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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.
| 301. |
A mass is supported on a frictionless horizontal surface. It Is attached to a string and rotates about a fixed center at an angular velocity `omega_(0)`.If the length of the string and angular velocity both are doubled, the tension in the string which was initially `T_(0)` is nowA. `T_(0)`B. `T_(0)//2`C. `4T_(0)`D. `8T_(0)` |
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Answer» Correct Answer - D `T_(0)=m omega^(2)l` `T=m(2omega)^(2)xx2l=8T_(0)` |
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| 302. |
A circular racetrack of radius 300 m is banked at an angle of `15^(@)` If the coefficient of friction between the wheels of a race car and the road is 0.2 what is the (a) optimum speed of the race car to avoid wear and tear on its tyres , and (b) maximum permissible speed to aviod slipping ? |
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Answer» We know that maximum permissible speed on a banked road to avoid slipping is , `v_("max")=[(rg(mu_(s)+tantheta))/(1-mu_(s)tantheta)]^(1//2)` Now,putting the values given in the question, `r=300m,theta=15^(@),g=9.8(~~10)ms^(-2)"and "mu_(s)=0.2` We obtion, `v_("max")=[(300xx9.8(0.2+tan15^(@)))/(1-0.2tantheta15^(@))]^(1//2)` `=[(300xx9.8(0.2+0.26))/(1-0.2xx0.26)]^(1//2)` After solve this, we get, `v_("max")=38.1ms^(-1)` |
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| 303. |
A partical of mass `m` oscillates along the horizontal diameter `AB` inside a smooth spherical `AB` inside a smooth sperical shell of radius `R`. At any instate `K.E.` of the partical is `K`. Then force applied by partical on the on the shell at this instant is: A. `(K)/(R)`B. `(2K)/(R)`C. `(3K)/(R)`D. `(K)/(2R)` |
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Answer» Correct Answer - C |
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| 304. |
A bridge is in the form of a semicircle of radius `40m`. The greatest speed with which a motorcycle can cross the bridge without leaving the ground at the highest point (frictional force is negligibly small)A. `40m//s`B. `20m//s`C. `30m//s`D. `15m//s` |
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Answer» Correct Answer - B |
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| 305. |
A particle of mass `m` moves in a circle of radius `R` with a uniform speed `v`. (i) the angular speed of particle is `v//R`. (ii) the time period of revolution is `2piR//v`. (iii) the acceleration of particle is `v^(2)//R`. (iv) the work done by the centripetal force in half revolution is `(mv^(2)//R)xxpiR`.A. `(i),(ii)`B. `(ii),(iii)`C. `(ii),(ii),(iii)`D. All options are correct |
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Answer» Correct Answer - C The work done by the centriple force is always zero. |
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| 306. |
A body is whirled in a horizontal circle of radius `20cm`. It has an angular velocity of `10rad//s`. What is its linear velocity at any point on the circular pathA. `10m//s`B. `2m//s`C. `20m//s`D. `sqrt(2)m//s` |
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Answer» Correct Answer - B |
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| 307. |
If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficient so that a scooter going at `18km//h` does not skid? |
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Answer» Correct Answer - B `v=sqrt(muRg)` `:. mu=(v^(2))/(gR)=((5)^(2))/(10xx10)` `=0.25` |
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| 308. |
A particle moves in a circle of radius `30cm`. Its linear speed is given by `v=2t`, where `t` in second and `v` in `m//s`. Find out its radial and tangential acceleration at `t=3s`.A. `220m/s^(2),50m//s^(2)`B. `100m//s^(2),5m//s^(2)`C. `120m//s^(2)`D. `110m//s^(2),10m//s^(2)` |
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Answer» Correct Answer - C `R=0.3m,v=2t` At `t=3s,v=6m//s` `a_(c)=(v^(2))/(R )=((6)^(2))/(0.3)=(36)/(0.3)=120m//s^(2)` `v=2t` `a_(t)=(dv)/(dt)=2m//s^(2)` |
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| 309. |
A paricle of mass `m` is tied to a light string of length `L` and moving in a horizontal circle of radius `r` with speed `v` as shown. The forces acting on the particle are A. `mg and T`B. `mg, T,(mv^(2))/(r )`directed in wardsC. `mg,T,(mv^(2))/(r )`directed outewardsD. `(mv^(2))/(r )`only |
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Answer» Correct Answer - A Centriple force is not a new kind of force, centripetal force, i.e., a force acting towards center. |
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| 310. |
A point on the rim of a disc starts circular motion from rest and after time t, it gains an angular acceleration given by `alpha=3t-t^(2)`. Calculate the angular velocity after 2 s. |
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Answer» Angular acceleration, `alpha=(domega)/(dt)=2t-t^(2)` `rArr" "int_(0)^(omega)domega=int_(0)^(t)(3t-t^(2))dtrArromega=(3t^(2))/(2)-(t^(3))/(3)` At `2s," "omega=(10)/(3)"rad s"^(-1)` |
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| 311. |
A particle is given an initial speed `u` inside a smooth spherical shell of radius `R=1` m such that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is A. `g sqrt(10)`B. `g`C. `g sqrt(2)`D. `3g` |
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Answer» Correct Answer - A |
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| 312. |
A park has a radis of 10 m. If a vehicle goes round it at an averge speed of 18 km/hr, what shoud be the proper angle of banking? |
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Answer» Correct Answer - A::D Radius of park r=10 S-eed of vehicle = 18 km/hr=5m/sec Angle of banking is given by, `tantheta=v^2/(rg)` `rarr theta=tan^-1 25/(rg)` `=tan^-1 25/100 =tan^-1 (1/4)` |
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| 313. |
A man of `50 kg` is standing at one end on a boat of length `25m` and mass `200kg`.If he starts running and when he reaches the other end, has a velocity `2ms^(-1)` with respect to the boat.The final velocity of the boat isA. `2/3ms^(-1)`B. `2/5ms^(-1)`C. `8/5ms^(-1)`D. `8/3ms^(-1)` |
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Answer» Correct Answer - B `m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)` |
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| 314. |
A body of mass `1kg` initially at rest explodes and breaks into three parts of masses in the ration ` 1: 1: 3.` If the two pieces of equal masses fly off perpendicular to each other with a speed of `30m//s` The speed of third piece will be .A. `sqrt2 v`B. `v//2`C. `v//sqrt2`D. `sqrt2v` |
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Answer» Correct Answer - A `vecP_(3)=-(vecP_(1)+vecP_(2)),V_(3)=|vecP_(3)|/m_(3)` |
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| 315. |
A force which is equal and opposite to centripetal force in uniform circular motion , is calledA. centrifugal forceB. restoring forceC. nuclear forceD. cohesive force |
| Answer» Correct Answer - A | |
| 316. |
A car is moving with constant speed on a road as shown in figure. The normal reaction by the road on the car is `N_(A),N_(B), N_(C)` when it is at the points A,B and C respectively. A. `N_(A)=N_(B)`B. `N_(A)gtN_(B)`C. `N_(A)ltN_(B)`D. `N_(C)=N_(A)` |
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Answer» Correct Answer - B for normal reaction at points A and B `mg-N=(mv^(2))/r` `N=mg=(mv^(2))/r` `rArr N_(A) gt N_(B)` and normal reaction at C is `N_(O)`=mg , so `N_(C) gt N_(A) gt N_(B)` |
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| 317. |
If a body of mass m is moving along a horizontalcircle of radius R, under the action of centripetal force equal to `K//R^(2)`where K is constant then the kinetic energy of the particle will beA. `K^(2)R`B. `K//2R`C. `K//R`D. `K//R^(3)` |
| Answer» Correct Answer - B | |
| 318. |
What is the angle of friction between two surfaces in contact , if coefficient of friction is `1//sqrt(3)` ?A. `30^(@)`B. `60^(@)`C. `45^(@)`D. `37^(@)` |
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Answer» Correct Answer - B `mu=tan theta` |
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| 319. |
The direction of velocity continuously changes inA. uniforme circular motionB. nonuniforme circular motionC. oscillatory motionD. uniforme and non uniforme circular motion |
| Answer» Correct Answer - D | |
| 320. |
In UCM, the radius vector hasA. variable magnitude and changes direction continouslyB. unique directionC. constant in magnitude and same directionD. constant magnitude but continuously changing direction |
| Answer» Correct Answer - D | |
| 321. |
The coefficient of friction between the ground and the wheels of a car between the ground and the wheels of acar moving on a horizontal road is 0.5 If the car starts from rest , what is the minimum distance in which it can acquire a speed of `72 km // h` ? take `g = 10 ms^(-2)` .A. `0m`B. `20m`C. `30m`D. `40m` |
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Answer» Correct Answer - D `v^(2)-u^(2)=2as,a=mu_(k)g` |
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| 322. |
In the previous question, the tension in the string isA. `(5)/(8)N`B. `(8)/(5)N`C. `(50)/(8)N`D. `(80)/(5)N` |
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Answer» Correct Answer - B `T=(mg)/(costheta)=(0.1xx10)/(5//8)=(8)/(5)N` |
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| 323. |
In the above question `T` a is positive at the lowest point forA. `u=sqrt(2gl)`B. `u=sqrt(5gl)`C. `u=sqrt(7gl)`D. any value of `u` |
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Answer» Correct Answer - D At lowest point `T` and `a` are always in same direction (towards centre). :. `T.a` is always positive. |
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| 324. |
A car is moving along a horizontal curve of radius 20 m , and coefficient of friction between the road and wheels of the car is `0.25` . If acceleration due to gravity is `(9.8 m//s^(2))`, then its maximum speed isA. `3 m//s`B. `5 m//s`C. `7 m//s`D. `9 m//s` |
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Answer» Correct Answer - C `V=sqrt(mu rg)=sqrt(0.25xx20xx9.8)=sqrt(49)` `=7 m//s` |
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| 325. |
A particle moves in a circle of radius 1 cm at a speed given `v = 2t`, where `v` is cm/s and t is in seconds. Total acceleration of the particle at `t=1` second is `2 sqrt(n) cm//s^(2)`. Find value of `n`. |
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Answer» Correct Answer - 5 |
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| 326. |
A particle is moving in a circular path in the vertical plane. It is attached at one end of a string of length `l` whose other end is fixed. The velocity at lowest point is `u` . The tension in the string `T` and acceleration of the partiocle is a any position. Then `T` a is zero at highest point ifA. `ugtsqrt(5gl)`B. `u=sqrt(5gl)`C. Both (a) and (b) are correctD. Both (a) and (b) are wrong |
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Answer» Correct Answer - B If `ugtsqrt(5gl)` , then `T` and `a` are in same direction. hance, `T` . `a` is positive. If `u=sqrt(5gl)` , then `T=0` :. `T.a=0` |
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| 327. |
If a particle moves with uniform speed its tangntial acceleration will beA. `v^(2)//r`B. `r omega ^(2)`C. zeroD. infinite |
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Answer» Correct Answer - C For uniform velocity tangential acceleration is zero. |
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| 328. |
Tension of a string is 6.4 N and load is applied to it at its lower end of a string is 0.1 kg .If the length of string is 6 m , then its angular velocity will be `A. `3 rad//s`B. `4 rad//s`C. `2 rad//s`D. `1 rad//s` |
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Answer» Correct Answer - A `T=m r omega ^(2)` `therefore omega ^(2)=(T)/(mr)=(6.4)/(0.1xx6)` `omega ^(2)=10` `omega =sqrt(10)=3 rad //s `. |
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| 329. |
A wheel is subjected to uniform angular acceleration about its axis with its angular velocity is zero . In the first two seconds , it rotates through an angle `theta _(1)` and in the next two seconds , it rotates through an angle `theta _(2)` then the ratio of `theta_(2)//theta_(1)`A. `1:1`B. `2:1`C. `3:1`D. `4:1` |
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Answer» Correct Answer - C `theta _(1)=omega_(0)t+(1)/(2)alphat_(1)^(2)" "thereforetheta_(1)=2alpha ` `theta_(2)=omega_(0)t+(1)/(2)alpha t_(2)^(2)" "thereforetheta_(2)=8alpha` `theta_(2)=theta_(2)-theta_(1)=8lpha-2alpha" "therefore (theta_(2))/(theta_(1))=(3)/(1)` |
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| 330. |
The angular velocity of a particle increases from O to `omega` as it completest x rotations .Then number of rotation completes by it when its angular velocity becomes 2 `omega`.A. `xx`B. `2xx`C. `3xx`D. `4xx` |
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Answer» Correct Answer - D `omega _(2)^(2)- omega_(1)^(2)=2alphatheta_(1)` `omega^(2)-0=2alpha theta_(1)` `omega _(2)^(2)-omega _(1)^(2)=2 alpha theta _(2)` `4 omega ^(2)-0=2alphatheta_(2)` `(theta_(2))/(theta_(1))=(4omega^(2))/(omega^(2))=4` `(x_(2))/(x_(1))=4` `x_(2)=4x` |
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| 331. |
The centripetal force required to hold 1 kg object in circular path by means of a string 1 m long, if the object is moving at constant speed of `2 m//s` will beA. 2 NB. 8 NC. 4 ND. 12 N |
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Answer» Correct Answer - C `F=(mv^(2))/(r)` |
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| 332. |
An object of mass 50 kg is moving in a horizontal circle of radius 8 m . If the centripetal force is 40 N, then the kinetic energy of an object will beA. 320 jB. 260 jC. 220 jD. 160 j |
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Answer» Correct Answer - D `K.E= (1)/(2)mv^(2)=(1)/(2)((mv^(2))/(r))r=(1)/(2)Fxxr` |
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| 333. |
A coin placed on a rotating table just slip when it is placed at a distance 4 r from the centre, on doubling the angular velocity of the table , the coin will just slip now the coin is at a distance from centre isA. 4 rB. `r//4`C. rD. 2 r |
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Answer» Correct Answer - C `r_(2) onega_(1)^(2)` `r_(2)=(r_(1) omega_(1)^(2))/(omega_(2)^(2))=(4r_(1) omega_(1)^(2))/(4 omega_(1)^(2)` `r_(2)=r` |
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| 334. |
An ant travels along a long rod with a constant velocity ii relative to the rod starting from the origin. The rod is kept initially along the positive x-axis. At t = 0, the rod also starts rotating with an angular velocity w (anticlockwise) in x-y plane about origin. ThenA. the position of the ant at any time t is `bar(r)=ut(cos omegathati+sinomegahatj)`B. the speed ofthe ant at any time t is `usqrt(1+omega^(2)t^(2))`C. the magnitude of the tangential acceleration of the ant at uny time t is `(omega^(2)tu)/(sqrt(1+omega^(2)t^(2)))`D. the speed of the ant at any time t is `sqrt(1+2omega^(2)t^(2)u)` |
| Answer» Correct Answer - A::B::C | |
| 335. |
On a circular turn, table rotating about its center horizontally with uniform angular velocity rad/s placed two blocks of mass 1kg and 2kg, on a diameter symmetrically about center. Their separation is 1m and friction is sufficient to avoid slipping. The spring between them as shown is stretched and applied force of SN. lf `f_(1) and f_(2)` are values of friction on 1 kg & 2kg block respectively:- A. `"For" omega=2rad//s, f_(1)=3N&F_(2)=1N`B. `"For" omega=3rad//s, f_(1)=0.5N&F_(2)=4N`C. `"For" omega=sqrt10rad//s, f_(1)=0&F_(2)=5N`D. `"For" omega=sqrt10rad//s, f_(1)=0&F_(2)=5N` |
| Answer» Correct Answer - A::B::C | |
| 336. |
When a body is acclerated : (i) its velocity always changes (ii) its speed always changes (iii) its direction always changes (iv) its speed may or may not change. Which of the following is correct ?A. `(i),(ii)`B. `(i),(iv)`C. `(ii),(ii)`D. `(ii),(iii)` |
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Answer» Correct Answer - B When the velocity (either magnitude or direction or both) changes, acceleration is produced. |
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| 337. |
A particle is moving along a circular path of radius of `R` such that radial acceleration of particle is proportional to `t^(2)` thenA. Speed of particle is constantB. Magnitude of tangential acceleration of particle is constantC. Speed of particle is proportional to timeD. Magnitude of tangential acceleration is variable |
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Answer» Correct Answer - B::C `a_(r)alpha t^(2),v^(2)=krt^(2),v=sqrt(kr)t` `(dv)/(dt)=sqrt(kr)="constant"` |
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| 338. |
A car runs around a curve of radius 10m at a constant speed of `10ms^(-1)`. Consider the time interval for which car covers a curve of `120^(@)` are:-A. Resultant change in velocity of car `10sqrt3 ms^(-1)`B. instantaneous acceleration of car is `10ms^(-2)`C. Average acceleration of car is `(5)/(24)ms^(-2)`D. Instantaneous and average acceleration are same for the given period of motion. |
| Answer» Correct Answer - A::B | |
| 339. |
In the previous problem, if `h=5R//2`, the speed of block at the highest point isA. `sqrt(2gR)`B. `sqrt(gR)`C. `zero`D. `sqrt((gR)/(2))` |
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Answer» Correct Answer - B `v_(D)^(2)=v^(2)-2gxx2_(R )=5gR-4gR` `v_(D)=sqrt(gR)` |
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| 340. |
A block of mass `4kg` is placed in contact with the front vertical surface of a lorry.The coefficient of friction between the vertical surface and block is `0.8`.The lorry is moving with an acceleration of `15m//s^(2)`.The force of friction between lorry and block is `(g=10ms^(-2))`A. `48N`B. `24N`C. `40N`D. Zero |
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Answer» Correct Answer - C `W= mg,f=mu ma,m lt f` |
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| 341. |
A particle is placed at rest inside a hollow hemisphere of radius `R`. The coefficient of friction between the particle and the hemisphere is `mu = (1)/sqrt(3)`. The maximum height up to which the particle can remain stationary isA. `R/2`B. `(1-sqrt3/2)R`C. `sqrt3/2R`D. `(3R)/8` |
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Answer» Correct Answer - B `h=R-R cos theta,mu=tan theta` |
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| 342. |
A horizontal force is applied on a body on a rough horizontal surface produces an acceleration `a`.If coefficient of friction between the body and surface which is `a` is reduced to `mu//3`,the acceleration increses by `2 units`.The value of `mu` isA. `2//3g`B. `3//2g`C. `3//g`D. `1//g` |
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Answer» Correct Answer - C `F_(r)=F-f,f=mu_(k) mg` |
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| 343. |
A bucket full of water is tied at the end of rope of length 1.6 m. It is rotated in vertical circle with constant speed around the other end . What should be the minimum speed of bucket so that water does not spill out at the highest position of the water circle?A. `16 m//s`B. `4 m//s`C. `8 m//s`D. `2 m//s` |
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Answer» Correct Answer - D `vsqrt(rg)` |
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| 344. |
A stone tied to a rope is rotated in a vertical circle with uniform speed. If the difference between the maximum and minimum tensions in the rope is 20 N, mass of the stone in kg is (take, g= 10 `ms^(-2)`)A. `0.75`B. `1.0`C. `1.5`D. `0.5` |
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Answer» Correct Answer - B We know that the difference between the maximum and minimum tensions in the rope `T_(max)-T_(min)=2mg` Here, `T_(max)-T_(min)=20` Now, mass of the store F = ma `m=(F)/(a)=(20)/(2xx10)=1kg` |
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| 345. |
A stone tied to a string is rotated with a uniform speed in a vertical plane. If mass of the stone is `m`, the length of the string is `r` and linear speed of the stone is `v` when the stone is at its lowest point, then the tension in the string will be (g= acceleration due to gravity)A. `(mv^(2))/r+mg`B. `(mv^(2))/r-mg`C. `(mv^(2))/r`D. `mg` |
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Answer» Correct Answer - A `T-mg=(mv^(2))/r` (centripetal force at lowest point) `T=(mv^(2))/r+mg` |
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| 346. |
A particle is moving in a circle:A. the resultant force on the particle must be towards the centreB. the cross product of the tangential acceleration and the angular velocity will be zeroC. the direction of the angular acceleration and the angular velocity must be the sameD. the resultant force may be towards the centre |
| Answer» Correct Answer - D | |
| 347. |
Which vector in the figures best represents the acceleration of a pendulum mass intemediate point in its saving??A. B. C. D. |
| Answer» Correct Answer - B | |
| 348. |
A particle of mass m is tied to light string and rotated with a speed v along a circular path of radius r. If T=tension in the string and `mg=` gravitational force on the particle then actual forces acting on the particle areA. mg and T onlyB. mg T and on additional force of `(mv^(2))/(r )` directed inwards.C. mg, T and an additional force of `(mv^(2))/(r )` directed outwards.D. only a force `(mv^(2))/(r )` directed outwards |
| Answer» Correct Answer - A | |
| 349. |
A mass of 5 kg is tied to a string of length 1.0 m is rotated in vertical circle with a uniform speed of `4 m//s` . The tension in the string will be 170 N when the mass is at `(g=10 m//s^(2))`A. highest pointB. mid wayC. bottomD. cannot be justified |
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Answer» Correct Answer - B `T=(mv^(2))/(r)+mg cos theta` `therefore mg cos theta =T-(mv^(2))/(r)` `=170-(5xx16)/(1)=170-80` `cos theta = (90)/(5xx10)=(9)/(5)` Can not be predicate . |
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| 350. |
A string of length `0.1` m cannot bear a tension more than 100 N. It is lied to a body of mass 100 g and rotated in a horizontal circle. The maximum angular velocity can beA. `100"rad s"^(-1)`B. `1000"rad s"^(-1)`C. `10000s^(-1)`D. `0.1"rad s"^(-1)` |
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Answer» Correct Answer - A `T=100N=momega_(m)^(2)r` `rArr" "100=100xx10^(-3)xx omega_(m)^(2)xx0.1` `rArr" "omega_(m)=100rads^(-1)` |
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