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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.
| 1. |
A uniform rod AB of length 2l and mass `m` is rotating in a horizontal plane about a vertical axis through A, with angular velocity `omega`, when the mid-point of the rod strikes a fixed nail and is brought immediately to rest. Find the impulse exerted by the nail. |
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Answer» Angular impulse `=` change in angular momentum `therefore(Jxxr_(bot))=Iomega` where `J=` linear impulse `=(Iomega)/(r_(t))` `=((m)((2l)/(3)).omega)/(l)` `=(4)/(3)m//omega` |
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| 2. |
A rod of mass 2 kg ad length 2 m is rotating about its one end `O` wth an angular velocity `omega=4rad//s`. Find angular momentum of the rod about the axis rotation. |
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Answer» In pure rotational motion of a rigid body, component of total angualr momentum about axis of rotation is given by `L=Iomega=((ml^(2))/(3))omega` `(I_(0)=(ml^(3))/(3))` Substituting the value we have, `L=((2)(2)^(2))/(3)(4)` `=(32)/(3)kg-m^(2)//s` direction of this component is perperdicular to paper inwards (from right hand rule), as the rotation clockwise. |
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| 3. |
a hoop is placed on the rough surface such that it has an angular velocity `omega=4rad//s` and an angular deceleration `alpha=5rad//s^(2)` also its centre has a velocity of `v_(0)=5m//s` and a decoleration `a_(0)=2m//s^(2)` determine the magnitude of acceleration of point B at this instant. |
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Answer» `a_(B)=a_(0)+a_(B//0)` here `a_(B//0)` has two components `a_(t)` (tangential acceleration) and `a_(n)` (normal acceleration) `a_(t)=ralpha=(0.3)(5)=1.5m//s^(2)` `a_(n)=romega^(2)=(0.3)(4)^(2)=4.8m//s^(2)` and `a_(0)=2m//s^(2)` `thereforea_(B)=sqrt((suma_(x))^(2)+(suma_(y))^(2))` sqrt((2+4.8cos45^(@)-1.5cos45^(@))^(2)+(4.8sin45^(@)+1.5sin45^(@))^(2))` `=6.21m//s^(2)` |
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| 4. |
If two circular disks of the weight and thickness are made from metals having different densities. Which disk, if either will have the larger moment of inertia about its central axis. |
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Answer» `I=(1)/(2)mR^(2)` or `IpropR^(2)` `m`and thickness are same therfore, radius `R` of the material having smaller density should be more, so the moment of inertia. |
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| 5. |
Let `vecA` be a unit vecrtor along the axis of rotation of a purely rotating body and `vecB` be a unit vector along the velocity of a particle P of the body away from the axis. The value of `vecA.vecB` isA. 1B. -1C. 0D. none of these |
| Answer» Correct Answer - C | |
| 6. |
A body is uniformly rotating bout an axis fixed in an inertial frame of reference. Let `vecA` be a unit vector along the axis of rotation and `vecB` be the unit vector along the resultant force on a particle P of te ody away from the axis. The value of `vecA.vecB` isA. 1B. -1C. 0D. none of these |
| Answer» Correct Answer - C | |
| 7. |
The axis of rotation of a purely rotating bodyA. must pass thruogh the centre of masB. may pas through the centre of massC. must pass through a particle of the bodyD. may pass through a paerticle of the body |
| Answer» Correct Answer - B::D | |
| 8. |
A string of negligible thicknes is wrapped several times around a cylinder kept on a rough horizontal surface. A man standing at a distance l from the cylinder holds one end of the sitting an pulls the cylinder towards him figure. There is no slipping anywhere. The length of the string passed through the hand of the man whicle the cylinder reaches his hands is A. lB. 2lC. 3lD. 4l |
| Answer» Correct Answer - B | |
| 9. |
A cylinder having radius 0.4 m initially rotating (at `r=0`) with `omega_(0)=54rad//s` is placed on a rough inclinded plane with `theta=37^(@)` having friction coefficient `mu=0.5` the time taken by the cylinder to start pure rolling is `(g=10m//s^(2))`A. 5.4 sB. 2.4 sC. 1.4 sD. none of these |
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Answer» `a=(mucostheta)+(gsintheta)` `=0.5xx10xx0.8+10xx0.6=10m//s^(2)` `alpha=(mumgcostheta)R)/((1)/(2)mR^(2))=(2mugcostheta)/(R)` `=(2xx0.5xx10xx0.8)/(0.4)=20rad//s^(2)` Pure rolling will start when `v=Romega` or at `=R(omega_(0)-alphat)` `therefore10t=0.4(54-20t)` Solving this equation we get `t=1.2s` |
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| 10. |
A force `F=(2hati+3hatj+4hatk)N` is acting at point `P(2m,-3m,6m)` find torque of this force about a point `O` whose position vector is `(2hati-5hatj+3hatk)` m. |
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Answer» `tau=rxxF` Here `r=r_(P)-r_(O)=(2hati+3hatj+6hatk)-(2hati-5hatj+3hatk)=(2hatj+3hatk)m` Now `tau=rxxF=|{:(hati,hatj,hatk),(0,2,3),(2,3,-4):}|=(-17hati+6hatj-4hatk)N-m` |
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| 11. |
A ball moves over a fixed track as shown in thre figure. From A to B the ball rolls without slipping. If surface BC is frictionless and `K_(A),K_(B)` and `K_(C)` are kinetic energies of the ball at A, B and C respectively then (a). `h_(A)gth_(C),K_(B)gtK_(C)` (b). `h_(A)gth_(C),K_(C)gtK_(A)` (c). `h_(A)=h_(C),K_(B)=K_(C)` (d). `h_(A)lth_(C),K_(B)gtK_(C)` |
| Answer» On smooth part BC, due to zero torque, angular velocity and hence the rotational kinetic energy remains constant. While moving from B to C translational kinetic energy `therefore` the correct option is (a). | |
| 12. |
A tennis ball, starting from rest, rolls down the hill in the drawing. At the end of the hill the ball becomes airborne, leaving at an angle of `37^(@)` with respect to the ground treat the ball as a thin-walled spherical shell. Q. The velocity of projection `v` isA. `sqrt(2gh)`B. `sqrt((10)/(7)gh)`C. `sqrt((5)/(7)gh)`D. `sqrt((6)/(5)gh)` |
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Answer» `K_("total")` after falling a height `h` is mgh `(K_(R))/(K_(T))=(2)/(3)` `thereforeK_(T)=(1)/(2)mv^(2)=((3)/(5))(mgh)` `thereforev=sqrt((6gh)/(5))` |
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| 13. |
A rectangular block of size `(bbxxh)` moving with velocity `v_(0)` enters on a rough surface where the coefficient of friction is `mu` as shown in figure. Identify the correct statement.A. Thenet torque acting on the block about its COM is `mumg(h)/(2)` (clockwise)B. the net torque acting on the block about its COM is zeroC. The net torque acting on the block about its COM is in the anticlockwise senseD. None of the above. |
| Answer» No rotational motion. | |
| 14. |
In the figure shown a solid sphere of mass 4kg and radius 0.25 m is placed on a rough surface. `(g=10ms^(2)`) (a). Minimum coefficient of friction for pure rolling to take place, (b). If `mugtmu_(min)` find linear acceleration of sphere. (c). if `mu=(mu_(min))/(2)`, find the linear acceleration of cylinder. Here `mu_(min)` is the value obtained part (a). |
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Answer» (a). `mu_(min)=(tantheta)/(1+(mR^(2)//I))` `=(tan30^(@))/(1+5//2)=(2)/(7sqrt(3))` (b). `a=(gsintheta)/(1+I//mR^(2))` `=((10)sin30^(@))/(1+2//5)` `=(25)/(7)m//s^(2)` (c) `a=gsintheta-mugcostheta` `=(10)sin30^(@)-((1)/(7sqrt(3)))(10)(cos30^(@))` `=5-(5)/(7)=(30)/(7)m//s^(2)` |
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| 15. |
A horizontal force `F` is applied at the centre of solid sphere placed over a plank. The minimum coefficient of friction between plank and sphere required for pure rolling is `mu_(1)` when plank is kept at rest ad `mu_(2)` when plank can move, then `mu_(2)ltmu_(1)` Reason: Work done by frictional force on the sphere in both cases is zero.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, but the reaction is false.D. If assertion is false but the reason is true. |
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Answer» In first case `a=Ralpha` `therefore(F-mu_(1)mg)/(m)=R((mu_(1)mR)/(2//5mR^(2)))` `=(5)/(2)mu_(1)g` or `mu_(1)=(2)/(7)(F)/(mg)` In second case, ,brgt `a-Ralpha=a_("plank")` `therefore((F-mu_(2)mg)/(m))-R((mu_(2)mg)/(2//5mR^(2)))=(mu_(2)mg)/(M)` `therefore(F)/(m)=mu_(2)g+(5)/(2)mu_(2)g+(m)/(M)mu_(2)g` `thereforemu_(2)=(F)/((7//2+(m)/(M))mg)` Net work done by friction in pure rolling is zero. |
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| 16. |
A solid sphere of masss 0.50 kg si kept on a horizontal surface. The coeffici8ent of static friction between the surfaces in contact is 2/7. What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface? |
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Answer» Correct Answer - C If we take moment about the center than `FxxR=Ialpha+fxxR` `rarr F=2/5mRalpha+mumg`…….1 `Again F=ma_c-m mg`…….. 2 `-[because a_c=ralpha]` `rarr a=((F+mumg))/m` Putting the value of `a_c` in equation 1 we get `=(2/5)(F+mug)/m+mu mg` `rarr F=2/5F+2/5+2/5xx2/7xx0.5xx10+2.7xx0.5xx10` `rarr (3F)/5=4/7+10/7=2` `rarr F=((5x2))/3=10/3=3.3N` |
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| 17. |
A solid sphere and a hollow sphere both of same mass and same radius are hit by a cue at a height `h` above the centre `C`. In which case, (a). Linear velocity will be more? (b). Angular velocity will be more? (c). rotational kinetic energy will be more? |
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Answer» (a). `v=(J)/(m),J` and `m` are same for both (b). `omega=(Jh)/(I)` or `omegaprop(1)/(I)` J and `h` are same but `I_(solid)ltI_(hollow)` `thereforeomega_(solid)ltomega_(hollow)` (c). `K_(R)=(1)/(2)Iomega^(2)` `omegaprop(1)/(I)` `thereforeK_(R)prop(1)/(I)` |
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| 18. |
A solid body rotates about a stationary axis so that the rotation angle `theta` varies with time as `theta=6t-2t^(3)` radian. Find (a) the angular acceleration at the moment when the body stops and (b) the average value of angular velocity and angular acceleration averaged over the time interval between `t=0` and the complete stop. |
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Answer» `theta=6t-2t^(3)` `omega=(dtheta)/(dt)` `=6-6t^(2)` `alphal=(domega)/(dt)=-12t` `omega=0` at `t=1` sec `(omega)_(0-1)=((int_(0)^(1)omegadt)/(1))=int_(0)^(1)(6-6t^(2))dt` `(alpha)_(0-1)=((int_(0)^(1))alphadt)/(1)=int_(0)^(1)(-12t)` `=-6rad//s^(2)` |
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| 19. |
A man standing on a platform holds weights in his outstretched arms. The system is rotated about a central vertical axis. If the man now pulls the weights inwards close to his body thenA. the angular velocity of the system will increaseB. the angular momentum of the system will remain constantC. the kinetic energy of the system will increaseD. all of the above |
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Answer» I will decrease `thereforeomega` will increase (as `L=Iomega=` constant) `KE=(L^(2))/(2I)` will increase as `I` is decreasing. |
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| 20. |
Figure shows smooth inclined plane fixed in a car acceleratiing on a horizontal road. The angle of incline `theta` is related to the acceleration a of the car as `a=gtantheta`. If the sphere is set in pure rolling on the incline A. it will continue pure rollingB. it will slip down the planeC. its linear velocity will increaseD. it linear velocity will slowlyidecrease. |
| Answer» Correct Answer - A | |
| 21. |
A shaft is turning at `65rad//s` at time zero. Thereafter, angular acceleration is given by `alpha=-10rad//s^(2)-5trad//s^(2)` Where `t` is the elapsed time (a). Find its angular speed at `t=3.0` s (b). How much angle does it turn in these `3s`? |
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Answer» (a). `intdomega=intalphadt` `thereforeint_(65)^(omega)domega=int_(0)^(3)(-10-5t)dt` `thereforeomega=65-[10t+2.5t^(2)]_(0)^(3)` `=12.5rad//s` (b). `int_(65)^(omega)domega=int_(0)^(t)(-10-5t)dt` `thereforeomega=64-10t-2.5t^(2)` `int_(0)^(theta)dtheta=int_(0)^(3)omegadt=int_(0)^(3)(65-10t-2.5t^(2))dt` `thereforetheta=195-45-22.5` `=127.5rad` |
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| 22. |
When tall buildings are constructed on earth, the duration of day night slightly increases. Is this statement true or false? |
| Answer» Moment of inertia `I` increase. `Therefore omega` decreases (as `Iomega=`constant) hence time period increases (as `T=(2pi)/(omega)`) | |
| 23. |
Particle `P` shown in figure is moving in a circle of radius `R=10` cm with linear speed `v=2m//s` Find the angular speed of particle about point O. |
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Answer» `omega_(c)=(v)/(R)=(2)/(0.1)=20rad//s` `omega` about any point on circumference. `=(omega_(c))/(2)=10rad//s` |
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| 24. |
A wheel of rdius r and moment of inertia I about its axis is fixed at top of an inclined plane of inclination `theta` as shownin figure. A string is wrapped round the wheel and its free end supports a block of mass M which can slide on the plane. INitialy, tehwheel is rotating at a speed `omega` in direction such that the block slides pu the plane. How far wil the block move before stopping? |
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Answer» Suppose the deceleration of the block is a. The linear decelerationof the rim of the wheel is also a. The angular decelratinof the wheel is `alpha=a/r`. I the tension in the string is T, the equations of motion are as follows: `Mgsintheta-T=Ma` and `Tr=Ialpha=Ia/r` ElimiN/Ating T from these equations `Mgsintheta-Ia/r^2=Ma` giving `a=(Mgr^2sintheta)/(I+Mr^2)` THe initial velocity of the block up the incline is v=omegar`. Thus the distance moved by the block before stopping is `x=v^2/(2a) =(omega^2r^2(I+Mr^2))/(2Mgr^2sintheta)=((I+Mr^2)omega^2)/(2Mgsintheta)` |
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| 25. |
Under forward slip condition, translational kinetic energy of a ring is greater than its rotational kinetic energy is this statement true of false? |
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Answer» Under forward up slip condition `vgtRomega` `K_(T)=(1)/(2)mv^(2) `,brgt `K_(R)=(1)/(2)Iomega^(2)=(1)/(2)(mR^(2))omega^(2)` `thereforeK_(T)gtK_(R)` |
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| 26. |
A small object of uniform density rolls up a curved surface with an initial velocity `v`. It reaches up to a maximum height of `(3v^(2))/(4g)` with respect to the initial position. The object is (a). Ring (b). solid sphere (c). hollow sphere (d). disc |
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Answer» `(1)/(2)mv^(2)+(1)/(2)I((v)/(R))^(2)=mg((3v^(2))/(4g))` `thereforeI=(1)/(2)mR^(2)` `therefore` body is disc The correct option is (d). |
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| 27. |
A uniform disc of mass `M` and radius R is pivoted about the horizontal axis through its centre `C` A point mass m is glued to the disc at its rim, as shown in figure. If the system is released from rest, find the angular velocity of the disc when `m` reaches the bottom point B. |
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Answer» From conservation of mechanical energy decrease in gravitaional `PE=` increase in rottional KE or `mg(R)=[(1)/(2)MR^(2)+mR^(2)]((1)/(2)omega^(2))` or `omega=sqrt((4mg)/((2m+M)R))` |
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| 28. |
The density of as rod graduallly decreases from one end to the other. It is pivoted at an end so that ilt can move about a vertical axius through the pivot. A horizontal force F is applied on the free end n a direction perpendicular to the rod. The quantities, that do not depend on which end of the rod is pivoted areA. angular accelerationB. angular velocity when the rod completes one rotationC. angular momentum when tehrod competes one rotationD. torque of the applied force |
| Answer» Correct Answer - D | |
| 29. |
Two point P and Q. diametrically opposite on a disc of radius R have linear velocities v and 2v as shown in figure. Find the angular speed of the disc. |
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Answer» `omega=-(v_(r))/(PQ)=(2v-v)/(2R)=(v)/(2R)` Rotation is clockwise So, `omega` is perpendicular to paper inward. |
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| 30. |
The figure represent two cases. In first case a block of mass `M` is attached to a string which is tightly wound on a disc of mass `M` and radius `R`. In second case `F=Mg` initially the disc is stationary in each case. if the same length of string is unwound from the disc, thenA. same amount of work is done on both discsB. angular velocities of both the discs are equalC. both the discs have unequal angular accelerationsD. All of the above |
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Answer» In first case `TltMg` In second case `T=Mg` torque are different, so angular acceleration are different. |
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| 31. |
A uniform rod of mass `m` and length `l` is applied pivoted at point `O`. The rod is initially in vertical position and touching a block of mass `M` which is at rest on a horizontal surface. The rod is given a slight jerk and it starts rotating about point `O` this causes the block to move forward as shown The rod loses contact with the block at `theta=30^(@)` all surfaces are smooth now answer the following questions. Q. The hinge reaction at `O` on the rod when it loses contact with the block isA. `(3mg)/(4)(hati+hatj)`B. `((mg)/(4))hatj`C. `((mg)/(4))hati`D. `(mg)/(4)(hati+hatj)` |
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Answer» `a_(C)=((3sqrt(3))/(8)cos60^(@)-(3)/(8)gcos30^(@))hati` `+((3sqrt(3)g)/(8)gcos30^(@)+(3)/(8)gcos60^(@))(-hatj)` `=-((3g)/(4))hatj` Now, `mg(-hatj)+F_("hinge")=ma_(C)` substituting the value we get `F_("Hinge")=((mg)/(4))hatj` |
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| 32. |
Linear mss density (mass/length) of a rod depends on the distanec from one end (say A) as `lamda_(x)=(alphax+beta)` here `alpha` and `beta` are constants, find the moment of inertia of this rod about an axis passing through `A` and perpendicular to the rod. Length of the rod is `l`. |
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Answer» `I=int_(0)^(I)dI=int_(0)^(l)(dm)x^(2)` `=int_(0)^(l)(lamda_(X)dx)x^(2)` `=int_(0)^(l)(alphax+beta)x^(2)dx` `=(alphal^(4))/(4)+(betal^(3))/(3)` |
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| 33. |
The motor of an engine is erotating about its axis with an angular velocity of 100 rev/minute. It comes to rest in 15 s, after being switched off. Assumgn cnstant angular decelertion, calculate the number of revolutions made by it before coming to rest. |
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Answer» The initila ngular velocity =100 rev/minute =(10pi/)rad/s` FiN/Al angular velocity =0 Time interval =15s ltbr. Let angular acceleration be `alpha`. Using the equation `omega=omega_0+alphtat` we obtain alpha=(-2pi/9) rad/s^2` The angle rotated by teh motor during this motion is `theta=omega_0t+1/2alphat^2` `=((10pi)/3 (rad)/s) (15s) -1/2 ((2pi)/9 (rad)/s)(15s)^2` Hence the motor rotates thrugh 12.5 revolutions before coming to rest. |
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| 34. |
A body rotates about a fixed axis with an angular acceleration `1rad//s^(2)` through what angle does it rotates during the time in which its angular velocity increases from `5rad//s` to `15rad//s`? |
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Answer» `omega^(2)=omega_(0)^(2)+2alphatheta` `thereforetheta=(omega^(2)-omega_(0)^(2))/(2alpha)=(225-25)/(2xx1)` `=100rad` |
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| 35. |
A flywheel of moment of inertia `5.0kg-m^(2)` is rotated at a speed of `10rad//s` because of the friction at the axis it comes to rest in 10s. Find the average torque of the friction. |
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Answer» `0=omega_(0)-alphat` `impliesalpha=(omega_(0))/(t)=(10)/(10)` `=1rad//s^(2)` `tau=Ialpha=5N-m` |
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| 36. |
A ody is in pure rotation. The linear speed v of a particle, the distance r of the particle from the axis and the angular velocity `omega` of the body are related as `omega=v/r`. ThusA. `omega prop1/r`B. `omegapropr`C. `omega=0`D. `omega` is independent of r |
| Answer» Correct Answer - D | |
| 37. |
Figure shows a small wheel fixed coaxially on a bigger one of double the radius. The system rotates about the comon axis. The strings supporting A and B do not slip on the wheels. If x and y be thedistances travelled by A and B in the same time interval, then A. `x=2y`B. `x=y`C. `y=2x`D. none of these |
| Answer» Correct Answer - C | |
| 38. |
The sphere shown in figue lies on a rough plane when a particle of mass m travbelling at a speed `v_0` collides and sticks with it. If the line of motion of the particle is at a distance h above the plane, find a. the linear speed o the combine dsystem just after teh collision b. the angular speed of the system about the centre of the sphere just the collistion c. the value of h for which the sphere starts pure rolling on the plane Assume that the mass M of the sphesre is large compared to teh mass of the partcle so that is large compared toteh mass of the particle so that teh centre of mass of the combined system is not appreciably shifted from the centre of the sphere. |
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Answer» Take the particle plus the sphere as the system a. Using conservation of linear momentum, the inear speed o the combined system v is given by `mv_0=(M+m)v or, v=(mv_0)/(M+m)` ……..i b. Next we shall use conservation angular momentum about the centre of mass, which is to be taken at the centre of the sphere `(Mgtgtm)` Angular momentum of the particle before collsion is `mv_0(h-R).` If the system rotates with angular speed `omega` after collision, the angular momentum of the system becomes `(2/5MR^2+mR^2)omega` Hence, `mv_0(h-R)=(2/5 M+m)R^2omega` or, ` omega=(mv_0(h-R))/(2/5 M+m))R^2` c. The sphere will start roling just after the colision if `v=omegaR i.e. (mv_0)/(M+m)=(mv_0(h-R))/(2/5M+m)R)` giving`h=((7/5M+2m)/(M+m))R=7/5R`. |
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| 39. |
A holow sphere of radius R lies on a smoth horizontal surface. It is pulle dby a horizontl force acting tasngentially from the highest point. Find the distance travelled by the sphere durint eh time it makes oen full rotation. |
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Answer» Correct Answer - C::D Taking moment about the centre of hollow shpere we will get `rarr FxxR=(2/3)MR^2alpha` `rarr alpha=(3F)/(2MR)` again `2pi=1/2at^2` `(From theta=omega_0t+1/2alphat^2)` `rarr t^2=(8piMR)/(3F)` rarr a_c=F/M` `-X=1/(2a_c)t^2` `=1/2xxF/Mxx((8piMR)/(3F))=(4piR)/3` |
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| 40. |
In backward slip condition translational kinetic energ of a disc may be equal to its rotational kinetic energy is this statement true of false? |
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Answer» `K_(T)=(1)/(2)mv^(2),K_(R)=(1)/(2)=(1)/(2)Iomega^(2)=(1)/(2)((1)/(2)mR^(2))omega^(2)` Since, in backward sli condition `vltRomega` `thereforeK_(R)` may be equal to `K_(T)`. |
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| 41. |
A bucket is being lowered down into a well through a rope passing over a fixed pulley of radius 10 cm. Assume that the rope does not slip on the pulley. Find the angular velocity and angular acceleration of the pulley at an instant when the bucket is going down at at speed of 20 cm/s adn has an accelertion of `4.0 m/s^2`. |
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Answer» Since the rope does not slip on the pulley, the linear speed v of therim of the pulley is same as the speed of the bucket. the anguklar velocity of the pulley is then `omega=v/r=(20cm/s)/(10cm)=2rad/s` and the angular acceleration of the pulley is `alphaa/r(4.0m/s^2)/(10cm)=40rad/s^2` |
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| 42. |
The angular velocity of the engine (and hence of the wheel) on a scooter is proportional to the petrol input per second. The scooter is moving on a frictionless road with uniform velocity. If the petrol input is increased by 10% the linear velocity of the scooter is increased byA. 0.5B. 10C. 0.2D. 0 |
| Answer» Correct Answer - D | |
| 43. |
The figure shows a thin ring of mass `M=1kg` and radius `R=0.4m` spinning about a vertical diameter (take `I=(1)/(2)MR^(2))` A small beam of mass `m=0.2kg` can slide without friction along the ring When the bead is at the top of the ring the angular velocity is `5rad//s` What is the angular velocity when the bead slips halfwat to `theta=45^(@)`? |
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Answer» `I_(1)omega_(1)=I_(2)omega_(2)` `thereforeomega_(2)=((I_(1))/(I_(2)))omega_(1)` `=(((1)/(2)MR^(2)))/([(1)/(2)MR^(2)+m((R)/(sqrt(2)))^(2)])omega_(1)` `=((M)/(M+m))omega_(1)=((1)/(1+0.2))^(5)` `=(25)/(6)rad//s` |
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| 44. |
A rod is supported horizontally by means of two strings of equal length as shown in figure. If one of the string is cut. Then tension in other string at the same instant will.A. remains unaffectedB. increaseC. decreaseD. become equal to weight of the rod. |
| Answer» `T_(i)=(mg)/(2)` and `T_(f)=(mg)/(4)` | |
| 45. |
A uniform cylider of mass `M` and radius R has a string wrapped around it. The string is held fixed and the cylinder falls vertically, as in figure. (a). Show that the acceleration of the cylinder is downward with magnitude `a=(2g)/(3)` (b). Find the tension in the string. |
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Answer» `a=(Mg-T)/(M)` ..(i) `alpha=(TR)/((1)/(2)MR^(2))=(2T)/(MR)` ..(ii) `a=Ralpha` Solving these three equation we get `a=(2g)/(3)` and `T=(Mg)/(3)` |
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| 46. |
In the above question, the frictional force on the cylinder isA. `F//3` towards rightB. `F//3` towards leftC. `2F//3` towards rightD. `2F//3` towards left |
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Answer» `F+f=ma=(4F)/(3)` `thereforef=(F)/(3)` (towards right) |
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| 47. |
A mass of 1 kg is placed at (1m, 2m, 0). Another mass of 2 kg is placed at (3m, 3m, 0). Find the moment of inertial of both the masses about z-axis |
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Answer» `I=(1)[sqrt((1)^(2)+(2)^(2))]^(2)+2[sqrt((3^(2))+(4)^(2))]^(2)` `=55kg-m^(2) |
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| 48. |
Two blocks of masses 400 g and 200 g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has as moment of inertia `1.6xx10^-4kg-m^2` and a radius 2.0 cm. Find a. the kinetic energy of the system as the 400 block falls thrug 50 cm b. the speed of the blcok at this instant. |
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Answer» Correct Answer - A::B::D According to the question `04g-T_1=00.4a`……..1 `T_2-0.2g=0.2a`……2 `(T_1-T_2)r=(la)/r`……..3 From eq. 1, 2, and 3 `rarr a=((0.4-0.2)g)/((0.4+0.2+1.6/0.4))=g/5` ltb. Therefoee b. `V=sqrt(2gh)=sqrt(2xgxx1/20` `rarr sqrt((g/5))=sqrt((9.8/5))=1.4m/s` a. total kinetic energy of the system `=1/2m_1V^2+1/2m_2V^2+1/2 15^` ` =(1/2xx0.4xx1.4^2)+(1/2xx0.2xx1.4^2)+ {1/2xx(1.6/4xx1.4^2)}` `=(0.2+0.1+0.2)(1.4)^2` `=0.54x1.96=0.98Joule` |
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| 49. |
Figure shows two blocks of masses m and M connected bky a string pasing over a pulley. The horizontal table over which the mass m slides is smooth. The pulley has a radius r and moment of inertia I about its axis and it can freely rotte about this axis. Find the accelerationof the mass M asuming that the string does not slip on the pulley. |
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Answer» Correct Answer - B Acording to the equation `Mg-T_1=Ma`…..1 `T_2=ma`………2 `rarr(T_1-T_2)=(I_a)/r^2` [because a=ralpha]……..e if we dd the equation 1 and 2 we will get `Mg+T_2-T_1=Ma+ma`…..4 `rarr Mg-Ia/r^2=Ma+ma` `rarr (M+m+I/r^2)a=Mg` `rarr a=(Mg)/(M+m+I/r^2)` |
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| 50. |
A cylinder is rolling down a rough inclined plane. Its angular momentum about the point of contact remains constant. Is this statement true or false? |
| Answer» Torque of `mgsintheta` is non-zero. So, angular momentum is not constant. | |