block is kept at rest on a rought ground as shown. Two force `F_(1)`and `F_(2)` are acting on it. If we increased either of the two force `F_(1)`and `F_(2)`, force of friction acting on the block will increase. By increasing `F_(1)`, normal reaction from ground will increase. A. If both Accertion and Reason are true and the reason is correct explanation of the Assertion.B. If both Accertion and Reason are true but reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D By increasing `F_(1)` limiting value of friction will increase. But it is not neccesary that actual value of friction acting on block will increase. |
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| 279. |
A block of mass `m` is at rest on a rought wedge as shown in figure. What is the force exerted by the wedge on the block? |
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Answer» Since, the block is permanently at rest, it is in equilibrium. Net force on it should be zero. In this case, only two forces are acting on the block (i)Weight `=mg` (downwards). Contact force (resultant of normal reation and friction forces) applied by the wedge on the block. For the block to be in equilibrium, these two forces should be equal and opposite. Therefore, force exerted by the wedge on the block is `mg` (upwards). |
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| 280. |
Figure shows (x,t) (y,t) diagram of a particle moving in 2-dimensions. If the particle has a mass of 500 g , find the force (direction and magnitude) acting on the particle .A. 1N along y-axisB. 1N along x-axisC. 0.5N along x-axisD. 0.5N along y-axis |
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Answer» Correct Answer - A Since the graph between x and t is a strainght line and passing through the origin. `therefore x=t` Since the graph between y and t is a parabola. `therefore y=t^(2)` `therefore v_(x)=(dx)/(dt)=1 and a_(x)=(dv_(x))/(dt)=0` `and v_(y)=(dy)/(dt)=2t and a_(y)=2ms^(-2)` The force acting on the partical is `F=ma_(y)=(0.5kg)(2ms^(-2))=1N "along y-axis"` |
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| 281. |
Two billiard balls A and B each of mass 50 g and moving in opposite directions with speed of `5 m//s` each, collide and rebound with the same speed if the collision lasts for `10^(-3)` s which of the following statement(s) is (are) true?A. The impulse imparted to each ball is `0.25 kg- ms^(-1)` and the force on each ball is 250 NB. The impulse imparted to each ball is `0.25 kg-ms^(-1)` and the force exerted on each ball is `25 xx 10^(-5) `NC. The impulse imparted to each ball is `0.5` N-sD. The impluse and the force on each ball are equal in magnitude and opposite directions |
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Answer» Correct Answer - C::D Given `" " m_(1) = m_(2) 50(g) = (50)/(1000) kg = (1)/(20) kg ` Initial velocity (u) = `u_(1) = u_(2) = 5m//s` Final velocity (v) = `v_(1) = v_(2) =- 5 m//s` Time duration of collision = `10^(-3) s ` Change in linear momentum = m (v-u) = `(1)/(20) [-5-5] = - 0.5 N-s`. Force = `("Impulse")/("Time") = ("Change in momentum")/(10^(-3) s)` = `(0.5)/(10^(-3)) = 500 N` Impulsive and force are opposite in directions. |
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| 282. |
A body weighing `0.4kg` is whirled in a verticle circle making `2rps` If the radius of the circle is `1.2m` find the tension in the stringat (i) top of the circle (ii) bottom of the circle . |
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Answer» Correct Answer - `T_(H) = 71.94 N, T_(L) = 79.78 N` . Here, `m = 0.4 kg , n = 2 rps, r = 1.2 m` `T_(H) = ? T _(L) = ? ` From `T_(H) + m g = m r omega^(2) = 4 pi^(2) m r n^(2)` `T_(H) = 4 pi^(2) m r n^(2) - m g` ` = 4 xx (22)/(7) xx (22)/(7) xx 0.4 xx 1.2 xx 4 xx 9.8` `T_(H) = 75.86 - 3.92 = 71.94 N` Again `T_(L) = 4 pi^(2) m r n^(2) + mg` `= 75.86 + 3.92 = 79.78 N`. |
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| 283. |
The motion of a particle of mass m is given by x=0 for `t lt 0` s, x (t) = `A sin 4pit for 0 lt t(1//4) s (A gt 0)` and x=0 for `t gt (1//4)` s Which of the following statement(s) is (are) true?A. The force at t = (1/8) s on the particle is `-16 pi^(2) A-m`B. The particle is acted upon by on impulse of magnitude `4pi^(2)A-m` at t = 0 s and t (1/4) sC. The particle is not acted upon by any forceD. The particle is not acted upon by a constant force |
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Answer» Correct Answer - A::B::D x = 0 for t `lt `0 s. x(t) = A sin `pit` , 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) = 4 pi A cos 4 pi t ` a(t) = acceleration = `(dv(t))/(dt) = -16 pi^(2) A sin 4 pi t ` At t = `(1)/(8) s , a(t) = - 16 pi^(2) A sin 4 pi xx (1)/(8) = - 16 pi^(2) A ` F = ma(t) = `-16 pi^(2) A xx m = - 16 pi^(2)mA` Impulse = Change in linear momentum =` F xx t = (-16 pi^(2) Am) xx (1)/(4)` = `4 pi^(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 , force is also not constant . |
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| 284. |
a 100kg gun fires a ball of 1kg horizontally from a cliff of height 500m. If falls on the ground at a distance of 400m from the bottom of the cliff. The recoil velocity of the gun is (Take g: `10ms^(-2)`A. `0.2"ms"^(-1)`B. `0.4"ms"^(-1)`C. `0.6"ms"^(-1)`D. `0.8"ms"^(-1)` |
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Answer» Correct Answer - B Here, Mass of the gun, M=100kg Mass of the ball, M=1kg Height of the cliff, h=500m g=`10ms^(-2)` Time taken by the ball to reach the ground is `t=sqrt((2h)/(g))=sqrt((2xx500m)/(10ms^(-2)))=10s` Horizontal distance covered=ut `therefore 400=u xx 10` Where u is the velocity of the ball, `u=40 ms^(-1)` According to the law of conservation of linear momentum, we get, 0=Mv+mu `v=-("mu")/(M)=-((1kg)(40ms^(-1)))/(100kg)=-0.4ms^(-1)` -ve sign shows that the direction of recoil of hte gun is oppositre to that of the ball. |
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| 285. |
Figure shows (x,t) (y,t) diagram of a particle moving in 2-dimensions. If the particle has a mass of 500 g , find the force (direction and magnitude) acting on the particle . |
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Answer» Clearly from diagram (a) , the variation can be related as x = t `implies (dx)/(dt) = 1` m/s `a_(x) = 0` From diagram (b) `" " y = t^(2)` `implies " " (dy)/(dt) = 2t` or `a_(y) = (d^(2)y)/(dt^(2)) = 2 m//s^(2)` Hence , `" " I_(y) = ma_(y) = 500 xx 10^(-3) xx 2 = 1N " " (because m = 500 g)` `F_(x) = ma_(x) = 0` Hence , net force , `" " F = sqrt(F_(x)^(2) + F_(y)^(2)) = F_(y) = 1N " " `(along y-axis) |
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| 286. |
Shows `(x,t),(y,t)` diagram of a particle moving in 2 dimensions If the particle has a mass of `500g` find the force (direction and magnitude) acting on the particle . |
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Answer» As `(x,t)` diagram of particle is a st line , motion along X-axis is unifrom From ` x = ut, u = (x)/(t) = (2)/(2) = 1m//s` Force along X-axis is zero The `(y,t)` diagram is a parabola. If a is unifrom acceleration along Y-axis then from ` y = (1)/(2)at^(2)` `4 = (1)/(2) a x 2^(2), a = 2 m//s^(2) ` As `F = ma :. F = (500)/(1000) xx 2 = 1 N` along Y-axis . |
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| 287. |
A stunt man jumps his car over a crater as shown (neglect air resistance) A. during the whole fight the driver experiences weightlessnessB. during the whole fight the driver never experiences weightlessnessC. during the whole flight the driver experience weightlessnessD. the apparent weight increases during upwards journey |
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Answer» Correct Answer - a |
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| 288. |
Two blocks, each having a mass M, rest on frictionless surfaces as shown in the figure. If the pulleys are light and frictionless and M on the incline is allowed to move down, then the tension in string will be A. `(2)/(3) Mg sin theta`B. `(3)/(2) Mg sin theta`C. `(Mg sin theta)/(2)`D. `2Mg sin theta` |
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Answer» Correct Answer - C ( c) Acceleration of system , `a=(" Net pulling force")/(" Total mass")=(Mg sin theta)/(2M)` `a=(1)/(2)g sin theta` Now, the block on ground is moving due to tension. Hence, `T=Ma=(Mg sin theta)/(2)` |
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| 289. |
A ladder AB is spported by a smooth vertical wall ad rough horizontal floor as shown. A boy tarts moving from A to B slowly. The ladder remains at rest, then pick up the correct statements (s):A. Normal reaction at A will increaseB. Normal reation at will decreaseC. Normal reaction at B will increaseD. |
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Answer» Correct Answer - a,c,d |
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| 290. |
A boy holding a spring balance in his hand suspends a weight of `1kg` from it The balance slips from his hands and falls down . What will be the reading of the balance while it is in air. ? |
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Answer» Correct Answer - `Zero` On slipping the balance falls vertically downwards with `a =g` `:. R = m (g- a) = m (g -g)` `zero` . |
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| 291. |
At certain of time velocities of `1` and `2` both are `1m//s` upwards. Find the velocity of `3` at that moment |
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Answer» In above solution ,we have found that `a_(2) + a_(3) + 2a_(1) = 0` Similarly, we can find `V_(2) + V_(3) + 2V_(1) = 0` Taking upward direction as positive we are given `V_(1) = V_(2) = 1m//s` ` :.V_(3) = - 3 m//s` i.e. velocity of block `3 is 3m //s` (downwards) |
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| 292. |
A man of mass m=60kg is standing on weighing machine fixed on a triangular wedge of angle `theta`=60^(@)` as shown in the figure. The wedge is moving up with an upward acceleration `alpha=2m//s^(2)`. The weight regiatered by machine is A. 600NB. 720NC. 360ND. 240N |
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Answer» Correct Answer - c |
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| 293. |
Assuming all the surfaces to be frictionless accelration of the block c shown in the figure is A. `5m//s^(2)`B. `7m//s^(2)`C. `3.5m//s^(2)`D. `4m//s^(2)` |
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Answer» Correct Answer - c |
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| 294. |
A block of mass m is kept on an inclined plane of a lift moving down with accelration of `2m//s^(2)` .what should be the coefficient of friction for the block to move down with constant velocity relative relative to lift? A. `mu=(1)/sqrt(3)`B. `mu=0.4`C. `mu=0.8`D. `mu=0.5` |
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Answer» Correct Answer - a |
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| 295. |
An athelete runs a certain distance before taking a long jump . Why ? |
| Answer» In a along jump , the horizontal range `R prop ("initial velocity")^(2)` An athelete runs a certain distance before taking a long jump because velocity acquired by running is added to the velocity of the athelete at the time of jump . Due to it , he can jump over a longer distance . | |
| 296. |
Tie a small piece of stone to one end of a string and whirl it in a circle with your hand The centripetal force required by the stone is being supplied by your hand through the string If the string breaks suddenly you observe that the stone flies off along the tangent to the circle at that instant . Read the above passage and answer the following questions: (i) Why does the stone fly off along the tangent to the circle at the instant the string breaks (ii) What lessons of life do you learn from this study ? |
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Answer» (i) At the instant the string breaks the centripetal force is no longer provided to the stone Therefore the stone is not being forced to move along the circle On account of inertia of direction the stone flies off along the tangent to the circle that intant. (ii) To change the straight line path of stone to circular path a continous force (= centripetal) is needed The moment this force ceases the stone reverts to its natural st line path The same is true in day to life Continuous motivation and efforts are needed to keep the young kids on the desired path The moment we stop motivating kids may revert to the path of their choice . |
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| 297. |
Figure shows the position-time (x-t) graph of one dimensional motion of a mass 500g. What is the time interval between two consecutive impulses received by the body? A. 2sB. 4sC. 6sD. 8s |
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Answer» Correct Answer - A Figure shows that slope of x-t graph changes from positive to negative at t=2s, and it changes from negative to positive at t=4 and so on. Thus direction of velocity is reversed after every tow seconds. Hence, the body must be receing consecutinve impulses after every two seconds. |
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| 298. |
A particle moves so that its position vector is given by `vec r = cos omega t hat x + sin omega t hat y`, where `omega` is a constant which of the following is true ?A. Velocity and acceleration both are parellel to rB. Velocity is perpendicular to r and acceleration is directed towards to origin.C. Velocity is perpendicular to r and acceleration is directed towards to origin.D. Velocity and acceleration both are perpendicular to r. |
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Answer» Correct Answer - B |
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| 299. |
A car having a mass of 1000 kg is moving at a seed of 30 metres/sec. Brakes are applied to bring the car to rest if the frictional force between the tyres and the road surface is 5000 newtons,the car will come to reast inA. 5sB. 10sC. 12sD. 6s |
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Answer» Correct Answer - D (d) Retardation, `a=(F)/(m)=5 ms^(-2) implies0=30-gt` `therefore" " t=6s` |
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| 300. |
In the diamram shown in figure , match the following `(g=10m//s^(2))` |
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Answer» Correct Answer - (A)R,( B) T,(C ) Q ,(D ) T |
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