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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
The angular displacement of the rod is defied a s`0=(3)/(20) t^(2)` where `theta` is in radian and t is in second. The collar B slides along the rod in such a way that its distance from O is `r=0.9-0.12t^(2)` where r is in metre and t is secnd The velocity of collar at `theta=30^(@)` is A. `0.45 m//s`B. `0.48 m//s`C. `0.52 m//s`D. `0.27 m//s` |
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Answer» (c) `:.v_(r)=(Dr)/(dt)=-024t` `v_(1)romega=r("d"theta)/(dt)=r((6)/(20)t)=(0.9-0.12t^(2))((6t)/(20))` `:." "v=(sqrt(v_(r)^(2)+v_(t^(2)))` `:.theta=(3)/(20)t^(2)or(pi)/(6)=(3)/(20)t^(2)` `:." " t=sqrt(((20pi)/(18)))` Putting the value of `theta, v=0.25m//s` |
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| 252. |
A body of mass `m` is rotated in a vertical circle with help of light string such that velocity of body at a point is equal to critical velocity at that point. If `T_(1), T_(2)` be the tensions in the string when the body is crossing the highest and the lowest positions then the following relation is correctA. `T_(2)-T_(1)=6 mg `B. ` T_(2)- T_(1)=4 mg `C. `T_(2)-T_(1)=3 mg`D. `T_(2)-T(1)=2 mg ` |
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Answer» Correct Answer - A `f_(1)=(mv^(2))/(r)-mg=0andT_(2)=(mv^(2))/(r)+mg` `=6 mg` `thereforeT_(2)-T_(1)=6 mg.` |
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| 253. |
A body of mass m is revolving along a vertical circle of radius r such that the sum of its kinetic speed of the body at the highest point is `sqrt2 rg` then the speed of the body at the lowest point isA. `sqrt4 gr `B. `sqrt 6 gr`C. `sqrt 2gr`D. `sqrt gr` |
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Answer» Correct Answer - B `v_(b)^(2)-v_(h)^(2)=4 gr` `v_(b)^(2)-2gr=4grimpliesv_(b)=sqrt(6 gr)`. |
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| 254. |
If a small babyof mass m is moving with angular velocity `omega` in a circle of radius r, then its K.E. will beA. `(momega r)/(2)`B. `(m omega ^(2) r )`C. `(m omega ^(2) r^(2))/(2)`D. `(m omega r ^(2))/(2)` |
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Answer» Correct Answer - C `KE=(1)/(2)mv^(2) (v=r omega)` `therefore KE=(1)/(2) mr^(2) omega ^(2) ` `=(m omega ^(2) r^(2))/(2)`. |
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| 255. |
A particle of mass M is moving in a horizontal circle of radius R with uniform speed V. When it moves from one point to a diametrically opposite point, itsA. K.E. changes by `mv^(2)//4`B. momentum does not changeC. momentum changes by 2mvD. K.E. changes by `mv^(2)` |
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Answer» Correct Answer - C `Delta p = 2p sin theta //2` `=2mv sin theta //2=2 mv sin 90 =2 mv`. |
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| 256. |
If a body revolves n times in a circle of radius `pi` cm in one minute, then its linear celocity will beA. `(60)/(2n)cm//s`B. `(2n)/(60)cm//s`C. `(2pi^(2)n)/(60)cm//s`D. `(60)/(2pi^(2))cm//s` |
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Answer» Correct Answer - C `v= r omega =(2pi r )/(T)= ((2 pi xx pi )/(60))n` |
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| 257. |
If a figure, the linear velocity `vecv` in terms of polar coordinate is A. `vecv=vecir cos omegat+vec jr sin omega t`B. `vecv=vecir sin omegat+vec jr cos omega t`C. `vecv=(+vecir sin omegat+vec jr cos omega t)omega`D. `vecv=-vecir omega sin omegat+vec jr omega cos omega t` |
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Answer» Correct Answer - D `vec r =r cos omega t vec I + vec j er omega t vec j ` ` (dvec r )/(dt)=- r omega sin omega t + vec j cos omega t ` |
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| 258. |
The circular orbit of two satellites have radii `r_(1)` and `r_(2)` respectively `(r_(1)ltr_(2))`. If angular velosities of satellites are same, then their centripetal accelerations are related asA. `a_(1)gta_(2)`B. `a_(1)=a_(2)`C. `a_(1)lta_(2)`D. Data insufficient |
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Answer» Correct Answer - C `a=romega^(2)" or "aproprrArr" so if "r_(1)ltr_(2),"then "a_(1)lta_(2)` |
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| 259. |
Two cars of mass ` m_(1) and m_(2) ` are moving in circle of radii `r_(1) and r_(2) ` , respectively . Their speeds are such that they make complete circles in the same time `t` . The ratio of their centripetal acceleration is :A. `m_(1)r_(1):m_(2)r_(2)`B. `m_(1):m_(2)`C. `r_(1):r_(2)`D. `1:1` |
| Answer» Correct Answer - C | |
| 260. |
A car moving on a horizontal road may be thrown out of the road in taking a turn.A. by the gravitational forceB. due to lack of centripetal forceC. due to rolling frictional force between tyre roadD. due to the reaction of the ground |
| Answer» Correct Answer - B | |
| 261. |
A car moving on a horizontal road may be thrown out of the road in taking a turn.A. By the gravitational forceB. Due to lack of sufficient centripetal forceC. Due to friction between road and the tyreD. Due to reaction of earth |
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Answer» Correct Answer - B Here required centripetal force provide by friction force. Due to lack of sufficient centripetal force car thrown out of the road in taking a turn. |
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| 262. |
A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equastion is `x^(2) =4ay`. The wire frame is fixed and the bead is released from the point `y=4a` on the wire frame from rest. The tangential acceleration of the bead when it reaches the position given by `y=a` is A. `g/2`B. `(sqrt3g)/(2)`C. `(g)/(sqrt2)`D. g |
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Answer» Correct Answer - C `x^(2)=4ay,(dy)/(dx)=1/(2a)x` at point `(2a,2a),(dy)/(dx)=1,theta=45^(@)` Component of weight along tangential direction is `mg sin theta` Tangential acceleration is `g sin theta` |
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| 263. |
A string can withstand a tention of 25 N. What is the greatest speed at which a body of mass 1 kg can be whiled in a horizontal circle using 1 m lengt of the string?A. `2.5m//s`B. `5m//s`C. `7.5m//s`D. `10 m//s` |
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Answer» Correct Answer - B `F=(mv^(2))/(r)` `thereforev^(2)=(Fr)/(m)=(25xx1)/(1)` `v^(2)=25` `v=sqrt25=5m//s`. |
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| 264. |
One end of a string of length 1 m is tied to a body of mass `0.5` kg. It is whirled in a vertical circle with angular velocity 4 rad `s^(-1)`. Find the tension in the string when the body is at the lower most point of its motion. (take, g = 10 `ms^(-1)`) |
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Answer» At lower most point, the tension in the string, `T=momega^(2)r+mg` `=0.5xx(4)^(2)xx1+0.5xx10` = 13 N |
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| 265. |
A body is acted on by a force given by `F=(10+2t)N`. The impulse received by by the body during the first four second isA. `40 N s`B. `56 N s`C. `72 N s`D. `32 N s` |
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Answer» Correct Answer - B `J=int F.dt` |
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| 266. |
A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows thatA. Its velocity is constantB. Its acceleration is constantC. Its kinetic energy is constantD. It moves in a strainght line |
| Answer» Correct Answer - C | |
| 267. |
A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows thatA. its velocity is contatntB. its acceleration is constantC. its kinetic energy is constantD. It does not move on a circular path |
| Answer» (c) Perpndicular force does not changes the magnitude of velocity hence the kinetic energy remain constant. | |
| 268. |
A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of radius 30 m. What horizontal force on the scooter is needed to make the turn possible?If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road what should be the proper angle of banking? |
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Answer» Correct Answer - A::C IN the diagram `Rcostheta=mg`…i `Rsintheta=((Mv^2)/r)`……..ii Dividing equation i with equation ii we get, `tantheta =((mv^2)/(rmg))=v^2/(rg)` Here `v=36 km/hr=10 m/sec, r=30m` `:. Tanthete= v^2/(rg)=100/(30xx10)=(1/3)` `rarr thetat=tan^-1 (1/3)` |
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| 269. |
A stone tied to a string of length `L` is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at lowest position and has a speed `u` . Find the magnitude of the change in its velocity as it reaches a position, where the string is horizontal.A. `sqrt u^(2)-2 gL`B. `sqrt2g L`C. `sqrt u^(2)- gL`D. `sqrt (2(u^(2)-gL))` |
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Answer» Correct Answer - D `v_(L)=u,v_(h)=sqrt(v_(L)^(2)-2gh)=sqrt((u^(2))-2gl` `v_(L) and v_(h)` are tan to each other then their resultant is , `v_(R)=sqrt(v_(L)^(2)+v_(h)^(2))=sqrt(u^(2)+u^(2)-2gl)` ` =sqrt(2u^(2)-2gl)` `=sqrt(2(u^(2)-gL))` |
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| 270. |
A particle is moving in a circle of radius 20 cm has a linear speed of `10 m//s` at a certain instant and linear speed is increases at the rate of `2m//s^(2)`. What is the rate at which its acceleration in U.C.M. is increasing at that instant?A. `400 m//s^(2)`B. `200m//s^(3)`C. `300m//s^(3)`D. `100m//s^(3)` |
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Answer» Correct Answer - B `(d)/(dt)((v^(2))/(r))=(2v)/(r) (dv)/(dt)` |
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| 271. |
The SI unit of angular acceleration isA. radian/sB. radian/sC. radian`/s^(2)`D. meter `r/s^(2)` |
| Answer» Correct Answer - C | |
| 272. |
The rate change of angular velocity w.r to t in uniform circular motion is aA. radial accelerationB. angular acceleration `(vec theta)`C. angular displacementD. angular displacement |
| Answer» Correct Answer - B | |
| 273. |
The angular displacement in circular motion isA. dimensional quantityB. dimensionless quantityC. unitless quantity and dimensionless quantityD. unitless quantity |
| Answer» Correct Answer - B | |
| 274. |
The SI unit of angular velocity isA. metre/sB. radian/sC. `radian/s^(2)`D. `s^(-1)` |
| Answer» Correct Answer - B | |
| 275. |
Finite angular displacement is not a vector becauseA. it do not obey the law of vector additionB. it obeys the law of addition of vectorsC. its direction is given by right hand ruleD. it changes with time |
| Answer» Correct Answer - A | |
| 276. |
The rate of change of angular displacement in uniform circular motion is .A. angular velocity `(vec omega)`B. angular speed `(vec omega)`C. angular acceleration `(vec theta)`D. radial acceleration |
| Answer» Correct Answer - A | |
| 277. |
If an automobile moves round a curve of radius 300 m at constant speed of `60 m//s`, then the change of velocity round a curve of `60^(@)` will beA. zeroB. `30m//s`C. `120 m//s`D. `60m//s` |
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Answer» Correct Answer - D `Deltav=2v "sin" theta//2` `=2xx60(1)/(2)=60m//s`. |
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| 278. |
Angular displacement is measured inA. meterB. timeC. radianD. steradian |
| Answer» Correct Answer - C | |
| 279. |
If a particle goes round the circle once in a time period T, then the angular velocity `omega` will beA. `2 pi t`B. `T//2 pi`C. `2 pi //T`D. `pi //T` |
| Answer» Correct Answer - D | |
| 280. |
A glass marble moves from one end of a semiciecular arc of radius R to the other end of the arc, The ratio of distance travelled by the marble to its displacement isA. `pi//R`B. `R//pi`C. `2pi`D. `pi//2` |
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Answer» Correct Answer - D `(d)/(D)=(piR)/(2R)=(pi)/(2)`. |
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| 281. |
The angular subtended at the centre of the circle by an arc of length equal to the radius of circle isA. one radianB. one degreeC. one steradianD. `90^(@)` |
| Answer» Correct Answer - C | |
| 282. |
Three blocks of equal masses (each `3kg`) are suspended by weightless strings as shown. If applied force is `100N`,then `T_(1)` is equal to `(g=10m//s^(2))` A. `130 N`B. `190 N`C. `100 N`D. `160 N` |
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Answer» Correct Answer - A `T_(1)=F+mg` |
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| 283. |
A small spherical ball is suspended through a string of length `l`. The whole arrangement is placed in a vehicle which is moving with velocity `v`. Now suddenly the vehicle stops and ball starts moving along a circular path. If tension in the string at the highest point is twice the weight of the ball thenA. `v=sqrt(5gl)`B. `v=sqrt(7gl)`C. velocity of the ball at higest point is `sqrt(gl)`D. velocity of the ball at the highest point is `sqrt(3gl)` |
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Answer» Correct Answer - B::D |
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| 284. |
Three equal masses `A,B` and `C` are pulled with a constant force `F`. They are connected to each other with strings. The ratio of the tension between `AB` and `BC` is A. `1:2`B. `2:1`C. `3:1`D. `1:1` |
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Answer» Correct Answer - B `F-T_(1)=m_(1)a,T_(1)-T_(2)=m_(2)a,T_(2)=m_(3)a` |
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| 285. |
In circular motion of a particle,A. particle cannot have uniform motionB. particle cannot have uniformly acceleration motionC. particle cannot have net acceleration equal to zeroD. particle cannot have any force in tangential direction |
| Answer» Correct Answer - A::B::C | |
| 286. |
A person applies a constant force `vecF` on a particle of mass m and finds tht the particle movs in a circle of radius r with a uniform speed v as seen from an inertial frame of reference.A. This is not possibleB. There are other forces on the particleC. The resultant of the other forces is `(mv^2)/r ` towards the centreD. The resultant of the other forces varies in magnitude as well as in direction. |
| Answer» Correct Answer - B::D | |
| 287. |
A particle of mass m is moving in a horizontal circle of radius r under a centripetal force given by`((-K)/(r^(2)))` where k is a constant then,A. the total energy of the particle is`((-K)/(2r))`B. the total energy of the particle is `((k)/(r))`C. the total energy of the particle is `((k)/(2r))`D. the total energy of the particle is `(-(k)/(r))` |
| Answer» Correct Answer - (a) | |
| 288. |
Kinetic energy of a particle moving along a circle of radisu R depends on the distance covered as `K =as^(2)` where a is a constant . Find the force acting on the particle as a function of s.A. `(2a)/(s)sqrt(1+((s)/(R))^(2))`B. `2assqrt(1+((R)/(s))^(2))`C. `2assqrt(1+((s)/(R))^(2))`D. `(2s)/(a)sqrt(1+((R)/(s))^(2))` |
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Answer» (c) According to given problem `(1)/(2)Mv^(2)=as^(2)` " " [Hence, M =mass] `rArr" " v=ssqrt((2a)/(M))" " (i)` So, `" " a_(g) =(v^(2))/(R)=(2as^(2))/(MR)` Further more as `a_(t)=(dv)/(dt)=(dv)/(ds),(ds)/(dt)=v(dv)/(ds)` (By chain rule) from Eq. (i) we get `i,e" " v=sqrt((2a)/(M))` yeilds `:." " a_(t)=[ssqrt((2a)/(M))][sqrt((2a)/(M))]=(2as)/(M) " " ( :.v/s=sqrt((2a)/(m)))` so that," " `a_("net")=sqrt(a_(R)^(2)+a_(t)^(2))` `=sqrt(((2as^(2))/(MR))^(2)+((2a)/(M))^(2))=(2as)/(M)(sqrt(1+((s)/(R))^(2)))` `:. " " F-Ma_("net")=2assqrt(1+(s//R)^(2))` |
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| 289. |
A mass `m_(1)` lies on a fixed smooth cylinder. An ideal chord attached to `m_(1)` passes over the cylinder and is conneced to `m_(2)`. The system is released at `theta = 30^(@)`. Acceleration of `m_(1)` just after releasing the system is `(5)/(n)m//s^(2)`. Find the value of `n`. |
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Answer» Correct Answer - 3 |
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| 290. |
If the radius of curvature of the path of two particles of same masses are in the ratio `1 : 2`, then in order to have same centripetal force, their velocity, should be in the ratio ofA. `1:4`B. `4:1`C. `sqrt2:1`D. `1:sqrt2` |
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Answer» Correct Answer - D The centripetal force , `F=(mv^(2))/(r) therefore =(mv^(2))/(F)`, ` therefore r alpha v^(2) or v alpha sqrtr ` (If m and F are constant), ` therefore(v_(1))/(v_(2))=sqrt((v_(1))/(v^(2))=sqrt(1)/(2)`. |
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| 291. |
A 500 kg crave takes a turne of radius 50 m wiyh velocity of ` 54 km//h.The centripetal force isA. 1200 NB. 2250 VC. 750 ND. 250 N |
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Answer» Correct Answer - B `F=(mv^(2))/(r)=(500xx225)/(50)=2250 N` |
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| 292. |
A block of mass m is kept on a horizontal ruler. The frilction coefficient between the ruler and the block is `mu`. The ruler is fied at one end and the block is at a distance L from the fied end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. a. What can the maximum angular speed of the ruler isuniformly increased from zero t an angular acceleration `alpha` , at what angular speed will the block slip? |
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Answer» Correct Answer - A::B::D The frictional force provides the necessary centripetal force. :. `mLomega^(2)=mumg` or `omega=sqrt((mug)/(L))` (b) Net force of circular motion will be provided by the friction `omega=alphat` ....(i) `F_(n et)=msqrt(a_(t)^(2)+a_(n)^(2))` :. `mumg=msqrt((Lalpha)^(2)+(Lomega^(2))^(2))` ...(ii) Here , `a_(t)=Lalpha` and `a_(n)=Lomega^(22)` Substituting `omega=alphat` in Eq. `(ii)` we have, :. `mug=sqrtr(L^(2)alpha^(2)+L^(2)alpha^(4)t^(4))` :. `t=((mu^(2)g^(2)-L^(2)alpha^(2))/(L^(2)alpha^(4)))^(1)/(4)` Substituting the value of `t` in Eq. `(i)` we have, o`omega=[((mug)/(L))^(2)-alpha^(2)]^(1)/(4)` |
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| 293. |
A stream of water flowing horizontally with a speed of `15 ms^(-1)` pushes out of a tube of cross sectional area `10^(-2)m^(2)` and hits a vertical wall near by what is the force exerted on the wall by the impact of water assuming.that it does not rebound? (Density of water `=1000 kg m^(3)`)A. `1250N`B. `2250N`C. `4500N`D. `2550N` |
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Answer» Correct Answer - B `F=(dp)/(dt)=A rhov^(2)` |
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| 294. |
A simple pendulum of length l has a maximum angular displacement `theta`. The maximum kinetic energy of the bob of mass m will beA. `mgl(1-costheta)`B. `mglcostheta`C. `mgl sintheta`D. None of these |
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Answer» Correct Answer - A Height, h = l `(1-costheta)` `v_(m)^(2)=2gh=2gl(1-costheta)` `therefore"Maximum kinetic energy, K"_(m)=(1)/(2)mv_(m)^(2)=mgl(1-cos theta)` |
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| 295. |
A particle is moving in a circular path with a constant speed. If `theta` Is the angular displacement, then starting from `theta=0` , the maximum and mimnimum change in the linear momentum will occur when nvalue of `theta` is respectivelyA. `45^(@)` and `90^(@)`B. `90^(@)` and `180^(@)`C. `180^(@)` and `360^(@)`D. `90^(@)` and `270^(@)` |
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Answer» Correct Answer - C At `theta=180^(@),|DeltaP|=2mv=` maximum At `theta=360^(@),|DeltaP|=0=` minimum |
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| 296. |
A particle is revolving in a circle with increasing its speed uniformly. Which of the following is constant ?A. Centripetal accelerationB. Tangential accelerationC. Angular accelerationD. None of these |
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Answer» Correct Answer - C Magnitude of tangential acceleration is constant but its direction keeps on changing. |
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| 297. |
A body of mass m= 20 g is attached to an elastic spring of length L=50 cm and spring constant k=2 `Nm^(-1)`. The syste is revolved in a horizontal plan with a frequency v=30 rev/min. Find the radius of the circular motion and the tension in the spring .A. `0.55 m, 0.1 1N`B. `0.5 m, 0.52 N`C. `0.55 m, 0.1 N`D. `0.9 m, 0.2 N` |
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Answer» (c) Angular velocity `omega=2pir=2pixx(30)/(60)=pi rad//s` For an elastic spring force `F=kx.` Where x is the extension Radius of circular motion `r=L+x` Centripetal force `=mromega^(2)=F` `rArr" " M(L+x)omega^(2)=kx` `rArr" " x=(mLomega^(2))/(k-momega^(2))=(0.02xx0.5xx(3.14)^(2))/(2-0.02xx(3.14)^(2))` Radius of the circular motion (r) `=L+x=0.5+0.05=0.55 M` Tension in the spring `T=kx =2xx0.05=0.1 N` |
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| 298. |
Assertion In circular motion, dot product of v and `omega` is always zero. Reason `omega` is always perpendicular to the plane of the circular motion.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 v is always perpendicular to `omega`. |
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| 299. |
The speed of a particle moving in a circle is increasing. The dot product of its acceleration and velocity isA. negativeB. zeroC. positiveD. may be positive or negative |
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Answer» Correct Answer - C Speed of particle is increasing, means tangential component of acceleration is positive. |
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| 300. |
An object is moving in a circle of radius 100 m with a constant speed of `31.4 m//s` . What is its average speed for one complete revolutionA. ZeroB. `31.4 ms^(-1)`C. `3.14 ms^(-1)`D. `sqrt(2)xx31.4 ms^(-1)` |
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Answer» Correct Answer - B As the speed is constant throughout the circular motion, therefore its average speed is equal to instantaneous speed. |
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