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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.
| 451. |
A bob of mass 1 kg is suspended from an inextensible string of length 1 m. When the string makes an angle `60^(@)` with vertical, speed of the bob is 4 m/s Upto what maximum height will the bob rise with respect to the bottommost point ?A. 1.6 mB. 1.8 mC. 1.3 mD. 2 m |
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Answer» Correct Answer - C |
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| 452. |
A bob of mass 1 kg is suspended from an inextensible string of length 1 m. When the string makes an angle `60^(@)` with vertical, speed of the bob is 4 m/s Net acceleration of the bob at this instant isA. `16 m//s^(2)`B. `20.4 m//s^(2)`C. `18.2 m//s^(2)`D. `10.4 m//s^(2)` |
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Answer» Correct Answer - C |
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| 453. |
A stone of mass of 1 kg is tied to the end of a string 1 m long. It is whirled in a vertical circle. The velocity of the stone at the bottom of the circle is just sufficient to take it to the top of circle without slackening of the string. What is the tension in the string at the top of the circle? (Take, g =10 `ms^(-2)`)A. zeroB. 1 NC. `sqrt(10)N`D. 10 N |
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Answer» Correct Answer - A In critical case, tension at topmost point is zero. |
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| 454. |
When a motor cyclist takes a `U-turn` in `4s` what is the average angular velocity of the motor cyclist.A. `pi//2 rad//s`B. `pi//4 rad//s`C. `pi//3 rad//s`D. `pi//5 rad//s` |
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Answer» Correct Answer - B `(T_(1))/(2)=4" "therefore T=8s` `omega=(2pi)/(T)=(2pi)/8=(pi)/(4)` |
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| 455. |
A motor cyclist moving with a velocity of 72 km/hour on a flat road takes a turn on the road at a point where the radius of curvature of the road is 20 meters . The acceleration due to gravity is `10m//sec^(2)`. In order to avoid skidding, he must not bend with respect to the vertical plane by an angle greater thanA. `theta=tan^(-1)(6)`B. `theta=tan^(-1)(2)`C. `theta=tan^(-1)(25.92)`D. `theta=tan^(-1)(4)` |
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Answer» Correct Answer - B To avoid skidding `tantheta=(v^(2))/(gr)=(20xx20)/(10xx20)=2` `rArr" "theta=tan^(-1)(2)` |
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| 456. |
A small coin of mass 5 g is placed at a distance 5 cm from centre on a flat horizontal turn table. The turn table is observed to make 3 revolutions in `sqrt(10)s`. The frictional force on the coin is `(n xx 10^(-3)) N`. Find value of `n`. |
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Answer» Correct Answer - 9 |
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| 457. |
A particle is acted upon by a force of constant magnitude which is always perpendiculr 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 straight line |
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Answer» Correct Answer - C When a force of constant magnitude acts on velocity of particle perpendicular, then there is no change in the kinetic energy of particle. Hence, kinetic energy remains constant |
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| 458. |
A small ball (of negligible size) is placed over a sphere of same mass and radius 1 m as shown. All surface are smooth. A slight push is given to the ball. When the radius vector joining the ball makes an angle of `30^(@)` with the vertical, speed of sphere is `v. (g = 10 m//s^(2))` Value of `v` in m/s is A. 0.9B. 1.6C. 2.4D. 0.4 |
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Answer» Correct Answer - A |
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| 459. |
If the magnitude of velocity in the previous question is decreasing with time, what is the direction of f angular acceleration `(oversetrightarrow(alpha))? `A. eastB. westC. northD. south |
| Answer» Correct Answer - C | |
| 460. |
A particle starts travelling on a circle with constant tangential acceleration. The angle between velocity vector and acceleration vector, at the moment when particle completes half the circular track. IsA. `tan^(-1)(2pi)`B. `tan^(-1)(pi)`C. `tan^(-1)(3pi)`D. `tan^(-1)(2)` |
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Answer» Correct Answer - A |
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| 461. |
A particle starts moving along a circle of radius `(20//pi)m` with constant tangential acceleration. If the velocity of the parthcle is `50 m//s` at the end of the second revolution after motion has began, the tangential acceleration in `m//s^(2)` is :A. `1.6`B. 4C. `15.6`D. `31.2` |
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Answer» Correct Answer - D As `v=omegar` `rArr" "50=omega((20)/(pi))` `rArr" "omega=(50pi)/(20)=(5pi)/(2)` `rArr" "(2pi)/(T)=(5pi)/(2)rArrT=(4)/(5)s` Given, `t=2T=(8)/(5)s` `therefore" "omega=alphat` `rArr" "(5pi)/(2)=alphaxx(8)/(5)rArr=(25pi)/(16)rads^(-2)` `therefore" "a_(t)=r(alpha)=((20)/(pi))((25pi)/(16))=31.2rads^(-2)` |
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| 462. |
A particle starts moving from rest at `t=0` with a tangential acceleration of constant magnitude of `pi m//s^(2)` along a circle of radius 6 m. The value of average acceleration, average velocity and average speed during the first `2 sqrt(3)` s of motion, are respectively :A. `3sqrt(2)m//s^(2),pi m//s,pisqrt(3)m//s`B. `pi m//s^(2),2 sqrt(3) m//s, pi sqrt(3) m//s`C. `pi sqrt(3) m//s^(2),2 sqrt(3) m//s, pi m//s`D. None of the above |
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Answer» Correct Answer - B |
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| 463. |
A bob is suspended from a crane by a cable of length `l=5 m`. The crane and load are moving at a constant speed `v_(0)`. The crane is stopped by a bumper and the bob on the cable swings out an angle of `60^(@)`. The initial speed `v_(0)` is `(g=9.8 m//s^(2))` A. 10 m/sB. 7 m/sC. 4 m/sD. 2 m/s |
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Answer» Correct Answer - B |
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| 464. |
A fighter plane is pulling out for a dive at a speed of `900 km//h`. Assuming its path to be a vertical circle of radius `2000 m` and its mass to be `16000 kg`, find the force exerted by the air on it at the lowest point. Take `g=9.8 m//s^(2)`A. `3.28xx10^(5)N`B. `6.56xx10^(5)N`C. `9.28xx10^(5)N`D. `12.56xx10^(5)N` |
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Answer» Correct Answer - A Given , radius of circular path = 2000 m Mass of plane = 16000 kg and speed = 900 `kmh^(-1)` `=(900xx10^(3))/(3600)ms^(-1)` = 250 `ms^(-1)` The required centripetal force, `f_(c)=(mv^(2))/(r)` (in upward directon) `=(16000xx250xx250)/(2000)=5xx10^(5)N` The gravitational force, `f_(g)=mg=16000xx9.8` `=1.568xx10^(5)N` Thus, force will act as reaction force in upward direction. ltbtgt `therefore` Net upward force `=5xx10^(5)-1568xx10^(5)` `=(5-16)xx10^(5)N=3.4xx10^(5)N` |
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| 465. |
A particle of mass `m` is fixed to one end of a light rigid rod of length `l` and rotated in a vertical ciorcular path about its other end. The minimum speed of the particle at its highest point must beA. zeroB. `sqrt(gl)`C. `sqrt(1.5gl)`D. `sqrt(2gl)` |
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Answer» Correct Answer - A Rod does not slack (like string). So, minimum velocity at topmost point may be zero also. |
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| 466. |
Assertion If a particle is rotating in a circle, then centrifugal force is acting on the particle in radially outward direction . Reason centrifugal force is equal and opposite to the centripetel force.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 Centripetal force acts when the particle is observed from a rotating non-inertial frame. |
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| 467. |
Which of the following quantities may remain constant during the motion of an object along acurved path ? (i) Velocity (ii) Apeed (iii) Acceleration (iv) Magnitude of acceleration |
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Answer» Correct Answer - A::C::D In uniform circular motion speed remains constant. In projectile motion (which is a curved path acceleration remain constant. |
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| 468. |
Assertion: When water in a bucket is whirled fast overhead, the water does not fall out at the top of the ciorcular path. Reason: The centripetal force in this position on water is more than the weight of water.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 - A Weight (plus normal reactioon) provides the necessary centripetal force or weight is used in providing the centripetal force. |
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| 469. |
Assertion Centripetel force `(mv^(2))//(R)` acts on a particle rotating in a circle. Reason Summation of net force acting on the particle is equal to `(mv^(2))//(R)` in the above case.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 `(mv^(2))/(R)` does not act. But this much force is required. |
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| 470. |
Water in a bucket is whirled in a vertical circle with a string attatched to it. The water does not fll down even when thebucket is inverted at the top of its path. We conclude that in this positionA. `mg=(mv^2)/r`B. mg is greater than `(mv^2)/r`C. mg is not greater than `(mv^2)/r`D. `mg is not less than (mv^2)/r |
| Answer» Correct Answer - C | |
| 471. |
Which of the following quantities may remain constant during the motion of an object along a curved path.A. speedB. velocityC. accelerationD. momentum |
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Answer» Correct Answer - A In curved path, may be circular or parabolic In circuit path speed and magnitude of acceleration are constant In parabolic path acceleration is constant. |
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| 472. |
A particle is moving in a curved path. Which of the following quantities may remain constant during its motion?A. AccelerationB. VelocityC. Magnitude of accelerationD. None of these |
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Answer» Correct Answer - C A particle moving in a curved path is experiencing an acceleration. Even if moving around the perimeter of a circle with a constant speed, there is still a change in velocity and subsequently an acceleration. This acceleration is directed towards the centre of the circle. The quantity which may remain constant here is magnitude of acceleration. |
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| 473. |
Assertion When water in a bucket is whirled fast overhead, the water does not fall out at the top of the circular path. Reason The centripetel force in this position on water is more than the weight of water.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. |
| Answer» Correct Answer - A | |
| 474. |
Assertion In vertical circular motion, speed of a body cannot remain constant. Reason In moving upwards, work done by gravity is negative.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. |
| Answer» Correct Answer - A | |
| 475. |
A stone is tied to a string and swings with uniform motion in a horizontal circle. The string breaks and at a time t, the stone is displaced `Deltar=3 hat(i)+4hat(j)-5hat(k)` metres. (The positive z-axis is vertically up) Select the correct alternative.A. the time t is 1 sB. the time t is 0.5 sC. the speed of the stone while in circular motion is 5 m/sD. the speed of the stone while in circular motion is `5 sqrt(2)m//s` |
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Answer» Correct Answer - A::C |
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| 476. |
A road is 10 m wide. Its radius of curvature is 50 m . The outer edge is above the lower edge by a distance of 1.5 m . This road is most suited for the velocityA. `2.5 m//s`B. `4.5 m//s`C. `6.5 m//s`D. `8.5 m//s` |
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Answer» Correct Answer - D `(v^(2))/(rg)=(h)/(l)` `implies v=sqrt((rgh)/(l))=sqrt((50xx1.5xx9.8)/(10)) =8.5 7 m//s.` |
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| 477. |
On a railway track of radius of curvature 1600 m. If the distance between two trackes is 1.8 m, then the elevation of the outer track above the inner track will be `(g=10m//s^(2))`A. `0.450 m`B. `0.0450 m`C. `4.50 m`D. `4.0 m` |
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Answer» Correct Answer - B ` tan theta =(v^(2))/(rg) therefore (h)/(l)=(v^(2))/(rg)` Where l is distance between two tracks. |
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| 478. |
The radius of curvature of railway line at a place is 200 m. If the distance between the rail is 1.6 m and the outer rail is raised by 0.08 m above the inner rail .The speed of the train for which there is no side pressure on the rails `(g=10 m//s^(2))`A. `5 m//s`B. `10 m//s`C. `15 m//s`D. `20 m//s` |
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Answer» Correct Answer - B `r=200m,l=1.6m,h=0.08 m ` `v^(2)=(rgh)/(l)=(200xx10xx0.08)/(1.6)` `v^(2)=(2xx80)/(1.6)=100` `v=10 m//s`. |
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| 479. |
A motorcycle is going on an overbridge of radius R. The driver maintains a constant speed. As the motorcycle is ascending on the overbridge, the normal force on itA. increasesB. decreasesC. remains the sameD. fluctuates |
| Answer» Correct Answer - A | |
| 480. |
A motorcycle is going on an overbridge of radius R. The driver maintains a constant speed. As the motorcycle is ascending on the overbridge, the normal force on itA. increaseB. decreaseC. remains the sameD. fluctuates |
| Answer» Correct Answer - A | |
| 481. |
On a railway curve, the outside rail is laid higher than the inside one so that resultant force exerted on the wheels of the rail car by the tops of the rails willA. have a horizontal component inwardsB. be verticle componentsC. equal to the centripetalD. be decreased |
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Answer» Correct Answer - A This horizontal inward component provides required centripetal force . |
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| 482. |
If in aeroplane is moving on a circular path with a uniform speed 300 km /h, then the change in velocity after quarter of the circle will beA. `150 km//h`B. zeroC. `600 km//h`D. `300sqrt2 km //h` |
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Answer» Correct Answer - D `Delta v = 2v sin theta//2` `=2v sin (90)/(2)` `=2xx300xx(1)/(sqrt2) = 300 sqrt2`. |
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| 483. |
If the overbridge is concave instead of being convex, the thrust on the road at the lowest position will beA. `mg+(mv^(2))/(r)`B. `mg-(mv^(2))/(r)`C. `(m^(2)v^(2)g)/(r)`D. `(v^(2)g)/(r)` |
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Answer» Correct Answer - A Thrust at the lowest point of concave bridge `mg+(mv^(2))/(r)`. |
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| 484. |
The work done by the centripetal force in quarter revolution would be,A. infinityB. increasesC. zeroD. decreases |
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Answer» Correct Answer - C `omega=vecF.vecS=FS cos 90=FSxx0=0` |
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| 485. |
Two racing cars of masses `m_(1)` and `m_(2)` are moving in circles of radii `r_(1)` and `r_(2)` respectively. Their speeds are such that each makes a complete circle in the same duration of time t. The ratio of the angular speed of the first to the second car isA. `m_(1_:m_(2)`B. `M_(1)r_(1):m_(2)r_(2)`C. `m_(1)r_(2):m_(2)r_(1)`D. `1:1` |
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Answer» Correct Answer - D `omega _(1): omega_(2)=(2pi)/(T_(1)):(2pi)/(T_(2)) " "As T_(1)=T_(2), omega _(1):omega_(2)= 1:1`. |
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| 486. |
A particle of mass `m` starts moving in a circular path of canstant radiur `r` , such that iss centripetal acceleration`a_(c)` is varying with time a=`t` as `(a_(c)=k^(2)r//t)` , where `K` is a contant. What is the power delivered to the particle by the force acting on it ? |
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Answer» Correct Answer - A::B::D As `a_(c)=(v^(2)//r)` so `(v^(2)//r)=k^(2)rt^(2)` :. Kinetic energy `K=(1)/(2)mv^(2)=(1)/(2)mk^(2)r^(2)t^(2)` Now, from work-energy theorem `W=DeltaK=(1)/(2)mk^(2)r^(2)t^(2)-0` So, `P=(dW)/(dt)=(d)/(dt)((1)/(2)mk^(2)r^(2)t^(2))=mk^(2)r^(2)t` Alternate solution: Given that `a_(c)=k^(2)rt^(2)` , so that `F_(c)=ma_(c)=mk^(2)rt^(2)` Now, as `a_(c)=(v^(2)/r)` , so `(v^(2)//r)=k^(2)rt^(2)` or `v=krt` So, that `a_(t)=(dv//dt)=kr` i.e. `F_(t)=ma_(t)=mkr` Now, as `F=F_(c)+F_(t)` So, `P=(dW)/(dt)=F.v=(F_(c)+F_(t))_(v)` In circular motion, `F_(c)` is perpendicular to `v` while `F_(t)` parallel to it, so `P=F_(t)v` `P=mk^(2)r^(2)t` |
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| 487. |
A particle moves with constant angular velocity in a circle. During the motion itsA. energy is conservedB. momentum is conservedC. energy and momentum both are conservedD. None of the above |
| Answer» Correct Answer - A | |
| 488. |
If the speed of a body in uniform circular motion and radius of circular path is doubled then the centripetal force will beA. halvedB. doubledC. quadrupledD. tripled |
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Answer» Correct Answer - B `F_(2)=2F_(1) " "(because v_(2)=2v_(1) and r_(2)=2r_(1))` |
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| 489. |
A particle moving along a circular path due to a centripetal force having constant magnitude is an example of motion with :A. constant speed and velocityB. variable speed and velocityC. variable speed and constant velocityD. constant speed and variable velocity. |
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Answer» Correct Answer - D Centripetal force is constant in magnitude that means speed is constant and due to change in direction velocity is variable. |
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| 490. |
A particle of mass 2 kg is moving along a circular path of radius 1 m. If its angular speed is `2pi" rad s"^(-1)`, the centripetal force on it isA. `4piN`B. `8piN`C. `4pi^(4)N`D. `8pi^(2)N` |
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Answer» Correct Answer - D Centripetal force `=mRomega^(2)=(2)(1)(2pi)^(2)=8pi^(2)N` |
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| 491. |
Two particles A and B are located at distances `r_(A)` and `r_(B)` from the centre of a rotating disc such that `r_(A) gt r_(B)` . In this case (Angular velocity `(omega)` of rotation is constant)A. both A and B do not have any accelerationB. both A and B have same accelerationC. A has greater acceleration than BD. B has greater acceleration than B |
| Answer» Correct Answer - C | |
| 492. |
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration `a_(c)` is varying with time t as `a_(c) = k^(2)rt^(2)`, where k is a constant. The power delivered to the particle by the forces acting on it is :A. ` m^(2)K^(2)r^(2)t^(2)`B. `mK^(2)r^(2)t^(2)`C. `mK^(2)rt^(2)`D. `mKr^(2)t^(2)` |
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Answer» Correct Answer - B `F=ma=mK^(2)rt^(2)` `P=(w)/(t)=(F.s)/(t)=(mK^(2)rt^(2))/(t)r` `=mK^(2)r^(2)t` |
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| 493. |
A particle moving along a circular path due to a centripetal force having constant magnitude is an example of motion with :A. constant speed and velocityB. variable speed and variable velocityC. variable speed and constant velocityD. constant speed and variable velocity |
| Answer» Correct Answer - D | |
| 494. |
A car of mass 1000 kg moves on a circular path with constant speed of 16 m/s. It is turned by 90º after travelling 628 m on the road. The centripetal force acting on the car is-A. 64 NB. 3 cmC. 1.5 cmD. 1 cm |
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Answer» Correct Answer - B `s=rtheta` `therefore =(s)/(theta)=(6.28xx100)/((pi)/(2)) ` `r=(2pixx100)/((pi)/(2))=400` `F=(mv^(2))/(r)=(1000xx16xx16)/(400)=640 N`. |
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| 495. |
The change in the centripetal force of a body moving in a circular path , if speed is made half and radius is made 4 times the origenal value , willA. increacese by `(16)/(15)`B. decrease by `(15)/(16)`C. decrease by `(8)/(15)`D. increase by `(8)/(15)` |
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Answer» Correct Answer - B `(F_(2))/(F_(1))=(v_(2))/(v_(1))^(2)xx(r_(1))/r_(2)` `=(1)/(4)xx(1)/(4)=(1)/(16)` `(F_(2))/(F_(1)-1=(1)/(16)-1` `(F_(2)-F_(1))/(F_(1))=(1-16)/(16)=(-15)/(16)` `(DeltaF)/(F_(1))=(-15)/(16)`. |
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| 496. |
A block of mass `m=1`kg has a speed `v = 4 m//s` at `theta = 60^(@)` on a circular track of radius R = 2 m as shown in figure. Size of the block is negligible. Coefficient of friction between block and the track is `mu = 0.5`. The force of friction between the two is A. 10 NB. 8.5 NC. 6.5 ND. 5 N |
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Answer» Correct Answer - C |
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| 497. |
A particle goes round a circular path with uniform speed v.After describing half the circle, what is the change in its centripetal acceleration?A. `(v^(2))/(r)`B. `(2v^(2))/(r)`C. `(2v^(2))/(pir)`D. `(v^(2))/(pir)` |
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Answer» Correct Answer - B `Delta a =2a "sin" (theta)/(2)=(2v^(2))/(r)xx1=(2v^(2))/(r) m//s`. |
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| 498. |
A block is released from height `(ne 0)` on a rouch track AB as shown in figure. Coefficient of friction between the block and the surface is `mu = 0.5`. Track BC is smooth. From C onwards there is a circular smooth track of radius R = 50 cm If the block is placed on the above calculate height, how many times will the block cross point C ?A. infinite number of timesB. 2 timesC. 4 timesD. 3 times |
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Answer» Correct Answer - A |
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| 499. |
A block is released from height `(ne 0)` on a rouch track AB as shown in figure. Coefficient of friction between the block and the surface is `mu = 0.5`. Track BC is smooth. From C onwards there is a circular smooth track of radius R = 50 cm For what value of `h` block does not leave contact with any surface ? Given, `tan theta = (3)/(4) and g = 10 m//s^(2)`A. h=2.5 mB. `h le 2.0` mC. h = 2.0 mD. `h le 1.5` m |
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Answer» Correct Answer - D |
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| 500. |
A cyslist goes round a circular path of length 4000 m in 20 second. Calculate the angle through which he bends from vertical in order to maintain the balanceA. `sin^(-1)(0.64)`B. `tan^(-1)(0.64)`C. `cos^(-1) (0.64)`D. None of these |
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Answer» (a) `400=2pir` `therefore" " tantheta=(v^(2))/(rg)=((400/20)/(400/(2pi)))^(2)g` `therefore" " tantheta=(400x2pi)/(400g)=(6.28)/(9.8)=0.64` `therefore" " theta=tan^(-1)(0.64)` |
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