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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.
| 1. |
A body rests a rough horizontal plane `A` force is applied to the body directed towards the plane at an angle `theta` with the vertical. The body can be moved along the plane .A. only if `theta` is greater than the angle of frictionB. only if `theta` is lesser than the angle of frictionC. only if `theta` is equal to the angle of frictionD. for all values of `theta` |
| Answer» Correct Answer - A | |
| 2. |
A particle of mass M rests on a rough inclined plane at an angle q to the horizontal`("sin" theta = (4)/(5))`. It is connected to another mass m as shown in fig. The pulley and string are light. The largest value of m for which equilibrium is possible is M. Find the smallest value of m for which equilibrium is possible |
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Answer» Correct Answer - 3M/6 |
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| 3. |
A tall elevator is going up with an acceleration of `a = 4 m//s^(2)`. A 4 kg snake is climbing up the vertical wall of the elevator with an acceleration of a. A 50 g insect is riding on the back of the snake and it is moving up relative to the snake at an acceleration of a. Find the friction force between the elevator wall and the snake. Assume that the snake remains straight. |
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Answer» Correct Answer - 73.1 N |
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| 4. |
A block has been placed on an inclined plane . The slope angle of `theta` of the plane is such that the block slides down the plane at a constant speed . The coefficient of kinetic friction is equal to :A. `sin theta`B. `cos theta`C. gD. `tan theta` |
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Answer» Correct Answer - D When the block slides down the plane with a constant speed then the inclination of the plane is equal to angle of repose `(theta)`. Coefficient of friction=tan of the angle of reposed `=tan theta`. |
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| 5. |
A block of mass m slides down an inclined plane of inclination `theta` with uniform speed The coefficient of friction between the block and the plane is `mu`. The contact force between the block and the plane is .A. `mg sin thetasqrt(1+mu^(2))`B. `sqrt((mg sin theta)^(2)+(mumgcostheta)^(2))`C. `mg sin theta`D. `mg` |
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Answer» Correct Answer - D Block slides down with constant velocity Hence net force on the block is zero `f =mg sin theta, N =mg cos theta ,R = sqrt(f^(2) +N^(2))` . |
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| 6. |
The force required to just move a body up an inclined plane is double the force the required prevent it from sliding down. If phi is angle of friction and theta is the angle which incline makes with the horizontal then,A. `tantheta=tanphi`B. `tantheta=2tanphi`C. `tantheta=3tanphi`D. `tanphi=2tantheta` |
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Answer» Correct Answer - C `mg(sintheta+mucostheta)=2mg(sintheta-mucostheta)` `tantheta=3mu` `tantheta=3tanphi` . `{ :. `mu=tanphi)` . |
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| 7. |
A boy standing on a weighing machine notices his wight as 400 N. When he suddenly jumps upward the weight shown by the machine becomes 600 N. The acceleration with which the boy jumps up is : (Take `g=10m//s^(2)`)A. `5 ms^(-2)`B. `3.4 ms^(-2)`C. `6 ms^(-2)`D. `9.8 ms^(-2)` |
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Answer» Correct Answer - A |
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| 8. |
A block slides down a frictionless plane inclined at an angle `theta`. For what value of angle `theta` the horizontal component of acceleration of the block is maximum? Find this maximum horizontal acceleration. |
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Answer» Correct Answer - `theta = 45^(@) , g//2` |
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| 9. |
A body of mass m starts sliding down an incline of `30^(@)` from rest. The body comes to rest just when it reaches the bottom. If the top half of the plane is perfectly smooth and the lower half is rough, find the force of friction :A. `(mg)/(4)`B. `(mg)/sqrt(3)`C. mgD. `(mg)/sqrt(2)` |
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Answer» Correct Answer - C |
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| 10. |
In fig., a man true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string a body of mass m. Assuming no friction, and negligible mass of cabin and weighing machine, the measured mass of man is (normal force between the man and the machine is proportional to the mass) A. Measured mass of man is `(Mm)/((M+m))`B. Acceleration of man is `(mg)/((M+m))`C. Acceleration of man is `(mg)/((M+m))`D. Measured mass of man is M |
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Answer» Correct Answer - A::C |
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| 11. |
A block m= 0.5kg slides down a frictionless inclined plane 2 m long as showi in figure-2.196. It then slides on a rough horizontal table surface of `mu = 0.3` for 0.5 m. It then leaves the top of the table,which is 1.0 m high. How far from the base of the table does the block land ? |
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Answer» Correct Answer - [1.85 m] |
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| 12. |
A block of mass `2 kg` is given a push for a moment horizontally and then the block starts sliding over a horizontal plane. The graph shows the velocity`-` time of the motion. The co`-` efficient of sliding friction between the plane and the block is `:` A. 0.02B. 0.2C. 0.04D. 0.4 |
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Answer» Correct Answer - B |
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| 13. |
The spring balance inside a lift suspends an object. As the lift begins to ascent, the reading indicated by the spring balance willA. IncreaseB. DecreaseC. Remain unchangedD. Depend on the speed of ascend |
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Answer» Correct Answer - A |
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| 14. |
A 60kg man stands on a spring scales in a lift. At some instant. He finds that the scale reading has changed from 60kg to 50 kg for a while and then comes back to original mark. What should be concluded?A. The lift was in constant motion upwardsB. The lift was in constant motion downwardsC. The lift while in constant motion upwards, is stopped suddenlyD. The lift while in constant motion downwards, is suddenly stopped |
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Answer» Correct Answer - C |
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| 15. |
A boy whose mass is 50 kg stands on a spring balance inside a lift. The lift starts to ascent with an acceleration of `2ms^(-2)`. The reading of the machine or balance `(g=10ms^(-2))` isA. 50 kgB. ZeroC. 49 kgD. 60 kg |
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Answer» Correct Answer - D |
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| 16. |
A body weighs 6gms when placed in one pan and 24gms when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty, the true weight of the body is :A. 13 gmB. 12 gmC. 15.5 gmD. 15 gm |
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Answer» Correct Answer - B |
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| 17. |
A spring balance and a physical balance are kept in a lift. In these balance equal masses are placed. If now the lift starts moving upward with constant acceleration, then.A. The reading of spring balance will increases increses and the equilibrium position of the physical balance will disturbB. The reading of spring balance will remain in unchanged and physical balance will remain in equilibriumC. The reading of spring balance will decrease and physical balance will remain in equilibriumD. The reading of spring balance will increases and the physical balance will ramain in equilibrium |
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Answer» Correct Answer - D The friction force will act downwards. So the reading of spring balance will increase. In case of physical balance, the fictitious force will act on both the pans, so the equilibrium is not affected. |
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| 18. |
A spring balance and a physical balance are kept in a lift. In these balance equal masses are placed. If now the lift starts moving upward with constant acceleration, then.A. The reading of spring balance will increase and the equilibrium position of the physical balance will disturbB. The reading of spring balance will remain unchanged and physical balance will remain in equilibriumC. The reading of spring balance will decrease and physical balance will remain in equilibriumD. The reading of spring balance will increase and the physical balance will remain in equilibrium |
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Answer» Correct Answer - D |
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| 19. |
If a body of mass `m` is carried by a lift moving with an upward acceleration `a `, then the forces acting on the body are (i) the reaction `R` on the floor of the lift upwards (ii) the weight mg of the body acting vertically downwards. The equation of motion will be given byA. `R=mg-ma`B. `R=mg+ma`C. `R=ma-mg`D. `R=mgxxma` |
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Answer» Correct Answer - B |
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| 20. |
Two equal masses are kept on the pans of a simple balance in a lift acceleration upward. Then.A. Pans will remain at the same levelB. Nothing can be said as data is incompleteC. left side pan will lower down.D. Right side pan will lower down. |
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Answer» Correct Answer - A Both masses will experience same force. |
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| 21. |
Let `a_(1) & a_(2)` are the acceleration of A & B. Let `b_(1) & b_(2)` the acceleration of C & D relative to the wedges A and B respectively, choose the right relation. (directions of `a_(1), a_(2), b_(1) & b_(2)` are shown in figure below) : A. `a_(1)-a_(2) +b_(1) -b_(2) =0`B. `a_(1)+a_(2) - b_(1) - b_(2) =0`C. `a_(1)+a_(2) + b_(1) + b_(2) =0`D. `a_(1)+b_(2) = a_(2) + b_(1)` |
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Answer» Correct Answer - B |
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| 22. |
Find the acceleration of block `B` realtive to the ground if the block A moves to the left with an acceleration `a_(0)` .A. `sqrt(31a_(0))`B. `sqrt(25a_(0))`C. `sqrt(30a_(0))`D. `30a_(0)` |
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Answer» Correct Answer - A `5Ta_(0)-Ta_(BA)=0,a_(BA)=5a_(0)` `a_(B)=sqrt(a_(0)^(2)+(5a_(0))^(2)+2a_(0)5a_(0)cos60^(@)),a_(B)=sqrt31a_(0)` |
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| 23. |
In the diagram shown, the acceleration of the block `B` as shown in relative to the block A and relative to ground is `a_(BA)` and `a_(BG)` respectively. If the block A is moving towards left with an acceleration `a_(0)` then .A. `a_(BA) =2a_(0)`B. `a_(BG) =3a_(0)`C. `a_(BA) =3a_(0)`D. `a_(BG) = a_(0) sqrt(10 + 6 cos theta)` |
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Answer» Correct Answer - C::D Let `T` be the tension, `3Ta_(0)-Ta_(BA) =0` `a_(BA)=3a_(0),a_(BG)=sqrt(a_(0)^(2)+(3a_(0))^(2)+2.a_(0).3a_(0)"cos"theta)` `a_(BG)=a_(0)sqrt(10+6c"co"stheta)` . |
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| 24. |
Figures I, II, III, and IV depict variation of force with time The impulse is hioghest in the case of situations depicted. Figure(s). , A. I and IIB. III and IC. III and IVD. IV only |
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Answer» Correct Answer - C |
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| 25. |
A rocket of mass 120kg. Is fired in the gravity free space is ejected gases with velocity 600 m/s at the rate of 1kg/s. what will be the initial accleration of the rocket:-A. `1m//s^(2)`B. `5m//s^(2)`C. `10m//s^(2)`D. `15m//s^(2)` |
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Answer» Correct Answer - B `F=1xx600` `therefore a=(600)/(120)=5 m//s^(2)` |
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| 26. |
A 54kg girl on ice skates on a frozen lake pulls with a constant force on all light rope that is tied to a 41 kg sled. The sled is initially 22m from the girl, and both the sled and the girl start from rest. Neglecting friction, determine the distance the girl travels to the point where she meets the sled. |
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Answer» Correct Answer - [9.5 N] |
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| 27. |
Consider the system ofpulleysasshown in figure-2.197. Find the acceleration of the three masses `m^(1),m^(2)` and `m^(3). (m_(1) = I kg,m_(2) = 2kg " and " m_(3) =3kg)` |
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Answer» Correct Answer - `[19g//29, 17g//29, 21g//29]` |
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| 28. |
An automobile travelling with a speed `60 km//h` , can brake to stop within a distance of `20 m` . If the car is going twice as fast i. e. , `120 km//h`, the stopping distance will beA. 20 mB. 40 mC. 60 mD. 80 m |
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Answer» Correct Answer - D |
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| 29. |
Two forces are such that the sum of their magnitudes is 18N and their resultant is 12 N which is perpendicular to the smaller force. Then the magnitude of the forces areA. 12 N, 6 NB. 13 N, 5NC. 10 N, 8 ND. 16 N, 2 N |
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Answer» Correct Answer - B |
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| 30. |
Two weights `w_(1)` and `w_(2)` are suspended from the ends of a light string passing over a smooth fixed pulley. If the pulley is pulled up at an acceleration g , the tension in the string will beA. `(4w_(1)w_(2))/(w_(1)+w_(2))`B. `(2w_(1)w_(2))/(w_(1)+w_(2))`C. `(w_(1)w_(2))/(w_(1)+w_(2))`D. `(w_(1)w_(2))/(2(w_(1)+w_(2))` |
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Answer» Correct Answer - A |
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| 31. |
Two forces with equal magnitudes F act on a body and the magnitude of the resultant force is F /3. The angle between the two forces isA. `cos^(-1)(-(17)/(18))`B. `cos^(-1)(-(1)/(3))`C. `cos^(-1)((2)/(3))`D. `cos^(-1)((8)/(9))` |
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Answer» Correct Answer - A |
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| 32. |
The resultant of two forces , one double the other in magnitude is perpendicular to the smaller of the two forces. The angle between the two forces is ________?A. `60^(@)`B. `120^(@)`C. `150^(@)`D. `90^(@)` |
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Answer» Correct Answer - B |
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| 33. |
A force of 10 Newton acts on a body of mass 20 kg for 10 seconds. Change in its momentum isA. `50 km-m//s`B. `100 kg-m//s`C. `300kg-ms`D. `1000kg-m//s` |
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Answer» Correct Answer - B Change in momentum=forcexxtime =100xx10=100` kg.m/s. |
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| 34. |
If two forces of 5 N each are acting along X and Y axes, then the magnitude and direction of resultant isA. `5sqrt2,pi//3`B. `5sqrt2,pi//4`C. `-5sqrt2,pi//3`D. `-5sqrt2,pi//4` |
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Answer» Correct Answer - B |
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| 35. |
A particle is moving under the influence of a force which is fixed in magnitude and acting at an angle `theta` in the direction of motion. The path described by the particle isA. a circleB. an ellipseC. a parabolaD. a straight line |
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Answer» Correct Answer - C If the fixed force acts at angle `theta` in the direction of motion, the path described by the particle is parabola. |
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| 36. |
A man is standing on a balance and his weight is measured. If he takes a step in the left side, then weightA. Will decreaseB. Will increaseC. Remains sameD. First decreases then increases |
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Answer» Correct Answer - C |
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| 37. |
A car C of mass `m_(1)`, rests on a plank of mass `m_(2)`. The plank rests on a smooth floor. The string and pulley are ideal. The car starts and moves towards the pulley with certain acceleration : A. If `m_(1) gt m_(2)`, the string will remain under tensionB. If `m_(1) lt m_(2)`, the string will become slackC. If `m_(1)=m_(2)`, the string will have no tension, and C and P will have accelerations of equal magnitudesD. C and P will have acceleration of equal magnitude if `m_(1) ge m_(2)` |
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Answer» Correct Answer - A::B::C::D |
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| 38. |
A smooth rod is fixed at an angle `alpha` to the horizontal. A small ring of mass m can slide along the rod. A thread carrying a small sphere of mass M is attached to the ring. To keep the system in equilibrium, another thread is attached to the ring which carries a load of mass `m_(0)`at its end (see figure). The thread runs parallel to the rod between the ring and the pulley. All threads and pulley are massless. (a) Find `m_(0)` so that system is in equilibrium. (b) Find acceleration of the sphere M immediately after the thread supporting `m_(0)` is cut. |
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Answer» Correct Answer - (a) `(M + m) g "sin"alpha` (b) `((M+ M) g sin^(2)alpha)/(m + M sin^(2) alpha)` |
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| 39. |
Two bodies of masses `5kg` and `4kg` are tied to a string as shown If the table and pulley are smooth, then acceleration of `5kg` mass will be .A. `19.5 m//s^(2)`B. `0.55 m//s^(2)`C. `2.72m//s^(2)`D. `5.45m//s^(2)` |
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Answer» Correct Answer - D `F =ma, a = (m_(2)g)/(m_(1)+m_(2))` |
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| 40. |
A block of mass `m_(1)` rests on a horizontal table. A string tied to the block is passed on a frictionless pulley fixed at the end of the table and to the other end of string is hung another block of mass `m_(2)`. The acceleration of the system isA. `(m_(2)g)/((m_(1)+m_(2)))`B. `(m_(1)g)/((m_(1)+m_(2)))`C. `g`D. `(m_(2)g)/(m_(1))` |
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Answer» Correct Answer - A |
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| 41. |
Two identical blocks each of mass 'M' are tied to the ends of a string and the string is laid over a smooth fixed pulley. Initially the masses are held at rest at the same level. What fraction of mass must be removed from one block and added to the other, so that is has an acceleration of `1//5^(th)` of the acceleration due to gravity ? .A. `1//10`B. `1//5`C. `2//5`D. `1//20` |
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Answer» Correct Answer - B `F =ma, a = ((m_(1)-m_(2))/(m_(1)+m_(2)))g` |
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| 42. |
A block A of mass 7 kg is placed on a frictionless table. A thread tied to it passes over a frictionless pulley and carries a body B of mass 3 kg at the other end. The acceleration of the system is (given `g=10ms^(-2)`) A. `100ms^(-2)`B. `3ms^(-2)`C. `10ms^(-2)`D. `30ms^(-2)` |
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Answer» Correct Answer - B |
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| 43. |
A smooth ring A of mass m can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley B and carries a block C of mass `M(=2m)` as shown in figure. At an instant the string between the ring and the pulley makes an angle `theta` with the rod. (a). Show that, if the ring slides with a speed v, the block descends with speed `v cos theta`, (b). With what acceleration will the ring starts moving if the system is released from rest with `theta= 30^0` |
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Answer» Correct Answer - `[6.78 m//s^(2)]` |
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| 44. |
A car is moving in a circular horizonta track of radius 10m with a constant speed of 10 m/s. A pendulum bob is suspended from the roof of the cat by a light rigid rod of length 1.00m. The angle made by the rod with track isA. `0^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)` |
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Answer» Correct Answer - C `tan theta = (v^(2))/(rg)` |
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| 45. |
A pendulum is suspended from the roof of a rall road car. When the car is moving on a circular track the pandulum inclines:A. ForwardB. BackwardC. Towards the centre of the pathD. Away from the centre of the path |
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Answer» Correct Answer - D Because of centrifugal forces the pendulum inclines away from centre. |
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| 46. |
A block of mass M = 4 kg is kept on a smooth horizontal plane. A bar of mass m = 1 kg is kept on it. They are connected to a spring as shown & the spring is compressed. Then what is the maximum compression in the spring for which the bar will not slip on the block when released if coefficient of friction between them is 0.2 & spring constant = 1000 N/m : (Take `g=10m//s^(2)`) |
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Answer» Correct Answer - A |
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| 47. |
The coefficients of static friction between contact surface of two bodies is 1. The contact surface of one body support the orther till the inclination is less thanA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - B `mu =tan theta` |
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| 48. |
Two blocks of masses `m_(1)` and `m_(2)` are placed in contact with each other on a horizontal platform. The coefficient of friction between the platform and the two blocks is the same. The platform moves with an acceleration. The force of interaction between the blocks is `:` A. Zero in all casesB. Zero only if `m_(1) = m_(2)`C. Non zero only if `m_(1) gt m_(2)`D. Non zero only if `m_(1) lt m_(2)` |
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Answer» Correct Answer - A |
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| 49. |
When a speeding bus stop suddenly, passengers are thrown forward from their seats becauseA. The back of seat suddenly pushes the passengers forwardB. Inertia of rest stops the train and takes the body forwardC. Upper part of the body continues to be in the state of motion whereas the lower part of the body in contact with seat remains at restD. Nothing can be said due to insufficient data |
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Answer» Correct Answer - C |
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| 50. |
Explain why? (a) a horse cannot pull a cart and run in empty space. (b) passengers are thrown forward from their seats when a speeding bus tops suddenly (c) It is easier to pull a lawn mower than to push it. (d)a cricketer moves his hands backwards while holding a catch |
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Answer» (a) The horse-cart system has no external force in empty space. The mutual forces between the horse and the cart cancel each other (Third Law). On the ground the contack force between the system and the ground (friction) causes their motion from rest. (b) Due to inertia of parts of the body which are not directly in contact with the seat. (c) A lawn mover is pulled or pushed by applying force at an angle. When we push, the normal force (N) must be more than its weight, for equilibrium in the vertical direction. This results in greater fricition `f(fpropN)` and, therefore, a greater applied force to move. Just the opposite happends while pulling. (d) To reduce the rate of change of momentum and hence to reduce the force necessary to stop the ball. |
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