513 + Interview Questions in CIRCULAR-MOTION IN GENERAL AWARENESS Page 9 InterviewSolution

401.	Which of then following statements are true for cream separators ?A. the particles of cream are lighter so that they will experience less centripetal force and will follow circular path of smaller radiusB. the particles of milk are heavier so thet they will experience large centripetal force and will follow the circular path of large radiusC. cream particle will collect near the axis of rotationD. all of the above
Answer» Correct Answer - D

Discussion

402.	A car moves with a constant speed on a road. The normal reaction exerted by the road on the car is `N_(A) and N_(B)` when it is at the points A and B respectively, then A. `N_(A)=N_(B)`B. `N_(A)gtN_(B)`C. `N_(A) lt N_(B)`D. insufficient information to decide the relation of `N_(A) and N_(B)`
Answer» Correct Answer - C

Discussion

403.	Centrifugal force is not a real force, but it arises due toA. accelerated frame of referanceB. mass of rotating bodyC. presence of centrifugal forceD. non accelerated frame of referance
Answer» Correct Answer - A

Discussion

404.	Centrifugal force is pseudo force becauseA. its magnitude is equal to centripetal forceB. origin can not be imaginaryC. its direction is outward along radiusD. it is not provided by any real force but it arises due to acceleration frame of reference
Answer» Correct Answer - A

Discussion

405.	A cyclist starts from the centre O of a circular park of radius 1km, reaches the edge P of the park, then cycles along the PQ cicumference and returns to the centre along OQ as shown in fig. If the round trip taken ten minute, the net displacement and average speed of the cylists (in kilometer and kinetic per hour) is A. 0, 1B. `(pi+4)/(2), 4`C. `21.4,(pi+4)/(2)`D. `0,21.4`
Answer» Correct Answer - D Average speed = `((2R+piR//2)km)/((1//6)h)` `=(12R+3piR)kmh^(-1)=(12+3pi)(km)/(h)" (as R = 1 km)"` `="21.4 kmh"^(-1)`

Discussion

406.	A circular curve of a highway is designed for traffic moving at 72 km/h. if the radius of the curved path is 100 m, the correct angle of banking of the road should be given by:A. `tan^(-1).(2)/(3)`B. `tan^(-1).(3)/(5)`C. `tan^(-1).(2)/(5)`D. `tan^(-1).(1)/(4)`
Answer» Correct Answer - C `V=sqrt(gR tan theta)rArr (20)^(2)=10xx100xx tan theta` `rArr tan theta=4/10=2/5rArr theta=tan^(-1) (2//5)`

Discussion

407.	A body is moving along a circular path with constant speed. If the direction of rotation is reversed and the speed is doubled,A. direction of centripetal acceleration is reversedB. direction of centripetal acceleration is remains unchangeC. magnitude of centripetal acceleration is doubledD. magnitude of centripetal acceleration is halved
Answer» Correct Answer - B

Discussion

408.	A circular curve of a highway is designed for traffic moving at 72 km/h. if the radius of the curved path is 100 m, the correct angle of banking of the road should be given by:A. `tan^(-1)((2)/(5))`B. `tan^(-1)((3)/(5))`C. `tan^(-1)((1)/(5))`D. `tan^(-1)((1)/(4))`
Answer» Correct Answer - A `v=72kmh^(-1)=72xx(5)/(18)ms^(-1)=20ms^(-1)` `theta=tan^(-1)((v^(2))/(Rg))=tan^(-1)((400)/(100xx10))=tan^(-1)((2)/(5))`

Discussion

409.	A particle of mass `m` is moving in a plane along a circular path of radius `r`. Its angular momentum about the axis of rotation is `L`. The centripetal force acting on the particle is.A. `(L^(2))/(mr)`B. `(L^(2))/(mr^(2))`C. `(L^(2))/(mr^(3))`D. `(L)/(mr^(2))`
Answer» Correct Answer - C `L=mvr thereforev=(L)/(mr)` `F=(mv^(2))/(r)=(L^(2))/(mr^(3))`

Discussion

410.	The angular velocity of a particle is given by `omega=1.5t-3t^(@)+2`, Find the time when its angular acceleration becomes zero.
Answer» `alpha=(domega)/(dt)=1.5-6t=0 Rightarrow t=0.25s`

Discussion

411.	A ball of mass `0.25kg` attached to the end of a string of length `1.96m` moving in a horizontal circle. The string will break if the tension is more than `25N`. What is the maximum speed with which the ball can be moved.A. `14m//s`B. `3m//s`C. `3.92m//s`D. `5m//s`
Answer» Correct Answer - A `T=(mv^(2))/(L)le25` `((0.25)(v^(2)))/(1.96)le25` `vle14` `v_(max)=14m//s`

Discussion

412.	A bullet of mass m moving with a horizontal velocity u strikes a stationary wooden block of mass M suspended by a string of length L = 50 cm. The bullet emerges out of the block with speed `(u)/(4)`. If M = 6 m, minimum value of u so that the block can complete the vertical circle (take, g = 10 `ms^(-2)`)A. 10 `ms^(-1)`B. 20 `ms^(-1)`C. 30 `ms^(-1)`D. 40 `ms^(-1)`
Answer» Correct Answer - D From mometum conservaion, `"mu"=m(u)/(4)+(6m)sqrt(5xx10xx0.5)` Solving we get, u = 40 `ms^(-1)`

Discussion

413.	In the shown figure inside a fixed hollow cylinder with vetical axis a pendulumm is rotating in a conical path with its axis same as that of the cylinder with uniform angular velocity. Radius of culinder is 30cm, length of string is 50cm and mass of bob is 400gm. The bob makes contact with the inner fricitonless wall of the cylinder while moving A. The minimum value of angular velocity of the bob so that it does not leave contact is s 5 rad/sB. Tension in the string is 5N for all values of angular velocityC. For angular velocity of 10 rad/s the bob pushes the cylinder with a force of 9ND. For angular velocity of 10 rad/s, tension in the string is 20N
Answer» Correct Answer - A::B::C

Discussion

414.	A block of mass `m` is placed inside a smooth hollow cylinder of radius `R` kept horizontally. Initially system was at rest. Now cylinder is given constant acceleration `2 g` in the horizontal direction by external agent. Find the maximum angular displacement of the block with the vertical. .A. `2 tan^(-1)(2)`B. `tan^(-1)(2)`C. `tan^(-1)(1)`D. `tan^(-1)((1)/(2))`
Answer» Correct Answer - A

Discussion

415.	On a train moving along east with a constant speed v, a boy revolves a bob with string of length l on smooth surface of a train, with equal constant speed v relative to train. Mark the correct option(s). A. Maximum speed of bob is 2 v in ground frameB. Tension in string connecting bob is `(4mv^(2))/(l)` at an instantC. Tension in string is `(mv^(2))/(l)` at all the moments.D. Minimum speed of bob is zero in ground frame
Answer» Correct Answer - A::C::D

Discussion

416.	Fing the maximum speed with which an automobile can round a curve of radius 8 m without slipping of the road is unbanked and he coefficient of friction between the orad an the tyres is `0.8 (g=10 m//^(2))`A. 8 m/sB. 10 m/sC. 20 m/sD. None of these
Answer» (a) Here centripetal force is provided by force of friction. `:." " f=mu_(s)N` `:." " f(mv^(2))/(r)` or `" " mu_(s)N=(mv^(2))/(r)` or `" " mu_(s) mg=(mv^(2))/(r)` `:.`" " v=(r)/sqrt(mugr)=sqrt(0.8xx108)=8 m//s`

Discussion

417.	A body is sliding down an inclined plane forming an angle `30^(@)` with the horizantal. If the coefficient of friction is`0.3` then acceleration of the body isA. `1.25.ms^(-2)`B. `2.35 ms^(-2)`C. `3.4ms^(-2)`D. `4.9ms^(-2)`
Answer» Correct Answer - B `a=g(sin theta-mu_(k) cos theta)`

Discussion

418.	A heavy uniform chain lies on horizantal table top. If the coefficient of friction between the chain and the table surface is `0.5`, the maximum percentage of the length of the chain that can hang over one edge of the table isA. `20%`B. `33.3%`C. `76%`D. `50%`
Answer» Correct Answer - B `l/Lxx100=mu/(mu+1)xx100`

Discussion

419.	A body of mass `2kg` moving on a horizantal surface with an initial velocity of comes to rest after `2` second. If one wants to keep this body moving on the same surface with a velocity of `4 ms^(-1)` the force required isA. zeroB. `2 N`C. `4 N`D. `8 N`
Answer» Correct Answer - C `v=u+at,F=ma`

Discussion

420.	A toy car of mass `m` can travel at a fixed speed. It moves in circle on a fixed horizantal table. A string is connected to the car and attached to a block os mass `M` that hangs as shown in figure (the portion of string below the table is always vertical). The coefficient of friction between the surface of table and tyres of the toy car is `mu`. Find the ratio of the maximum radius to the minimum radius for which the toy car can move in a circular path with center `O` on table. `(Given M = 3kg,m=2kg,mu=1/2)`
Answer» Correct Answer - B If `mg gt (mv^(2))/r`,friction force acts outwards `T-mumg=(mv^(2))/r_(max)` If `mg gt (mv^(2))/r` fractional froce on car towards centre `T+mu mg=(mv^(2))/r_(min)`

Discussion

421.	An unbanked curve has a radius of 60m. The maximum speed at which a car can make a turn if the coefficient of static friction is `0.75` , isA. `2.1 m//s`B. `14 m//s`C. `21 m//s`D. `7 m//s`
Answer» Correct Answer - C `v_("max")=sqrt mu rg=sqrt(0.75xx60xx9.8)=21 m//s`

Discussion

422.	A block of weight `100N` is pushed by a force `F` on a horizantal rough plane moving with an acceleration `1m//s^(2)` when force is doubled its acceleration becomes `10m//s^(2)`. The coefficient of friction is `(10m//s^(-2))`A. `0.4`B. `0.6`C. `0.5`D. `0.8`
Answer» Correct Answer - D `F_(R)=F-f,f=mu_(k)mg`

Discussion

423.	A block of mass `5 kg` is lying on a rough horizontal surface. The coefficient of static and Kinetic friction are `0.3 and 0.1` and `g=10m//s^(-2)` If a horizontal force of `50N` is applied on the block, the frictional force isA. `25N`B. `5N`C. `10N`D. Zero
Answer» Correct Answer - B `f_(s)=mu_(S)N,f_(k)=mu_(k)N,N=mg`

Discussion

424.	When a train starting from rest is uniformly accelerating, a plumb bob hanging from the roof of a compartment is found to be inclined at an angle of `45^(@)` with the vertical.The time taken by the train to travel a distance of `1//2km` will be nearlyA. `7s`B. `10s`C. `15s`D. `25s`
Answer» Correct Answer - B `alpha cos theta= g sin theta, s=ut+1/2 at^(2)`

Discussion

425.	Two bodies of masses 8 kg and 4 kg are moving in concentric circular orbits of radii `r_(1) and r_(2) ` respectively . If their time periods are same , the ration of their centripetal accelerations isA. `r_(1):r_(2)`B. `2r_(1):r_(2)`C. `r_(1):2r_(2)`D. `r_(1): r_(2)`
Answer» Correct Answer - A `a_(1):a_(2)=r_(1)omega_(1)^(2):r_(2)omega_(2)^(2)=r_(1):r_(2)" " (therefore omega _(1): omega_(2))`

Discussion

426.	Two buses A and B are moving around concentric circular pathe of radii `r_(A)` and `r_(B)` If the two buses complete the circular paths in the sme time. The ratio on their linear speeds isA. 1B. `r_(A)//R_(B)`C. `r_(B)//r_(A)`D. None
Answer» Correct Answer - (b)

Discussion

427.	A fan makes 2400 rpm. If after it is switched off, it comes to rest in 10 s, then find the number of times it will rotate before it comes to rest after it is switched off.A. 400B. 100C. 200D. 50
Answer» Correct Answer - C Given, `f=2400"rpm"=(2400)/(60)=40Hz` `therefore" " omega=2pif=80pi"rad s"^(-2)` Now, `theta=[(omega_(0)+omega)/(2)]t` `rArr" theta=[(80pi+0)/(2)]xx10=400pirad` `therefore` Fan will turn angle `400pi` after it is swtiched off `rArr` The number of rotation `=(theta)/(2pi)=(400pi)/(2pi)=200`

Discussion

428.	The motor of an engine is rotating about its axis with an angular velocity of 100 rpm. It comes to rest in 15 s after being switched off. Assuming constant angular deceleration, calculate the number of revolution made by it before coming to rest.A. `12.5`B. 40C. `32.6`D. `15.6`
Answer» Correct Answer - A From equation of motion (angular), `0=omega_(0)-alphatrArralpha=(omega_(0))/(t)=((100xx2pi)//60)/(15)=0.69"rad s"^(-2)` Now, angle rotated before coming to rest is given by `theta=(omega_(0)^(2))/(2alpha)=((100xx2pi)/(60))^(2)/(2xx0.7)=78.25rad` `therefore" Number of revolutions"=(theta)/(2pi)=12.5`

Discussion

429.	A wheel is rotating at 900 rpm about its axis. When the power is cut off, it comes to rest in 1 min. The angular retardation (in rad `s^(-2)`) isA. `(pi)/(2)`B. `(pi)/(4)`C. `(pi)/(6)`D. `(pi)/(8)`
Answer» Correct Answer - A As, `omega=omega_(0)+alphat` Here, `omega_(0)=900"rpm"=((2pixx900))/(60)"rad s"^(-1),omega=0` and `t=60s, 0=(2pixx900)/(60)+alphaxx60` `alpha=(2pixx900)/(60xx60)=(pi)/(2)`

Discussion

430.	A wheel having moment of interia `2 kgm^(-2)` about its axis, rotates at 50 rpm about this axis. The angular retardation that can stop the wheel in one minute isA. `(pi)/(36)"rad s"^(-2)`B. `(pi)/(18)"rad s"^(-2)`C. `(pi)/(72)"rad s"^(-2)`D. `(pi)/(9)"rad s"^(-2)`
Answer» Correct Answer - A The initial angular velocity `="50 rpm"=(5pi)/(3)"rad s"^(-1)` Using, `omega=omega_(0)+alphat` `alpha=(omega-omega_(0))/(t)=(0-(5pi)/(3))/(6)"rad s"^(-2)=-(pi)/(36)"rad s"^(-2)`

Discussion

431.	The angular velocity of a wheel is `70 rad//sec` . If the radius of the wheel is 0.5 m , then linear velocity of the wheel isA. `10 m//s`B. `20 m//s`C. `35 m//s`D. `70 m//s`
Answer» Correct Answer - C `V=r omega =0.5 xx70=35 m//s`

Discussion

432.	The wheel of a toy car rotates about a fixed axes. It slows down from 400 rps to 200 rps in 2 s. Then, its angular retardation in `"rad s"^(-2)` is (rps = revolutions per second)A. `200pi`B. `100pi`C. `400pi`D. None of these
Answer» Correct Answer - A `alpha=(Deltaomega)/(Deltat)"but "omega=2pif` `therefore"Angular acceleration",alpha=(2piDeltaf)/(Deltat)=(2pixx200)/(2)` `=(200pi)rads^(-2)`

Discussion

433.	A wheel starts from rest and acquires an angular velocity of `60 rad//s` in half a minute . Then its angular acceleration isA. `4 rad//s^(2)`B. `2 rad//s^(2)`C. `1 rad//s^(2)`D. `0.5 rad//s^(2)`
Answer» Correct Answer - B `alpha=(omega_(2)-omega_(1))/(t)=(60)/(30)=2 rad//sec^(2)`.

Discussion

434.	The wheel of a motor rotates with a constant acceleration of `4" rad s"^(-1)`. If the wheel starts form rest, how many revolutions will it make in the first 20 second?
Answer» The angular displacement in the first 20 s is given by `theta=omega_(0)t+(1)/(2)alphat^(2)=(1)/(2)("4 rads"^(-2)(20s)^(2))` `(because "Angular velocity" omega=0)` = 800 rad As, the wheel turns by `2pi` radian in each revolution, the number of revolutions in 20 s is `N=(theta)/(2pi)=(800)/(2pi)=128`

Discussion

435.	The car of a wheel rotating with certain angular velocity is stopped in 7 seconds and before it stops , it makes 35 revolutions . Then initially it was rotating with the frequancy .A. `10 Hz`B. `20 Hz`C. `15 Hz`D. `30 Hz`
Answer» Correct Answer - A `N=(n_(1)+n_(2))/(2)xxt` `thereforen_(1)=(2N)/(t)-n_(2)=(2xx35)/(7) -0=10 Hz`.

Discussion

436.	A body starts rotating from rest and completes 10 revolutions in 4 s .Find its angular accelerationA. `2.5 pi rad//s^(2)`B. `5 pi rad//s^(2)`C. `7.5 pi rad //s^(2)`D. `10 pi rad //s^(2)`
Answer» Correct Answer - A `theta=omega_(1)t+(1)/(2)alphat^(2)` `therefore 20 pi= 0+(1)/(2)xxalphaxx16` `therefore alpha=(20pi)/(8)=2.5 rad//sec^(2)`.

Discussion

437.	A mass 2 kg is whirled in a horizontal circle by means of a string at an initial speed of 5 revolutions per minute . Keeping the radius constant the tension in the string is doubled. The new speed is nearlyA. 14 rpmB. 10 rpmC. 2.25 rpmD. 7 rpm
Answer» Correct Answer - D Tension In the string `T=m omegaa ^(2) r=4pi^(2)n^(2)` mr `therefore T alpha n^(2)implies(n_(2))/(n_(1))=sqrt((T_(2))/(T_(1)))impliesn_(2)=5sqrt((2T)/(T))=7"rpm"`.

Discussion

438.	As mosquito is sitting on an L.P. record disc rotating on a trun tabel at `33 1/3` revolutions per minute. The distance of the mosquito from the centre of the turn table is 10 cm. Show that the frictioncoefficient between the record and the mosquito is greater than `pi^2/81`. Take `g=10m/s^2`
Answer» Correct Answer - A A mosquito is sitting on an L.P record disc and rotating on a turn table at `33 1/2` rpm. `n=33/1/3 rpm` `100(3xx60)rps` `:. omega==2pixx100/180` `=(10pi)/9 rad/sec` `r=10cm=0.4m` `g=10 m/sec^2` `mumg?=mromega^2` `:. mugt(romega^2)/g.=(0.1xx(10pi/9)^2)/10` `rarr mugt(pi^2/81)`

Discussion

439.	A sphere of mass 200 g is attached to an inextensible string of length 130 cm whose upper end is fixed to the ceilling . The sphere is made to describe a horizontal circle of radius 50 cm Calculate the periodic time of this conical pendulum and the tension in the string .A. `1.2 s`B. `2.2 s`C. `1.5 s`D. `3 s`
Answer» Correct Answer - B `m=0.2 kg , l_(1)=130 cm, T=2 pi sqrt((l cos theta)/(g))` `h^(2)=l^(2)-r^(2)` `=1.69-0.25` `h^(2)=1.44` `h=1.2` `T=2pi sqrt((h)/(g))` `2 pi sqrt((1.2)/(9.8))=2xx1.1=2.2 s `

Discussion

440.	A `6.0 kg` object is suspended by a vertical string from the ceilling of an elevator which is acceleration upward at a rate of `2.2ms^(-2)`. the tension in the string isA. `11N`B. `72N`C. `48N`D. `59N`
Answer» Correct Answer - B `T=m(g+a)`

Discussion

441.	A rod of length L is pivoted at one end and is rotated with as uniform angular velocity in a horizontal plane. Let `T_1 and T_2` be the tensions at the points L//4 and 3L//4 away from the pivoted ends.A. `T_1gtT_2`B. `T_2gtT_1`C. `T_1=T_2`D. `The relation between `T_1 and T_2` depends on whether the rod rotates clockwise or anticlockwise.
Answer» Correct Answer - A

Discussion

442.	(a) How many revolutions per minute must the apparatus shown in figure make about a vertical axis so that the cord makes an angle of `45^(@)` with the vertical ? (b)What is the tension in the cord then? Given, `l=sqrt(2)m,a=20 cm` and `m=5.0 kg`?
Answer» Correct Answer - C `(a) r=a+l sin 45^(@)=(0.2)+(sqrt(2))(1/(sqrt(2)))=1.2 m` Now, `T cos 45^(@)=mg.....(i)` and `T sin 45^(@)=mr omega^(2).....(ii)` From eqn. (i) and (ii), we have `omega=2n pi =sqrt(g/r)` `:. n=1/(2pi)sqrt(g/r)=60/(pi)sqrt(9.8/1.2)rpm=27.3 r pm ` (b). From eq. (i), we have `T=sqrt(2)mg=(sqrt(2))(5.0)(9.8)` `=69.3 N`

Discussion

443.	A particle moves in a circular path of radius R with an angualr velocity `omega=a-bt`, where a and b are positive constants and t is time. The magnitude of the acceleration of the particle after time `(2a)/(b)` isA. `(a)/(R)`B. `a^(2)R`C. `R(a^(2)+b)`D. `Rsqrt(a^(4)+b^(2))`
Answer» Correct Answer - D `alpha=(domega)/(dt)=-b` `a_(t)=Ralpha=-Rb` At `t=(2a)/(b),omega=-a` `a_(n)=Romega^(2)=Ra^(2)` Now, `a=sqrt(a_(t)^(2)+a_(n)^(2))=Rsqrt(a^(4)+b^(2))`

Discussion

444.	A particle moves in a circle of radius ` 4.0 cm` clockwese at constant speed of ` 2cm S^(-1)` . If ` hat x` and ` hat y` ar unit accleration vectors along X- asis and Y-axis respectively, find the accleration of the particle at the instant half way between ` PQ. Fig. 2 ( d) . 38. .A. `-4(hatx+haty)`B. `4(hatx+haty)`C. `-(hatx+haty)1/sqrt2`D. `(hatx+haty)4`
Answer» Correct Answer - A::B Acceleration vetor `bara=v^(2)/R(-hatR)` `=-((Rcos45hatx+Rsin45haty))/R=-(hatx+haty)1/sqrt2`

Discussion

445.	A pendulum of length `l=1m` is released from `theta_(0)=60^(@)` . The rate of change of speed of the bob at `theta=30^(@)` is. A. `5 sqrt(3) m//s^(2)`B. `5 m//s^(2)`C. `10 m//s^(2)`D. `2.5 m//s^(2)`
Answer» Correct Answer - B

Discussion

446.	A pendulum of mass 1 kg and length  = 1m is released from rest at angle  = 60º. The power delivered by all the forces acting on the bob at angle  = 30º will be: (g = 10 m/s2)A. 13.4 WB. 20.4 WC. 24.6 WD. zero
Answer» Correct Answer - A

Discussion

447.	A stone of a mass m tied to a string of length /is rotated in a circle with the other end of the string as the centre. The speed of the stone is V. If the string breaks, the stone willA. move towards the centreB. move away from the centreC. move along a tangentD. stop
Answer» (c) When the string break, threr is not tension in the string hence there is no radial acceleration to change the direction of the stone the required (tentison provide centriperal force ) hence the stone will move along the tangent,

Discussion

448.	A particle of mass `m` is attached to one end of a light inextensible string and the other end of the string is fixed in vertical plane as shown. Particle is given a horizontal velocity `u = sqrt((5)/(2)gl)` The maximum angle made by the string with downward vertical isA. `cos^(-1)((1)/(4))`B. `sin^(-1)((1)/(4))`C. `(pi)/(2)+cos^(-1)((1)/(4))`D. `pi-cos^(-1)((1)/(4))`
Answer» Correct Answer - D

Discussion

449.	A ball of mass `0.6 kg` attached to a light inextensible string rotates in a vertical circle of radius `0.75 m` such that it has speed of `5 ms^(-1)` when the string is horizontal. Tension in the string when it is horizontal on other side is `(g = 10 ms^(-2))`.A. 30 NB. 26 NC. 20 ND. 6 N
Answer» Correct Answer - C Tension `T = (mv^(2))/(r)=((0.6)(25))/(0.75)=20N`.

Discussion

450.	A ball of mass `0.6 kg` attached to a light inextensible string rotates in a vertical circle of radius `0.75 m` such that it has speed of `5 ms^(-1)` when the string is horizontal. Tension in the string when it is horizontal on other side is `(g = 10 ms^(-2))`.
Answer» Tension in the string when it makes angle `theta` with the vertical `T=(mv^(2))/(r)+"mg cos"theta` When the string is horizontal, `theta=90^(@)` `therefore" "T=(mv^(2))/(r)+mgxx0=(mv^(2))/(r)` `=(0.6xx(5)^(2))/(0.75)=20N`

Discussion

Explore topic-wise InterviewSolutions in .

A car moves with a constant speed on a road. The normal reaction exerted by the road on the car is `N_(A) and N_(B)` when it is at the points A and B respectively, then A. `N_(A)=N_(B)`B. `N_(A)gtN_(B)`C. `N_(A) lt N_(B)`D. insufficient information to decide the relation of `N_(A) and N_(B)`

Centrifugal force is not a real force, but it arises due toA. accelerated frame of referanceB. mass of rotating bodyC. presence of centrifugal forceD. non accelerated frame of referance

Centrifugal force is pseudo force becauseA. its magnitude is equal to centripetal forceB. origin can not be imaginaryC. its direction is outward along radiusD. it is not provided by any real force but it arises due to acceleration frame of reference

A circular curve of a highway is designed for traffic moving at 72 km/h. if the radius of the curved path is 100 m, the correct angle of banking of the road should be given by:A. `tan^(-1).(2)/(3)`B. `tan^(-1).(3)/(5)`C. `tan^(-1).(2)/(5)`D. `tan^(-1).(1)/(4)`

A circular curve of a highway is designed for traffic moving at 72 km/h. if the radius of the curved path is 100 m, the correct angle of banking of the road should be given by:A. `tan^(-1)((2)/(5))`B. `tan^(-1)((3)/(5))`C. `tan^(-1)((1)/(5))`D. `tan^(-1)((1)/(4))`

A particle of mass `m` is moving in a plane along a circular path of radius `r`. Its angular momentum about the axis of rotation is `L`. The centripetal force acting on the particle is.A. `(L^(2))/(mr)`B. `(L^(2))/(mr^(2))`C. `(L^(2))/(mr^(3))`D. `(L)/(mr^(2))`

The angular velocity of a particle is given by `omega=1.5t-3t^(@)+2`, Find the time when its angular acceleration becomes zero.

A ball of mass `0.25kg` attached to the end of a string of length `1.96m` moving in a horizontal circle. The string will break if the tension is more than `25N`. What is the maximum speed with which the ball can be moved.A. `14m//s`B. `3m//s`C. `3.92m//s`D. `5m//s`

Fing the maximum speed with which an automobile can round a curve of radius 8 m without slipping of the road is unbanked and he coefficient of friction between the orad an the tyres is `0.8 (g=10 m//^(2))`A. 8 m/sB. 10 m/sC. 20 m/sD. None of these

A body is sliding down an inclined plane forming an angle `30^(@)` with the horizantal. If the coefficient of friction is`0.3` then acceleration of the body isA. `1.25.ms^(-2)`B. `2.35 ms^(-2)`C. `3.4ms^(-2)`D. `4.9ms^(-2)`

A heavy uniform chain lies on horizantal table top. If the coefficient of friction between the chain and the table surface is `0.5`, the maximum percentage of the length of the chain that can hang over one edge of the table isA. `20%`B. `33.3%`C. `76%`D. `50%`

A body of mass `2kg` moving on a horizantal surface with an initial velocity of comes to rest after `2` second. If one wants to keep this body moving on the same surface with a velocity of `4 ms^(-1)` the force required isA. zeroB. `2 N`C. `4 N`D. `8 N`

An unbanked curve has a radius of 60m. The maximum speed at which a car can make a turn if the coefficient of static friction is `0.75` , isA. `2.1 m//s`B. `14 m//s`C. `21 m//s`D. `7 m//s`

A block of weight `100N` is pushed by a force `F` on a horizantal rough plane moving with an acceleration `1m//s^(2)` when force is doubled its acceleration becomes `10m//s^(2)`. The coefficient of friction is `(10m//s^(-2))`A. `0.4`B. `0.6`C. `0.5`D. `0.8`

A block of mass `5 kg` is lying on a rough horizontal surface. The coefficient of static and Kinetic friction are `0.3 and 0.1` and `g=10m//s^(-2)` If a horizontal force of `50N` is applied on the block, the frictional force isA. `25N`B. `5N`C. `10N`D. Zero

When a train starting from rest is uniformly accelerating, a plumb bob hanging from the roof of a compartment is found to be inclined at an angle of `45^(@)` with the vertical.The time taken by the train to travel a distance of `1//2km` will be nearlyA. `7s`B. `10s`C. `15s`D. `25s`

Two bodies of masses 8 kg and 4 kg are moving in concentric circular orbits of radii `r_(1) and r_(2) ` respectively . If their time periods are same , the ration of their centripetal accelerations isA. `r_(1):r_(2)`B. `2r_(1):r_(2)`C. `r_(1):2r_(2)`D. `r_(1): r_(2)`

Two buses A and B are moving around concentric circular pathe of radii `r_(A)` and `r_(B)` If the two buses complete the circular paths in the sme time. The ratio on their linear speeds isA. 1B. `r_(A)//R_(B)`C. `r_(B)//r_(A)`D. None

A fan makes 2400 rpm. If after it is switched off, it comes to rest in 10 s, then find the number of times it will rotate before it comes to rest after it is switched off.A. 400B. 100C. 200D. 50

The motor of an engine is rotating about its axis with an angular velocity of 100 rpm. It comes to rest in 15 s after being switched off. Assuming constant angular deceleration, calculate the number of revolution made by it before coming to rest.A. `12.5`B. 40C. `32.6`D. `15.6`

A wheel is rotating at 900 rpm about its axis. When the power is cut off, it comes to rest in 1 min. The angular retardation (in rad `s^(-2)`) isA. `(pi)/(2)`B. `(pi)/(4)`C. `(pi)/(6)`D. `(pi)/(8)`

A wheel having moment of interia `2 kgm^(-2)` about its axis, rotates at 50 rpm about this axis. The angular retardation that can stop the wheel in one minute isA. `(pi)/(36)"rad s"^(-2)`B. `(pi)/(18)"rad s"^(-2)`C. `(pi)/(72)"rad s"^(-2)`D. `(pi)/(9)"rad s"^(-2)`

The angular velocity of a wheel is `70 rad//sec` . If the radius of the wheel is 0.5 m , then linear velocity of the wheel isA. `10 m//s`B. `20 m//s`C. `35 m//s`D. `70 m//s`

The wheel of a toy car rotates about a fixed axes. It slows down from 400 rps to 200 rps in 2 s. Then, its angular retardation in `"rad s"^(-2)` is (rps = revolutions per second)A. `200pi`B. `100pi`C. `400pi`D. None of these

A wheel starts from rest and acquires an angular velocity of `60 rad//s` in half a minute . Then its angular acceleration isA. `4 rad//s^(2)`B. `2 rad//s^(2)`C. `1 rad//s^(2)`D. `0.5 rad//s^(2)`

The wheel of a motor rotates with a constant acceleration of `4" rad s"^(-1)`. If the wheel starts form rest, how many revolutions will it make in the first 20 second?

The car of a wheel rotating with certain angular velocity is stopped in 7 seconds and before it stops , it makes 35 revolutions . Then initially it was rotating with the frequancy .A. `10 Hz`B. `20 Hz`C. `15 Hz`D. `30 Hz`

A body starts rotating from rest and completes 10 revolutions in 4 s .Find its angular accelerationA. `2.5 pi rad//s^(2)`B. `5 pi rad//s^(2)`C. `7.5 pi rad //s^(2)`D. `10 pi rad //s^(2)`

A mass 2 kg is whirled in a horizontal circle by means of a string at an initial speed of 5 revolutions per minute . Keeping the radius constant the tension in the string is doubled. The new speed is nearlyA. 14 rpmB. 10 rpmC. 2.25 rpmD. 7 rpm

As mosquito is sitting on an L.P. record disc rotating on a trun tabel at `33 1/3` revolutions per minute. The distance of the mosquito from the centre of the turn table is 10 cm. Show that the frictioncoefficient between the record and the mosquito is greater than `pi^2/81`. Take `g=10m/s^2`

A `6.0 kg` object is suspended by a vertical string from the ceilling of an elevator which is acceleration upward at a rate of `2.2ms^(-2)`. the tension in the string isA. `11N`B. `72N`C. `48N`D. `59N`

(a) How many revolutions per minute must the apparatus shown in figure make about a vertical axis so that the cord makes an angle of `45^(@)` with the vertical ? (b)What is the tension in the cord then? Given, `l=sqrt(2)m,a=20 cm` and `m=5.0 kg`?

A particle moves in a circular path of radius R with an angualr velocity `omega=a-bt`, where a and b are positive constants and t is time. The magnitude of the acceleration of the particle after time `(2a)/(b)` isA. `(a)/(R)`B. `a^(2)R`C. `R(a^(2)+b)`D. `Rsqrt(a^(4)+b^(2))`

A pendulum of length `l=1m` is released from `theta_(0)=60^(@)` . The rate of change of speed of the bob at `theta=30^(@)` is. A. `5 sqrt(3) m//s^(2)`B. `5 m//s^(2)`C. `10 m//s^(2)`D. `2.5 m//s^(2)`

A pendulum of mass 1 kg and length  = 1m is released from rest at angle  = 60º. The power delivered by all the forces acting on the bob at angle  = 30º will be: (g = 10 m/s2)A. 13.4 WB. 20.4 WC. 24.6 WD. zero

A stone of a mass m tied to a string of length /is rotated in a circle with the other end of the string as the centre. The speed of the stone is V. If the string breaks, the stone willA. move towards the centreB. move away from the centreC. move along a tangentD. stop

A ball of mass `0.6 kg` attached to a light inextensible string rotates in a vertical circle of radius `0.75 m` such that it has speed of `5 ms^(-1)` when the string is horizontal. Tension in the string when it is horizontal on other side is `(g = 10 ms^(-2))`.A. 30 NB. 26 NC. 20 ND. 6 N

A ball of mass `0.6 kg` attached to a light inextensible string rotates in a vertical circle of radius `0.75 m` such that it has speed of `5 ms^(-1)` when the string is horizontal. Tension in the string when it is horizontal on other side is `(g = 10 ms^(-2))`.