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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.
| 201. |
A particle revolves round a circular path with a constant speed. (i) the velecity of the particle is along the tangent. (ii) the acceleration of the particle of the particle is always towards center. (iii) the magnetic of acceleration is constant. (iv) The work done by the centripetal force is always zero.A. `(i),(ii)`B. `(i),(ii),(iii)`C. `(ii),(iii),(iv)`D. All options are correct |
| Answer» Correct Answer - D | |
| 202. |
Let `bar(v)(t)` be the velocity of a particle al lime t. Then :A. `|dbar(v)(t)//dt|and d|bar(v)(t)|//dt|"are always equal"`B. `|dbar(v)(t)//dt|and d|bar(v)(t)|//dt|",may always equal"`C. `|dbar(v)(t)//dt |"can be zero while"and d|bar(v)(t)|//dt|",is not zero"`D. `|dbar(v)(t)//dt|ne0"implies " |d bar(v)(t)//dt|ne0 ` |
| Answer» Correct Answer - B::C::D | |
| 203. |
A uniform rod of mass `m` and length `l` rotates in a horizontal plane with an angular velocity `omega` about a vertical axis passing through one end. The tension in the rod at a distance x from the axis isA. `(1)/(2)momega^(2)x`B. `(1)/(2)momega^(2)(x^(2))/(l)`C. `(1)/(2)momega^(2)l(1-(x)/(l))`D. `(1)/(2)(momega^(2))/(l)[l^(2)-x^(2))` |
| Answer» Correct Answer - D | |
| 204. |
A simple pendulum is vibrating with an angular amplitude of `90^(@)` as shown in figure. For what value of `alpha` (angle between string and vertical) during its motion, the total acceleration is directed horizontally ? A. `tan^(-1)((1)/(2))`B. `tan^(-1)((1)/(sqrt(2)))`C. `cos^(-1)(1//sqrt(3))`D. `sin^(-2)(1//sqrt(3))` |
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Answer» Correct Answer - C |
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| 205. |
A car is moving on a circular road of diameter 50 m with a speed of 5 m/s. It is suddenly accelerated at a rate of `1 m//s^(2)`. If the mass of he car is 500 kg, then the net force acting on the car isA. 5 NB. 1000 NC. `500sqrt(2) N`D. `(500)/(sqrt(2))` |
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Answer» (c) Given ,r =25 m, v =5 m/s , m 500 kg `a_(t)=1m//s,a_(r)=(v^(2))/(r)=(5xx5)/(25)=1m//s^(2)` `a_("net")=sqrt(a_(t)^(2)+a_(r)^(2))` `sqrt(1^(2)+1^(2))=sqrt(2) m//s ^(2)` `F=ma_("net")=500sqrt(2))N` |
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| 206. |
Assertion: Speed of a particle moving in a circle varies with time as, `u=(4t-12)` . Such type of circular motion is not possible. Reason: Speed cannot change linearly with time.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, true but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - C For `tlt3sec` , speed is negative which is not possible., |
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| 207. |
A particle moves along a circle if radius (20 //pi) m with constant tangential acceleration. If the velocity of the particle is ` 80 m//s` at the end of the second revolution after motion has begun the tangential acceleration is .A. `40 m//s^(2)`B. `20 m//s^(2)`C. `10 m//s^(2)`D. `5 m//s^(2)` |
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Answer» Correct Answer - A `r=20/(pi)m, a_(1)` =cosntant `n=2^(nd)`=revolution v=80 m/s `omega_(0)=0, omega_(f)=v/r=80/(20//pi)=4pi ` rad/sec `theta=2pixx2=4pi` from `3^(rd)` `omega^(2)=omega_(0)^(2)+2 alpha theta` `rArr (4pi^(2))=0^(2)+2xxalphaxx(4pi)` `alpha=2pi rad//s^(2)` `a_(1)=alphar=2pixx20/(pi)` `=40 m//s^(2)` |
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| 208. |
A particle moves in a circle of radius `2 cm` at a speed given by `v = 4t`, where `v` is in `cm s^-1` and `t` is in seconds. (a) Find the tangential acceleration at `t = 1 s` (b) Find total acceleration at `t = 1 s`. |
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Answer» Given, v = 4t Radial acceleration, `a_(r)=(v^(2))/(R)=((4t)^(2))/(R)" or "a_(r)=(16r^(2))/(2)=8t^(2)` At t = 1 s, `a_(r)=8cms^(-2)` (i) Tangential acceleration, `a_(t)=(dv)/(dt)` or `a_(t)=(d)/(dt)(4t)=4cms^(-2)` i.e., `a_(t)` is constant or tangential acceleration at t = 1 s is `4cms^(-2)`. Total acceleration, `a=sqrt(a_(t)^(2)+a_(r)^(2))` or `a=sqrt((4)^(2)+(8)^(2))=sqrt(80)=sqrt3cms^(-2)` |
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| 209. |
A heavy & big sphere is hang with a string of length l. This sphere move in a horizontal circular path making an angle `theta` with vertical then its time period isA. `T=2pisqrt(l/g)`B. `T=2pisqrt((l sin theta)/g)`C. `T=2pisqrt(( l cos theta)/g)`D. `T=2pisqrt(l /(g cos theta))` |
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Answer» Correct Answer - C `T=2pisqrt((l cos theta)/(g)` |
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| 210. |
A car accelerates uniformly from rest to a speed of `10 m//s` in a time of 5 s .The number of revolutions made by one of its wheels during this motion if the radius of the wheel is `1//pi` m .A. 50B. 25C. 12.5D. 6.25 |
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Answer» Correct Answer - C `n_(2)=0` `n_(2)=(omega_(2))/(2 pi)=(V)/(2pi xxr)=(10xxpi)/(1xx2pi)=2 rps`. `N=((n_(1)+n_(2))/(2))t=(5)/(2)xx5=12.5`. |
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| 211. |
A `x and y` co-ordinates of a particle are `x=A sin (omega t) and y = A sin(omegat + pi//2)`. Then, the motion of the particle is A. circular anticlockwiseB. circular clockwiseC. elliptical clockwiseD. rectilinear |
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Answer» Correct Answer - B |
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| 212. |
If a car is to travel with a speed `v` along the frictionless, banked circular track of radius `r`, the required angle of banking so that the car does skid isA. `theta=tan^(-1)((v^(2))/(rg))`B. `theta=tan^(-1)((v)/(rg))`C. `theta=tan^(-1)((r^(2))/(rg))`D. `theta=tan^(-1)((v^(2))/(2rg))` |
| Answer» Correct Answer - A | |
| 213. |
A body moves along a circle of radius 1 m with a constant kinetic energy 25 j . Then force acting on it is `A. 25 NB. 50 NC. 10 ND. 5 N |
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Answer» Correct Answer - B `KE=(1)/(2) F r ` `therefore F =(2KE)/(r)=(2xx25)/(1)=50N`. |
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| 214. |
A block of mass m at the end of a string is whirled round in a vertical circle of radius R. The critical speed of the block at top of its swing below which the string would slacken before the block reaches the bottom is?A. `sqrt(5Rg)`B. `sqrt(3Rg)`C. `sqrt(2Rg)`D. `sqrt(Rg)` |
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Answer» Correct Answer - D `mg=(mv_(min)^(2))/(R)"or v"_(min)=sqrt(Rg)` (at topmost point) |
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| 215. |
A stone of mass `1//2` kg is whriled round the end of a piece of string in a horizontal circle of radius half a meter with a uniform speed of `1 m//s`. The tension in the string isA. `25xx10^(3)` dynesB. `10^(5)` dynesC. `5xx10^(4)` dynesD. `4xx10^(5)` dynes |
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Answer» Correct Answer - B `T=(mv^(2))/(r0=(1)/(2)xx(1)/(0.5)=1N` `=10^(5)` dyne. |
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| 216. |
The figure shows th velocity and acceleration of a point like body at the initial moment of its motion. The acceleration vector of the body remain constant. The minimum radius of curvature of trajectory of the body is A. `2 meter`B. `4 meter`C. `8 meter`D. `16 meter` |
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Answer» Correct Answer - D Acceleration vector shall change component of velocity `v costheta` along acceleration vector `r=v^(2)/a_(n)` `r` is min when it is min this is possible when component of velocity parallel to acceleration vector is zero i.e., `v_(0) cos theta=0` |
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| 217. |
If a particle covers half the circle of radius R with constant speed thenA. momentum change is mvrB. chanfe in K.E is `1//2mv^(2)`C. change in K.E. is `mv^(2)`D. change in K.E. is zero |
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Answer» Correct Answer - D As momentum is a vector quantity `therefore` Change in momentum `Delta P=2mv sin (theta //2)` `=2mv sin (90)=2mv` But kinetic energy remains always constant so change in kinetic energy is zero. |
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| 218. |
The average acceleration vector for a particle having a uniform circular motion isA. a constant vector of magnitude `v^(2)//r`B. a vector of magnitude`v^(2)//r` directed normal to the planeC. equal to the instantaneous acceleration vector at the start of the motionD. a null vector |
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Answer» Correct Answer - D In complete revolution change in velocity becomes zero. Hence, average acceleration is zero . |
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| 219. |
A particle moves in a circular orbit under the action of a central attractive force inversely proportional to the distance `r`. The speed of the particle isA. propotional to `r^(2)`B. Independent of rC. Proportional to rD. Proportional to 1/r |
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Answer» Correct Answer - B i. e . Speed of the particle is independent of r. |
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| 220. |
A particle moves in a circular orbit under the action of a central attractive force inversely proportional to the distance `r`. The speed of the particle isA. propotional to `r^(2)`B. independent of `r`C. propotional to `r`D. propotional to `1//r` |
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Answer» Correct Answer - B `Fprop(1)/(r )impliesF=(k)/(t ),k:` constant `F=(k)/(r )=(mv^(2))/(r )impliesv=sqrt((k)/(m))` `vpropr^(0)` |
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| 221. |
If a and b are two vectors.then the value of `(a+b)xx(a-b)` isA. `2(bar(b)xx bar(a))`B. `-2(bar(b)xx bar(a))`C. `bar(b)xx bar(a)`D. `bar(a)xxbar(b)` |
| Answer» Correct Answer - A | |
| 222. |
The value of n so that vectors `2hati+3hatj-2hatk, 5hati+nhatj+hatk and -hati+2hatj+3hatk` may be coplanar.will beA. 18B. 28C. 9D. 36 |
| Answer» Correct Answer - A | |
| 223. |
What can be the angle between `(vec(P)+vec(Q))` and `(vec(P)-vec(Q))` ? |
| Answer» Correct Answer - B | |
| 224. |
The area of the parallelogram represented by the vectors `vecA=2hati+3hatj` and `vecB=hati+4hatj` isA. 14 unitB. 7.5unitC. 10unitD. 5unit |
| Answer» Correct Answer - D | |
| 225. |
A particle moves in `x-y` plane according to the law `x = 4 sin 6t and y = 4 (1-cos 6t)`. The distance traversed by the particle in 4 second is (`x & y` are in meters)A. 96 mB. 48 mC. 24 mD. 108 m |
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Answer» Correct Answer - A |
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| 226. |
A moter-cyclist moving with a velocity of 144 `kmh^(-1)` on a flat road takes a turn on the road at a point where the radius of curvature of the road is 40 m. The acceleration due to gravity is 10 `ms^(-2)`. In order to avoid sliding, he must bend with respect to the vertical plane by an angleA. `theta=tan^(-1)(4)`B. `theta=45^(@)`C. `theta=tan^(-1)(2)`D. `theta=tan^(-1)(6)` |
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Answer» Correct Answer - A `v=144kmh^(-1)=144xx(5)/(18)ms^(-1)=40ms^(-1)` `tantheta=(v^(2))/(Rg)` `rArr" "tantheta=((40)^(2))/(40xx10)=(1600)/(400)=4` |
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| 227. |
A point moves along a circle with a velocity `v=t//2`. Find the acceleration of the point at the moment when it has covered a quarter circle from the beginning of motion. |
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Answer» `v=(t)/(2)` `a_(t)=(dv)/(dt)=(1)/(2)` (constant) Time taken by the particle to complete a half-circle `s=ut+(1)/(2)a_(t)t^(2)` `(piR)/(2)=0+(1)/(2)xx(1)/(2)t^(2)impliest^(2)=2piR` `a_(c)=(v^(2))/(R )=((t//2)^(2))/(R )=(t^(2))/(4R)=(2piR)/(4R)=(pi)/(2)` the resultant acceleration when the particle has covered a quarter circle `a=sqrt(a_(c)^(2)+a_(t)^(2))=sqrt(((pi)/(2))^(2)+((1)/(2))^(2))=(1)/(2)sqrt(pi^(2)+1)` |
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| 228. |
A particle is suspended from a string of length R. It is given a velocity `u = 3 sqrt(gR)` at the bottom Match the following `{:(,"Table-1",,"Table-2"),("(A)","Velocity at B","(P)","7 mg"),("(B)","Velocity at C","(Q)",sqrt(5gR)),("(C)","Tension in string at B","(R)",sqrt(7gR)),("(D)","Tension in string at C","(S)","5 mg"),(,,"(T)","None"):}` |
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Answer» Correct Answer - (A)R,(B)Q,(C)P,(D)T |
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| 229. |
A particle of mass `m` is suspended by a string of length `l` from a fixed rigid support. A sufficient horizontal velocity `=sqrt(3gl)` is imparted to it suddenly. Calculate the angle made by the string with the vertical when the accekleration of the particle is inclined to the string by `45^(@)` . |
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Answer» Correct Answer - A::B `h=l(l-costheta)` `v_^(2)v_(0)^(2)-2gh=3gl-2gl(l-costheta)=gl(1+2costheta)` At `45^(@)` means radial and tangential components of acceleration are equal. :. `(v^(2))/(l)=gsintheta` or `1+2costheta=sintheta` Solving the equation we get, `theta=90^(@)or(pi)/(2)` |
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| 230. |
A particle is suspended from a string of length R. It is given a velocity `u = 3 sqrt(gR)` at the bottom Match the following `{:(,"Table-1",,"Table-2"),("(A)","Velocity at B","(P)","7 mg"),("(B)","Velocity at C","(Q)",sqrt(5gR)),("(C)","Tension in string at B","(R)",sqrt(7gR)),("(D)","Tension in string at C","(S)","5 mg"),(,,"(T)","None"):}` |
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Answer» Correct Answer - A::B::C::D `v_(B)^(2)=u_(A)^(2)-2gh_(AB)=(9gR)-(2gR)=7gR` `therefore" "v_(B)=sqrt(7gR)` Further, `T_(B)=(mv_(B)^(2))/(R)=7mg` Again, `v_(C)^(2)=v_(A)^(2)-2gh_(AC)=(9gR)-2(2R)=5gR` `therefore" "v_(C)=sqrt(5gR)` Further, `T_(C)+mg=(mv_(C)^(2))/(R)` `therefore" T_(C)=4mg` |
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| 231. |
When a particle is recolving around a circular path with uniform angular velocity , then its acceleration isA. along the tangent drawn at any poitB. along the circumference of the circleC. along the radius towards the centre of the circleD. zero |
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Answer» Correct Answer - C Particle experiences only radial acceleration. |
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| 232. |
A particle is moving with a constant angular acceleration of `4rad//s^(2)` in a circular path. At time `t=0` , particle was at rest. Find the time at which the magnitudes of centripetal acceleration and tangential acceleration are equal. |
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Answer» Correct Answer - A::B `(Romega^(2))=Ralpha` :. `omega^(2)=alpha` or `(alphat)^(2)=alpha` `t=(1)/(sqrt(alha))=(1)/(sqrt(4))` `=0.5a` |
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| 233. |
Two particles of masses in the ratio 2:1 are moving in circular paths of radii in the ratio 3:2 with time period in the ratio 2:3 .The ratio of their centripetal forces isA. `9:2`B. `27:4`C. `4:3`D. `27:16` |
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Answer» Correct Answer - B `(F_(1))/(F_(2))=(m_(1))/(m_(2))xx(r_(1))/(r_(2))xx(T_(2)^(2))/(T_(1)^(2))=(2)/(1)xx(3)/(2)xx(9)/(4)` `=27:4`. |
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| 234. |
A gramophone disc rotates at 60 rpm . A coin of mass 18 g is placed at a distance of 8 cm from the centre . Calculate centrifugal force on the coin . Take `pi^(2) = 9. 87` . |
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Answer» Since, angular velocity, `omega=60xx(2pi)/(60)=2pi" rad s"^(-1)`. `therefore" Centrifugal froce", F=momega^(2)r=18xx(2pi)^(2)xx8` `=18xx4xx(3.14)^(2)xx8` `rArr" "F=5685.12"dyne"` `~~0.06N` |
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| 235. |
A coin placed on a gramophone record at `100//3` rpm flies off when it is place at a distance grater than 16 cm from the axis of rotation. If the record is rotating at `200//3`A. 3 cmB. 4 cmC. 2 cmD. 1 cm |
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Answer» Correct Answer - B Condition for the stability of the coin is `mu(mg)=mr omega ^(2)=mr(4 pi ^(2)n^(2))` `therefore r_(1) n_(1)^(2)=r_(2)n_(2)^(2) therefore (r_(1))/(r_(2))=(n_(2)^(2))/(n_(1)^(2)` `therefore (16)/(r_(2))=(200xx200)/(100xx100)=4 therefore r_(2)=4 cm `. |
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| 236. |
A particle is going with constant speed along a uniform helical and spiral path separately as shown in figure A. The velocity of the particle is constant in both casesB. The magnitude of acceleration of the particle is constant in both casesC. The magnitude of acceleration is constant in (a) and decreasing in (b)D. The magnitude of acceleration is decreasing continuously in both the cases |
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Answer» Correct Answer - C `a_(c)=(v^(2))/r`, radius is constant in case (a) and increases in case(b). So that magnitude of acceleration is constant in case(a) and decreases in case(b) |
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| 237. |
A train has to negotiate a curve of radius 400 m .The speed of the train is 72 km//h. If the distance between the two rails is 1 m ,then the outer rail will be raised over the inner rail by `(g=10m//s^(2))`A. 15 cmB. 10 cmC. 5 cmD. 2.5 cm |
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Answer» Correct Answer - B `(h)/(l)=(v^(2))/(rg)` |
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| 238. |
A particle is moving in a circle with a constant speed, the acceleration of the particle hasA. constant magnitude andB. constant magnitude and directionC. both magnitude and direction changesD. neither magnitude nor direction changes |
| Answer» Correct Answer - A | |
| 239. |
Assertion: Circular and projectile motions both are two idimensional motion.But In circular motion, we cannot apply `v=u+at` directly, whereas in projectile motion we can. Reason: Projectile motion takes place under gravity, while in circular motion gravity has no role.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, true but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - C In circular motion, `acancel(=)` constant, so we cannot apply `v=u+` at directly. In vertical circular motion, gravity plays an important role. |
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| 240. |
In case of a uniform circular motion, velocity and acceleration areA. perpendicularB. in same directionC. in opposite directionD. not related to each other |
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Answer» Correct Answer - A Velocity is tangential and acceleration is radial |
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| 241. |
When a particle moves in a circle with unifoem speed then,A. velocity and acceleration are constantB. both velocity and acceleration changeC. acceleration is constant but celocity changeD. magnitude of angular celocity constant but its direction change |
| Answer» Correct Answer - B | |
| 242. |
In a non uniform circular motion, the acceleration on the particle isA. centripetal acceleration onlyB. tangential acceleration onlyC. the resultant centripetal and tangential accelerationD. centrifugal acceleration |
| Answer» Correct Answer - C | |
| 243. |
The relation between linear velocity and angular velocity of a body moving in circle in vector from isA. `vec(a)_(r)=vecomega.vec(v)`B. `vec(a)_(r)=vecomega xx vec(v)`C. `vec(a)_(r)= vec(v)xxvec omega`D. `vec(a)_(r)= vec(v).vec omega` |
| Answer» Correct Answer - B | |
| 244. |
The uniform circular motion is accelerated motion, becauseA. the motion accelerates due to the change in velocityB. the motion accelerates due to the change in angular velocityC. the motion accelerates due to the forceD. all of these |
| Answer» Correct Answer - A | |
| 245. |
Assertion uniform circular motion is uniformly accelerated motion. Reason acceleration in uniform circular motion is always towards centre.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 Direction of acceleration continuously changes. In uniform circular motion, acceleration always acts towards the centre. Therefore, assertion is false but reason is true. |
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| 246. |
The force acts ona particle perfoming uniform circular motion and which is directed towards centre along radius of circle isA. centrifugal forceB. centripetal forceC. pseudo forceD. gravitational force |
| Answer» Correct Answer - B | |
| 247. |
Is a body in uniform circular motion in equilibrium ? |
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Answer» `acancel(=)0 implies F_(n et)=macancel(=)0` Hence, particle is not in equilibrium. |
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| 248. |
The direction of angular acceleration of a body moving in a circle in the plane of the paper is .A. perpendicular to radius and velocityB. towards the radiusC. tangentialD. at right angle to angular velocity |
| Answer» Correct Answer - A | |
| 249. |
`"|"oversetrightarrow(a).oversetrightarrow(b)"|"=sqrt3"|"oversetrightarrow(a)xxoversetrightarrow(b)"|"`, then the angle between `oversetrightarrow(a)andoversetrightarrow(b)` is:A. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(2)` |
| Answer» Correct Answer - A | |
| 250. |
A body of mass 5 kg is moving in a circle of radius 1m with an angular velocity of 2 radian/sec . The centripetal force isA. 40 NB. 20 NC. 30 ND. 10 N |
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Answer» Correct Answer - B `F=mr omega^(2)` |
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