InterviewSolution
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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 boy is deated on the top of a hemispherical mound of ice of radius R . He is given a little pucsh and starts sliding down the ice. If ice is frictionless , the boy will leave the ice at a point whose height isA. ` (3R)/(4)`B. `(2R)/(sqrt3)`C. `(2R)/(3)`D. `(R)/(3)` |
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Answer» Correct Answer - C `h=(2)/(3) R` |
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| 2. |
Velocity of a particle moving in a curvilinear path varies with time as `v=(2t hat(i)+t^(2) hat(k))m//s`. Here t is in second. At `t=1` sA. acceleration of particle is `8 m//s^(2)`B. tangential acceleration of particle is `(6)/(sqrt(5)) m//s^(2)`C. radial acceleration of particle is `(2)/(sqrt(5))m//s^(2)`D. radius of curvature to the path is `(5 sqrt(5))/(2)m` |
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Answer» Correct Answer - B::C::D |
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| 3. |
In all the four situations depicted in Table-1, a ball of mass `m` is connected to a string. In each case, find the tension in the string and match the appropriate entries is Table-2 |
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Answer» Correct Answer - (A)QRS,(B)PS,(C)QS,(D)R |
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| 4. |
The position vector of a particle in a circular motion about the origin sweeps out equal areal in equal time. ItsA. velocity remains constantB. speed remains constantC. accelertion remains constantD. tangential accelertion remains constant |
| Answer» Correct Answer - B::D | |
| 5. |
The position vector of a particle in a circular motion about the origin sweeps out equal area in equal time. ItsA. velocity remains constantB. speed remains constantC. acceleration remains constantD. tangential acceleration increases |
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Answer» Correct Answer - B |v|=constant |
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| 6. |
The position vector of a particle in a circular motion about the origin sweeps out equal area in equal time. ItsA. `(i),(ii)`B. `(ii),(iiI)`C. `(iii),(iv)`D. `(ii),(iv)` |
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Answer» Correct Answer - D `a_(t)=0`,i.e., constant |
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| 7. |
A road is banked with an angle 0.01 radian .If the radiaus of the road is ` 80m (g=10 m//s^(2))`then the safe velocity for the drive will beA. `4.8 m//s`B. `2.8 m//s`C. `3.8 m//s`D. `5.8m//s` |
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Answer» Correct Answer - B `theta =0.1 rad,r=80 m,g=10 m//s^(2)` `v=sqrt(rg tan theta) ` `=sqrt(80xx10xx0.01)=sqrt8=2.828 m//s` |
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| 8. |
A cyclist with combined mass 80 kg going around a curved road with a uniform speed `20 m//s`. He has to bend inward by an angle ` theta = tan^(-1)`(0.50) with the verticle , then the force of friction between road surface and tyres will be `(g=10m//s^(2)`A. 300 NB. 400 NC. 800 ND. 250 N |
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Answer» Correct Answer - B `(mv^(2))/(r)=(mrg tan theta )/(r)` |
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| 9. |
A block is released from rest at the top of an inclined plane which later curves into a circular track of radius `r` as shown in figure. Find the minimum height `h` from where it should be released so that it is able to complete the circle. A. `(R )/(2)`B. `(3R)/(2)`C. `zero`D. `(5R)/(2)` |
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Answer» Correct Answer - D speed of the block at `B,v=sqrt(2gh)` top complete the vertical circle, `vgesqrt(5gR)` `sqrt(2ghgesqrt(5gR)` `hge(5R)/(2),h_(min)=(5R)/(2)` |
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| 10. |
A block is released from rest at the top of an inclined plane which later curves into a circular track of radius `r` as shown in figure. Find the minimum height `h` from where it should be released so that it is able to complete the circle. |
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Answer» When the block reaches at `B`,its velocity `v=sqrt(2gh)`. To complete the verticle circle, the velocity of block should be at least `sqrt(5gR)` `vgesqrt(5gR)` `sqrt(2gh)gesqrt(5gR)` `hge(5R)/(2)` `h_(min)=(5R)/(2)` |
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| 11. |
A motor cyclist rides along a horizontal circle on the certicle cylindrical wall of a metal cylinder the radius of the cylinder is 10 m .If the speed is `20 m//s ` and accele ration due to gravity is `10 m//s ^(2)` then the least value of the coefficient of friction will be ,A. `0.25`B. `0.45`C. `0.35`D. `0.15` |
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Answer» Correct Answer - a `F= mu N ` ` therefore mu= (F)/(N)=(mgr)/(mv^(2))=(gr)/(v^(2))=(10xx10)/(20xx20)=0.25` |
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| 12. |
A motor cycle rides in a hollow sphere in a verticle circle of radius 30 m .What will be the minimum speed required so that he does not lose contact with the surface of sphere at the highest point?A. `5.442 km//s`B. `17.422 cm//s`C. `17.32 m//s`D. `54.22 m//s` |
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Answer» Correct Answer - C `v=sqrt(rg)=sqrt((30xx10))` `=17.32` |
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| 13. |
A can filled with water is revolved in vertical circle of radius 16 m and water just does not fall down. The time period of revolution will beA. 1 sB. 10 sC. 8 sD. 4 s |
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Answer» Correct Answer - D `T=2pisqrt((r)/(g))` |
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| 14. |
The kinematical equation of motion are applied to solve the problems of circular motion , because ofA. the acceleration is non uniformB. the acceleration is uniformC. the acceleration and velocity are uniformD. the motion is circular |
| Answer» Correct Answer - B | |
| 15. |
A cane filled with water is revolved in a vertical circle of radius 4 m and water just does not fall down. The time period of revolution will be –A. 4sB. 2sC. 1sD. 6s |
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Answer» Correct Answer - A Given, R= 4m `v=sqrt(gR)=sqrt(9.8xx4)` Now, time period `=(2piR)/(v)=(2xx3.14xx4)/(sqrt(9.8xx4))=4 s` |
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| 16. |
In circular motionA. radial acceleration is non-zeroB. radia velocity is zeroC. body is in equallibriumD. All of the above |
| Answer» (d) In cicrular motion rdial aceleration `a_(1)=(v^(2))/(r)` and the particle does not move along the radius of he circular path hence radial velocity is zero radial acceleration only changes the direction of motion of the particle . | |
| 17. |
Assertion A car moves along a road with uniform speed. The path of car lies in vertical plane and shown in fighre. The radius of curvature (R) of the path is same everywhere. If the car does not loose contact with road at the highest point, it can travel the shown path without loosing contact with road anywhere else. Reason For car to loose contact with road, the normal reaction between car and road should be zero. A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - C The normal reaction is not least at topmost point, hence assertion is false. |
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| 18. |
A simple pendulum has a length l . What minimum velocity should be imparted to its bob at the mean position so that bob reaches a height equal to l above the point of suspension?A. `sqrt gr`B. `sqrt(5gl)`C. `sqrt(2gl)`D. `sqrt((l)/(g))` |
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Answer» Correct Answer - B `v_(L)=sqrt(5 gr)`. |
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| 19. |
Assertion One end of a massless rod of length l is hinged so that it is free to rotate in vertical plane about a horizontal axis. If a particle is attached to the other end of the rod, then the minimum speed at lower most position of the particle is `sqrt(5gl)` to complete the circular motion. Reason Work done by cnetripetal force on the particle is always zero.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - D In case of massless rod, `v_("min")=sqrt(4gl)`. Work done by centripetal force on the particle is always zero. |
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| 20. |
A ball of mass `m` is attached to one end of a light rod of length `l`, the other end of which is hinged. What minimum velocity `v` should be imparted to the ball downwards, so that it can complete the circle ? A. `sqrt(gl)`B. `sqrt(5gl)`C. `sqrt(3gl)`D. `sqrt(2gl)` |
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Answer» Correct Answer - D |
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| 21. |
A rod of length L is hinged from one end. It is brought to horizontal position and released. The angular velocity of the rod when it is in vertical position. IsA. `sqrt(((2g)/(L)))`B. `sqrt(((3g)/(L)))`C. `sqrt(((g)/(2L)))`D. `sqrt(((g)/(L)))` |
| Answer» Correct Answer - (b) | |
| 22. |
A block of mass `m` slides 0n a frictionless table. It is constrained to move inside a ring of radius `R` . At time `t=0` , block is moving along the inside of the ring (i.e. in the tangential direction) with velocity `v_(0)` . The coefficient of friction between the block and the ring is `mu` . Find the speed of the block at time `t` . |
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Answer» Correct Answer - A `N=(mv^(2))/(R)` `f_(max)=muN=(mumv^(2))/(R)` :. Retardation `a=(f_(max))/(m)` `=(muv^(2))/(R)` :. `(-(dv)/(dt))=(muv^(2))/(R)` or `int_(v_(0))^(v)(dv)/(v^(2))=-(mu)/(R)int_(0)^(t)dt` or `v=(v_(0))/(1+(muv_(0)t)/(R))` |
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| 23. |
Two beads A and B of equal mass `m` are connected by a light inextensibe chord. They are constrined to move on a frictionless ring in vertical plane. The blocks are released from rest as shown in figure. The tension in the chord just after the release is A. `(mg)/(4)`B. `sqrt(2) mg`C. `(mg)/(2)`D. `(mg)/(sqrt(2))` |
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Answer» Correct Answer - D |
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| 24. |
Pulleys and strings are massless.The horizontal surface is smooth.What is the acceleration of the block A. `F/(2m)`B. `F/m`C. `(2F)/m`D. `m/(2F)` |
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Answer» Correct Answer - A `T=ma,F=2T` |
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| 25. |
The relation between tangential or linear acceleration and angular acceleration of a body moving in circle is given byA. `vec a=vec r // vec alpha`B. `vec a= vec alpha xx vec r`C. `vec a= vec alpha // vec r`D. `vec a= vec r xx alpha vec` |
| Answer» Correct Answer - B | |
| 26. |
The phisical quantites which reamin constant for a particle performing uniform circular motion in horizontal plane areA. kinetic energyB. torque is zeroC. angular momentumD. 1 and 3 |
| Answer» Correct Answer - D | |
| 27. |
A particle is performing uniforme circular motion, has constantA. velocityB. kinetic energyC. momentumD. acceleration |
| Answer» Correct Answer - C | |
| 28. |
In a uniform circular motion , the direction of linear velocity is along theA. work done is zeroB. torque is zeroC. angular speed constantD. all of the above |
| Answer» Correct Answer - D | |
| 29. |
A particle performing UCM, the particle is acted upon byA. gravitational accelerationB. radial accelerationC. the resultant accelerationD. angular acceleration |
| Answer» Correct Answer - B | |
| 30. |
The work done on a particle performing UCM isA. constant but non zeroB. `(mv^(2))/(r)xx2pir`C. zeroD. infinity |
| Answer» Correct Answer - C | |
| 31. |
For a particle performing UCM, the phisical quantities are constantA. speed and angular velocityB. kinetic energy and radius vectorC. angular velocity and kinetic energyD. a and c |
| Answer» Correct Answer - B | |
| 32. |
In `1.0 s`, a particle goes from point `A` to point `B` , moving in a semicircle of radius `1.0 m ` (see figure ). The magnitude of the average velocity A. `3.14ms^(-1)`B. `2 ms^(-1)`C. `1ms^(-1)`D. zero |
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Answer» Correct Answer - B `"Average velocity"=("Total displacement")/("Total time")=(2m)/(1s)=2ms^(-1)` |
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| 33. |
Two masses of 1kg and 5kg are attached to the ends of a massless string passing over a pulley of negligible weight. The pulley itself is attached to a light spring balance as shown in figure. The masses start moving during this interval, the reading of spring balance will be: A. `6 kg`B. less than `6 kg`C. more than `6 kg`D. may be more or less than `6kg` |
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Answer» Correct Answer - B `m_(1)g-T=m_(1)a,T-m_(2)g=m_(2)a,T^(1)=2T` |
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| 34. |
A rough horizontal plate rotates with angular velocity `omega` about a fixed verticle axis. A particle or mass `m` lies on the plate at a distance `(5a)/(4)` from this axis. The coefficient of friction between the plate and the particle is `(1)/(3)` . The largest value of `omega^(2)` for which the oparticle will continue to be at rest on the revolving plate isA. `(g)/(3a)`B. `(4g)/(5a)`C. `(4g)/(9a)`D. `(4g)/(15a)` |
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Answer» Correct Answer - D `mumg=nRomega^(2)` :. `omega^(2)=(mug)/(R)=((1//3)g)/(5a//4)=(4g)/(15a)` |
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| 35. |
Assertion A small block of mass m is rotating in a circle inside a smooth cone as shown in figure. In this case the normal reaction, `N ne "mg cos"theta` Reason In this case, acceleraion of the block is not along the surface of cone. It is horizontal. A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - A In vertical direction, ` Ncostheta=mg` In horizontal direction, `Nsintheta=(mv^(2))/(R)` |
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| 36. |
A ball of mass `m` is rotating in a circle of radius `r` with speed `v` inside a smooth cone as shown in figure. Let `N` be the normal reaction on the ball by the cone, then choose the correct option: A. `N = mg cos theta`B. `g sin theta = (v^(2))/(r)cos theta`C. `N sin theta - (mv^(2))/(r) =0`D. None of these |
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Answer» Correct Answer - A |
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| 37. |
A body of mass `10 kg` is lying on a rough inclined plane of inclination `37^(@)` and `mu = 1//2` , the minimum force required to pull the body up the plane isA. `5.4N`B. `10.8N`C. `2.7N`D. `18N` |
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Answer» Correct Answer - B `F=ma,F=mg(sin theta-mu_(k) cos theta)` |
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| 38. |
A ball of mass `m` is rotating in a circle of radius `r` with speed `v` inside a smooth cone as shown in figure. Let `N` be the normal reaction on the ball by the cone, then choose the correct option: A. `N=mg cos theta`B. `g sin theta=(v^(2))/(r ) cos theta`C. `N sin theta-(mv^(2))/(r )=0`D. none of these |
| Answer» Correct Answer - B::C | |
| 39. |
A spaceman in training is rotated in a seat at the end of a horizontal arm of length 5 m. If he can with stand acceleration upto 9 g, then what is the maximum number of revolution per second permissible? (Take, g =`10ms^(-2)`) |
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Answer» In circular motion, necessary centripetal force to the man is provided by effective weight of man. `therefore" "mxx9g=mromega^(2)=mrxx4pi^(2)n^(2)` `rArr" "n=sqrt((9g)/(4pi^(2)r))rArrn=sqrt((9xx10)/(4xx(3.14)^(2)xx5))` `=0.675"rev s"^(-1)` or hertz(Hz) |
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| 40. |
A pilot of mass 81 kg loops the loop with steady speed of `300 km//h`. If the radius is 0.5 km then the force with which the pilot is pressed into the seat at the highest point of the loop ,is `(g=10m//s^(2))`A. `3.15xx10^(2) N`B. zeroC. `8.10xx10^(2)N`D. `19.35xx10^(2)N` |
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Answer» Correct Answer - A `v=sqrt(gh)` `=sqrt(9.8xx9.8xx10^(-1))` `0.98 ms^(-1)` |
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| 41. |
In the above problem, the force exerted by the seat on the pilot at lowest point is ,A. zeroB. `1.25xx10^(3) N`C. `8.10xx10^(2) N`D. `19.35xx10^(2) N` |
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Answer» Correct Answer - D `F=(mv^(2))/(r)+mg cos theta " " here theta =0^(@)` |
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| 42. |
The pilot of an aircraft, who is not tied to his seat, can loop a verticle circle in air without falling out at the top of the loop . What is the minimum speed required so that he can successfully negotitate a loop of radius 4 km ? `(g=10 m//s^(2))`A. `100 m//s`B. `200 m//s`C. `300 m//s`D. `400 m//s` |
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Answer» Correct Answer - B `v=sqrt(rg)` `=sqrt(40000xx10)=sqrt(4xx10^(4))=2xx10^(2)` `=200 m//s` |
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| 43. |
A circular road of radius 50 m has the angel of banking equal to `30^0`. At what speed should a vehicle go on this road so that the friction is not used? |
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Answer» Correct Answer - A Angle ofbanking`=theta=30^0` `Radius =r=50 m` ` tantheta=(v^2/rg)tan30^0` `gt (1/sqrt3)=(v^2/(rg)` `rarr v^2=(rg)/sqrt3=(50x10)/sqrt3` `rarr v=sqrt(500/sqrt3)=17 m/sec` |
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| 44. |
Keeping the banking angle same , to increase the maximum speed with which a vehicle can traveln on the curve road by `10 %` , the radius of curvature of the road has to be changed from 20 m toA. 16 mB. 18 mC. 24.2 mD. 30.5 m |
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Answer» Correct Answer - C `v_(1) alpha sqrte_(1) and v_(2) alpha sqrtr_(2)` |
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| 45. |
Banking of roads at curve is necessary so as to avoidA. the dependence of centripetal force on the force of frictionB. overurning of vehicle moving with maximum safe speedC. rough nature of road surface which increases the force of friction and causes the wear and tears of tyre of vhicleD. can not be predicted |
| Answer» Correct Answer - b | |
| 46. |
The maximum safe speed of a vehicle on a circulartrack is `15km//h` . When the track becomes wet, the maximum safe speed is `10 km//h` .The ratio of coefficient of friction of dry track to that o0f the wet track isA. `2:3`B. `3:2`C. `9:4`D. `1:1` |
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Answer» Correct Answer - C `(v_(1))/(v_(2))=sqrt((mu_(1))/(mu_(2)))` `(v_(1)^(2))/v_(2)^(2)=(mu_(1))/(mu_(2))` `(15xx15)/(10xx10)=(mu_(1))/(mu_(2))` `(9)/(4)=(mu_(1))/(mu_(2))` |
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| 47. |
The banking angle is independednt ofA. velocity of vehicleB. mass of vehicleC. radius of curvature of roadD. height of inclination |
| Answer» Correct Answer - B | |
| 48. |
Maximum safe speed does not depended upon,A. radius of curvatureB. angle of inclination with the horizontalC. mass of the vehicleD. acceleration due to gravity |
| Answer» Correct Answer - C | |
| 49. |
When a cyclist turns on a circular path,the necessary centripetal force is provided by friction between the tyres and the road. If centripetal force is not provided by friction, then for the vehicle to move on circular path, the track is banked. A cyclist going straight suddenly turns on wet road, thenA. the cyclist is likely to skidB. the cyclist will skid only if his weight is less than the weight of cycle.C. the cyclist will skid if his weight is more than weight of cycle.D. cyclist will not skid at all. |
| Answer» Correct Answer - A | |
| 50. |
When a cyclist turns on a circular path,the necessary centripetal force is provided by friction between the tyres and the road. If centripetal force is not provided by friction, then for the vehicle to move on circular path, the track is banked. The correct angle of banking for a curved smooth road of radius `120 m` for a speed of `108 km//h(g=10 ms^(-2))` isA. `30^(@)`B. `37^(@)`C. `45^(@)`D. `60^(@)` |
| Answer» Correct Answer - B | |