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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.
| 51. |
When a vehicle is moving along the horizontal curve road, centripetal force is provided byA. vertical component of normal reactionB. horizontal component of normal reactionC. frictional component road surface and tyresD. all of these |
| Answer» Correct Answer - C | |
| 52. |
A vehicle of mass `10 kg` is moving with a velocity of `5ms^(-1)`. To stop it in `1//10` sec the required force in opposite direction isA. `500N`B. `5000N`C. `50N`D. `1000N` |
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Answer» Correct Answer - A `F=(dp)/(dt),F=(m(v-u))/t` |
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| 53. |
A particle is moving along a circular along a circular path of radius 5 m with a uniform speed `5ms^(-1)`. What will be the average acceleration when the particle completes half revolution?A. ZeroB. `10ms^(-2)`C. `10pims^(-2)`D. `(10)/(pi)ms^(-2)` |
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Answer» Correct Answer - D `a_(av)=(Deltav)/(Deltat)=(2xx5)/(pi)=(10)/(pi)ms^(-2),T=2pi` |
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| 54. |
A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is `5ms^(-1)` and the speed is increasing at a rate of `2ms^(-2)`. At this instant, the magnitude of the net acceleration will beA. `3.2ms^(-2)`B. `2ms^(-2)`C. `1.2ms^(-2)`D. `4.3ms^(-2)` |
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Answer» Correct Answer - A `a=sqrt(a_(t)^(2)+a_(n)^(2))` `a_(t)` = rate of change of speed =`2 ms^(-2)` `a_(n)=(v^(2))/(R)=((5)^(2))/(10)=2.5ms^(-2)` `therefore" "a=sqrt(a_(t)^(2)+a_(n)^(2))=sqrt((2)^(2)+(2.5)^(2))=3.2ms^(-2)` |
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| 55. |
Three particles A, B and C move in a circle of radius `r=(1)/(pi)`m, in anti-clockwise direaction with speed `1ms^(-1), 2.5ms^(-1)` and `2ms^(-1)` respectively. The initial position of A, B and C are as shown in figure. The ratio of distance travelled by B and C by the instant A, B and C meet for the first time isA. `3:2`B. `5:4`C. `3:5`D. `3:7` |
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Answer» Correct Answer - B `(d_(B))/(d_(C))=(v_(B^(t)))/(v_(C^(t)))=(v_(B))/(c_(C))=(2.5)/(2)=(5)/(4)` |
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| 56. |
Two particles `A` and `B` separated by a distance `2R` are moving counter clockwise along the same circular path of radius `R` each with uniform speed `v`. At time `t=0`, `A` is given a tangential acceleration of magnitude `a = (32v^(2))/(25piR)` in the same direction of initial velocityA. The time lapse for the two bodies to collide is `(6piR)/(5V)`B. The angle coverd by A is `(9pi)/(4)`C. Angular velocity of A is `(11V)/(5R)`D. Radial acceleration of A is `(289upsilon^(2))/(5R)` |
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Answer» Correct Answer - B As when they collide `1/2((32v^(2))/(25piR))t^(2)=piR,t(5piR)/(4V)` angle covered by `A` `theta=pi+(vt)/R,theta=(9pi)/4` |
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| 57. |
A body is sliding down an inclined plane `(mu = (1)/(2))`. If the normal reaction is twice that of the resultant downward force along the incline , the inclination of plane isA. `tan^(-1)[1/2]`B. `tan^(-1)(2)`C. `tan^(-1)(2/3)`D. `tan^(-1)(3/2)` |
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Answer» Correct Answer - C `F=mg(sin theta-mu_(k) cos theta),N= mg cos theta` |
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| 58. |
With what minimum speed `v` must a small ball should be pushed inside a smooth vertical tube from a heoght `h` so that it may reach the top of the tube? Radius of the tube is `R` . |
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Answer» Correct Answer - B `v_(t op)=sqrt(v^(2)-2g(2R-h))` To just complete the verticle circle `v_(t op)` may be zero. `:. 0=sqrt(v^(2)-2g(2R-h))` or `v=sqrt(2g(2R-h))` |
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| 59. |
A disc of radius 0.1 m starts from rest with an angular acceleration of ` 4.4 rad//s^(2)` .Then linear velocity of the point on its after 5 s isA. `0.22//s`B. `2.2 m//s`C. `4.4 m//s`D. `1.1 m//s` |
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Answer» Correct Answer - B `omega_(1)=0,omega_(2)=?` ` alpha = (omega _(2) - omega _(1))/(t)" " therefore omega _(2)= omega_(1)+alphat` `=22` `v_(2)=r omega _(2) =0.1xx22=2.2 m//s` |
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| 60. |
Starting from rest, a particle rotates in a circle of radius R 2 m with an angular acceleration = /4 rad/s2.The magnitude of average velocity of the particle over the time it rotates quarter circle is:A. 1.5 m/sB. 2 m/sC. 1 m/sD. 1.25 m/s |
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Answer» Correct Answer - C |
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| 61. |
The circle of the particle along the circumference of the circle with variable speed is aA. uniform circular motionB. non uniform circular motionC. accelerated motionD. rotational motion |
| Answer» Correct Answer - B | |
| 62. |
An automobile engine starting from rest is given an angular acceleration of `20 rad //s ^(2)` for 10 s .Find the angle turned during this periodA. `10 rad`B. `100 rad`C. ` 1000 rad`D. `0.1 rad` |
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Answer» Correct Answer - C ` alpha =(omega_(2)-omega_(1))/(t)" " thereforew_(2)=200 rad//sec.` `n_(2)=(omega_(2))/(2 pi)=(200)/(2pi)=(100)/(pi)` `N=(n_(1)+n_(2))/(2)xxt=(50)/(pi)xx10=(500)/(pi)` ` therefore theta =(500)/(pi)xx2pi=1000 rad` |
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| 63. |
The speed of a body moving in a circle of radius 15 cm changes from ` 180 rev//min to 600 rev //min "in" 11 s` . Then the angular acceleration of the body will beA. `1 rad//s^(2)`B. `2 rad//s^(2)`C. `3 rad //s^(2)`D. `4 rad//s^(2)` |
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Answer» Correct Answer - D `alpha=(omega_(2)-omega_(1))/(t)=(2pi(n_(2)-n_(1)))/(t)` |
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| 64. |
The motion of the particle along the circumference of the circle with constant angular speed isA. uniforme circular motionB. projectile motionC. non accelerated motionD. non uniform circular motion |
| Answer» Correct Answer - A | |
| 65. |
In the above problem No. 96, if angle turned is `270^(@)` then the change in velocity will beA. zero `km//h`B. `300 km//h`C. `600 km//h`D. `300sqrt 2 km//h` |
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Answer» Correct Answer - D `Deltav =2v "sin" (270)/(2)` `=2v si 135` `2xx300xx sin (90+45)` `=2xx300 sin 45` `=2xx300xx(1)/(sqrt2)` `=sqrt2xx300`. |
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| 66. |
To simulate the acceleration of large rockets, the astronauts are spun at the end of long rotating beam of radius 9.8 m . What will be angular velocity required for generating centripetal acceleration 8 times the acceleration due to gravity?A. `2.828 rad//s`B. `28.28 rad//s`C. `282.8 rad//s`D. zero |
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Answer» Correct Answer - A `C.A=r omega^(2)` `therefore8g=romega^(2)` `therefore omega=sqrt(8g)/(r)=sqrt((8xx9.8)/9.8)=2sqrt2=2.828` |
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| 67. |
If a poit on the circumference of a wheel having diameter 4 m has velocity of 1600 cm/s, then the angular velocity of the wheel will beA. `8 rad //s`B. ` 4 rad//s`C. `6 rad//s`D. `3 rad//s` |
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Answer» Correct Answer - A `omega=(v)/(r)=(16)/(2)=8m//s`. |
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| 68. |
In the above problem, the linear acceleration will beA. `0.6 m//s^(2)`B. `0.5 m//s^(2)`C. `0.4 m//s^(2)`D. `0.2 m//s^(2)` |
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Answer» Correct Answer - A `a_(r)=r alpha=0.15xx4=0.6m//s^(2)` |
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| 69. |
The angular speed of second hand in a watch isA. `(pi)/(60) rad//s`B. `(pi)/(30) rad//s`C. `(pi)/(120) rad//s `D. `(pi)/(2) rad//s` |
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Answer» Correct Answer - B `omega=(2pi)/(T)=(2pi)/(60)=(pi)/(30) rad//s.` |
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| 70. |
The ratio of angular speeds of minutes hand and seconds hand in a watch isA. `60:1`B. `30:1`C. `1:30`D. `1:60` |
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Answer» Correct Answer - D ` omega _("minuteshand")/( omega _("secondhand"))=(pi//1800)/(pi//30)=1:60`. |
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| 71. |
A stone of mass of 16 kg is attached to a string 144 m long and is whirled in a horizontal circle. The maximum tension the string can withstand is 16 Newton . The maximum velocity of revolution that can be given to the stone without breaking it, will beA. `20m//s`B. `16m//s`C. `14m//s`D. `12m//s` |
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Answer» Correct Answer - D Maximum tension `=(mv^(2))/(r)=16N` `therefore (16xxv^(2))/(144)=16 therefore v=12m//s`. |
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| 72. |
A stone of mass of 16 kg is attached to a string 144 m long and is whirled in a horizontal circle. The maximum tension the string can withstand is 16 Newton . The maximum velocity of revolution that can be given to the stone without breaking it, will beA. `20ms^(-1)`B. `16ms^(-1)`C. `14ms^(-1)`D. `12ms^(-1)` |
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Answer» Correct Answer - D Maximum tension `=(mv^(2))/(r)=16NrArr(16xxv^(2))/(144)=16` `therefore" "v=12ms^(-1)` |
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| 73. |
A body of mass m tied to a string of lrngth r is at its lowest position . What should be the minimum speed given to it so as just to complete one revolution ?A. `sqrtgr`B. `sqrt3gr`C. `sqrt5gr`D. `sqrt7gr` |
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Answer» Correct Answer - C `v_(L)=sqrt(5rg)` |
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| 74. |
The maximum velocity at the lowest point, so that the string just slack at the highest point in a vertical circle of radius l.A. `sqrt(gl)`B. `sqrt(3gl)`C. `sqrt(5gl)`D. `sqrt(7 gl)` |
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Answer» Correct Answer - C `V_("min")=sqrt(5gl)` |
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| 75. |
When the car is turning round a curve, the person sitting in the will experienceA. tangential forceB. centrifugal forceC. frictional forceD. centripetal foce |
| Answer» Correct Answer - B | |
| 76. |
A body of mass 1 kg is moving in a vertical circular path of radius 1 m. The difference between the kinetic energies at its highest and lowest position isA. 20 JB. 10 JC. `4sqrt5` JD. `10(sqrt5-1)` J |
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Answer» Correct Answer - A Difference between the kinetic energies at its highest and lowest positions is `2mgr=2xx1xx10xx1=20J` |
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| 77. |
A body of mass 1 kg is rotating in a verticle circle of radius 1 m .What will be the difference in its kinetic energy at the top and bottom of the circle ? `(g=10m//s^(2))`A. 10 jB. 30 jC. 20 jD. 50 j |
| Answer» Correct Answer - C | |
| 78. |
In the above problem, the centripetal acceleration experienced by the bob will beA. `17.3 m//s^(2)`B. `5.8 m//s^(2)`C. `10 m//s^(2)`D. `5 m//s` |
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Answer» Correct Answer - A `(v^(2))/(r)=g tan theta ` `=10xx tan 60 =10xx1.732` `a_(r)=17.32 m//s^(2)`. |
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| 79. |
A small particle of mass `036g` rests on a horizontal turntable at a distance `25cm` from the axis of spindle. The turntable is acceleration at rate of `alpha=(1)/(3)rads^(-2)` . The frictional force that the table exerts on the particle `2s` after the startup isA. `40muN`B. `30muN`C. `50muN`D. `60muN` |
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Answer» Correct Answer - C `f=ma=msqrt(a_(t)^(2)+a_(r)^(2))` `=msqrt((Ralpha)^(2)+(Romega)^(2))` `=msqrt((Ralpha)^(2)+[R(alphat)^(2)]^(2))` `=0.36xx10^(-3)sqrt((0.25xx(1)/(3))^(2)+[0.25((1)/(3)xx2)^(2)]^(2)` `=50xx10^(-6)N` `=50muN` |
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| 80. |
A table with smooth horizontal surface is turning at an angular speed `omega` about its axis. A groove is made on the surface along a radius and a particle is gently placed inside the groove at a distance a from the centre. Find the speed of the speed of the particle as its distance from teh centre becomes L. |
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Answer» Correct Answer - A::B At distance `x` from centre, Centrifugal force `a=xomega^(2)` ltbr. Or `v(dv)/(dx)=xomega^(2)` or `int_(0)^(v)vdv=omega^(2)int_(a)^(L)xdx` or `(v^(2))/(2)=(omega^(2))/(2)(L^(2)-a^(2))` or `v=omegasqrt(L^(2)-a^(2))` |
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| 81. |
A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity `omega` in a circular pathof radius R. A smooth groove AB of length `L(ltltR)` is made on the surface of the table. The groove makes an angle` theta` with the radius OA of the circle in which the cabin rotates. A small particle is kept at the point A in the groove and is released to move along AB. Find the time taken by tehparticle to reach the point B. |
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Answer» Correct Answer - A::B::C The cabin rotates with angular velocity `omegas` and radius R. `:.` The particle experiences a force `mRomega^2`. The component of `mRomega^2` alogn the provided force to the particle to move along AB. `:. mRomega^2costheta=ma` `rarr a=Romega^2 cos theta` Length of groove L` L=ut+1/2at^2` `rarr L=1/2 Romega^2 cos theta t^2` `rarr t^2=(2L)/(Romega^2costheta)` `rarr t=sqrt((2L)/(Romega^2-costheta))` |
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| 82. |
The bob of a 0.2 m pendulum describs an arc of circle in a vertical plane. If the tension in the cord is `sqrt(3)` times the weight of the bob when the cord makes an angle `30^(@)` with the vertical, the acceleration of the bob in that position is `g//n`. Find value of `n`. |
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Answer» Correct Answer - 1 |
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| 83. |
A particle performs verticle circular motion along the circular path . If the ratio of kinetic energy to potential energy of a particle at any position is (If the particle makes an angle `theta ` wityh vertical at the position .)A. `(3+2 cos theta)/(1- cos theta)`B. `(1)/(2)((3+2 cos theta ))/((1- cos theta ))`C. `(1- cos theta )/(3+2 cos theta )`D. `(1+ cos theta )/(3-2 cos theta )` |
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Answer» Correct Answer - B `(KE)/(PE)=((1)/(2)mv_(p)^(2))/(mgh)=(1)/(2)[(3+2 cos theta )/(1- cos theta)]` |
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| 84. |
Particles are released from rest A and side down the smooth surface of hight h to a conveyor B. The correct angular veleocity `omega` of the coneyor pulley of readius r to prevent any siding on the beit as the particles transfer to the conveyou is A. `sqrt(2gh)`B. `(2gh)/(r)`C. `sqrt(2gh)/(r)`D. `(2gh^(2))/(r^(2))` |
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Answer» (c) First of all, we have to calculate the velocity of particles at the point B. `:.` Loss in PE =gain in KE or mgh `=(1)/(2)mmv^(2)` `:.` To Prevent sliding , `v=r_(omega)` `:.` `omega=(v)/(r)=sqrt(2gh)/(r)` |
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| 85. |
Length of a simple pendulum is 2 m and mass of its bob is 0.2 kg .If the tension in the string exceeds 4 N, it will break. If `g=10 m//s^(2)` and the bobis whirled in a horizontal plane , the maximum angle through which the sting can make with vertical during rotation isA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - C `L=2m ,m=0.2 kg,T=4N` `T=mg(3-2 cos theta)` `4=0.2xx10[3-2 cos theta ]` `2=3-2 cos theta` `2-3=-2 cos theta` `+(1)/(2)=cos theta ` `theta = cos ((1)/(2))=60^(@)` |
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| 86. |
A simple pendulum of length 1 m the bob performs circular motion in horizontal plane if its string making an angle ` 60 ^(@)` with the verticle , then the period of rotation of the bob will be `(g=10 m//s^(2)`A. `2 s`B. `1.4 s`C. `1,98 s`D. `4 s` |
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Answer» Correct Answer - B `T=2pi sqrt((l cos theta)/(g))` `=2pi sqrt((1xx cos 60)/(10))` `=2pisqrt((1)/(2)xx(1)/(10))=sqrt2=1.48`. |
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| 87. |
A pendulum bob on a 2 m string is displaced `60^(@)` from the verticle and then released . What is the speed of the bob as it passes through the lowest point in its path?A. `sqrt2 m//s`B. `sqrt 9.8 m//s`C. `4.43 m//s`D. `1//sqrt2 m//s` |
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Answer» Correct Answer - C `v=sqrt(2gl(1- cos theta))` `=sqrt(2xx9.8xx2(1- cos 60^(@)))=4.43 m//s`. |
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| 88. |
The radius of the curved road on a national highway is `R`. The width of the road is `b`. The outer edge of the road is raised by `h` with respect to the inner edge so that a car with velocity `v` can pass safe over it. The value of `h` isA. `(v^(2)b)/(Rg)`B. `(v)/(Rgb)`C. `(v^(2)R)/(g)`D. `(v^(2)b)/(R )` |
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Answer» Correct Answer - A `tan theta=(h)/(b)=(v^(2))/(Rg)` `h=(v^(2)b)/(Rg)` |
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| 89. |
A body of mass m is on the top point of a smooth hemisphere of radius 5 m It is released to slide down the surface when its velocity is `5 m//s` . At this instant the angle made by the radius vector of the body with verticle is `( g=10 m//s^(2))`A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - C `mv^(2)/(r)=mg cos theta ` `v^(2)=r g cos theta ` `therefore cos theta =(v^(2))/(rg)=(25)/(5xx10)=(1)/(2)` ` theta =60^(@)` |
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| 90. |
A Particle is kept at rest at the top of a sphere of diameter `42m`.when disturbed slightly, it slides down. At what height `h` from the bottom, the particle will leave the sphereA. 14 mB. 28 mC. 35 mD. 7 m |
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Answer» Correct Answer - C As we know for hemisphere the pqrticle will leave the sphere at height `h=2r//3` `h=(2)/(3)xx21=14` but from the bottom `H=h+r=14+21=35` |
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| 91. |
A particle describes a horizontal circle in a conical funne whoses inner surface is smooth with speed of `0.5m//s` . What is the height of the plane of circle from vertex the funnel?A. 0.25 cmB. 2 cmC. 4 cmD. 2.5 cm |
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Answer» Correct Answer - D `v=sqrtgh` |
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| 92. |
A small block of mass m projected horizontally from the top of the smooth and fixed hemisphere of radius r with speed `u` as shown. For values of `u ge u_(0)(u_(0)=sqrt(gr))` it does not slide on the hemisphere. `l` i.e., leaves the surface at the top itself. In the above problem find its net acceleration at the instant it levels the hemisphere.A. zeroB. g/2C. gD. g/3 |
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Answer» Correct Answer - C |
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| 93. |
A small block of mass m projected horizontally from the top of the smooth and fixed hemisphere of radius r with speed `u` as shown. For values of `u ge u_(0)(u_(0)=sqrt(gr))` it does not slide on the hemisphere. `l` i.e., leaves the surface at the top itself. For `u=u_(0)//3`, Find the height from the ground at which it leaves the hemisphereA. `(19 r)/(9)`B. `(19 r)/(27)`C. `(10 r)/(9)`D. `(10 r)/(27)` |
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Answer» Correct Answer - B |
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| 94. |
A mass 2 kg describes a circle of radius 1 m on a smooth horizontal table at a uniform speed .If is joined to the centre of the circle by a string, which can just withstand 32 N, then the greatest number of revolution per minute , perfomed by the mass would beA. 38B. 4C. 76D. 16 |
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Answer» Correct Answer - A `T=mr omega^(2)=m r 4 pi ^(2)n^(2)` ` therefore n=sqrt((T)/(mr4pi^(2)))=sqrt(((2xx16))/(2xx1xx4xx10))` `=(2)/(sqrt10)rps` `=(120)/(3.16)=38rpm`. |
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| 95. |
A small block of mass `m` is released from rest from point `A` inside a smooth hemisphere bowl of radius `R`, which is fixed on group such that `OA` is horizontal. The ratio `(x)` of magnitude of centripetal force and normal reaction on the block at any point `B` varies with `theta` as : A. B. C. D. |
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Answer» Correct Answer - A |
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| 96. |
A stone is fastened to one end of a string and is whirled in a verticla circle of radius R. Find the minimum speed the stone can have at the highest point of the circle. |
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Answer» At highest point of vertical circle, `(mv^2)/R=mg` `rarr v^2=Rg` `rarr v=sqrt(Rg)` |
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| 97. |
A particle is moving with speed `v= 2t^(2)` on the circumference of a circle of radius R. Match the quantities given in Table-1 with corresponding results in Table-2 `{:(,"Table-1",,"Table-2"),("(A)","Magnitude to tangential acceleration of particle","(P)","decreases with time"),("(B)","Magnitude of centripetal acceleration of particle","(Q)","increases with time"),("(C)","Magnitude of angular speed of particle with respect to centre of circle","(R)","remains constant"),("(D)","Angle between the total acceleration and centripetal acceleration of partilce","(S)","depends upon the value of radius R"):}` |
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Answer» Correct Answer - (A)Q,(B)QS,(C)QS,(D)PS |
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| 98. |
A gramphone record is revolving with an angular velocity omega. A coin is placed at a distance `r` from the centre of the record. The static coefficient of friction is mu. The coin will revolve with the record if.A. `rge(mug)/(omega^(2))`B. `r=mugomega^(2)`C. `rlt(omega^(2))/(mug)`D. `rle(mug)/(omega^(2))` |
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Answer» Correct Answer - D `momega^(2)rlemumg` `rlarr(mug)/(omega^(2))` |
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| 99. |
A vehicle moves on a horizontal curved road of radius of curvature 50 m height of centre of gravity 1.5 m the distance between the two wheels 2 m and acceleration due to gravity `9.8 m//s ^(2)` . Then the maximum velocity with which it can travel on the road will beA. `18 m//s`B. `20 m//s`C. `19 m//s`D. `17 m//s` |
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Answer» Correct Answer - A `v_("max")=sqrt((rgd)/(2h))` |
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| 100. |
Sometimes the overturning of vehicle on horizontal curved road takes place , it is due toA. centripetal forceB. centrifugal forceC. heavy weight of vehicleD. all of these |
| Answer» Correct Answer - B | |