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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.
| 51. |
A uniform rod of mass `m` is rotated about an axis passing through point `O` as shown. Find angular momentum of the rod about rotational law. |
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Answer» `I=(m(3I)^(2))/(12)+m((l)/(2))^(2)=ml^(2)` `L=Iomega=ml^(2)omega` |
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| 52. |
A prticle oves on a straigh lie with uniform velocity. It sangular momentumA. is always zeroB. is zero about a point on the straight lineC. is not zero about a piont away from the straight lineD. about anyi given point remains constant |
| Answer» Correct Answer - B::C::D | |
| 53. |
Assertion: If linear momentum of a particle is constant, then its angular momentum about any point will also remain constant. Reason: Linear momentum remains constant if `F_(n et)=0` and angular momentum remains constant if `tau_(n et)=0`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. |
| Answer» constant linear momentum means moving with constant speed in straight line. Therefore, in `L=mvr_(bot)` all three m,v and `r_(bot)` are constant. | |
| 54. |
Let `I_(A)` and `I_(B)` be moments of inertia of a body about two axes `A` and `B` respectively. The axis `A` passes through the centre of mass of the body but `B` does not. ThenA. `I_AltI_B`B. `If I_AltI_B` the axes pareparallelC. `If the axes are parallel I_AltI_B`D. `If the axes are not parallel I_AgeI_B` |
| Answer» Correct Answer - C | |
| 55. |
If there is no external force acting on a nonrigid body, which of the followhng quantities must remain constant?A. angular momentumB. linear momentumC. kinetic energyD. moment of inertia |
| Answer» Correct Answer - A::B | |
| 56. |
Suppose the smaller pulley of the previous problem has its radius 5.0 cm and moment of inertia 0.10 kg-m^2. Find the tension in the part of the string joiningk the pulleys. |
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Answer» Correct Answer - C `m=2kg, I_1=0=0.10 kg-m^2` r_1=5 cm=0.05m` `I_2=2.20 kg-m^2` `r_2=10xm=0.1m` Therefore `mg-T_1=ma`…..1 `(T_1-T_2)r_1=I_1alpha`………..2 `T_2r_2=I_2alpha`…….3 substituting the value of `T_2` in the equation 2 we get `rarr (T_1-I_2alpha/r_1)r_2=I_1alpha` `(T_1-I_2a/r_1^2)=I_1a/r_2^2` `rarr T_1={(I_1/r_1^2)+(I_2/r_2^2)}a` LSubstituting the value of `T_1` in the equation 1 we get `rarr mg-{(I_1/r_1^2)+(I_2/r_2^2)}a=ma` `rarr (mg)/({(I_1/r_1^2)+(I_2/r_2^2)}+m+=a` `rarr a=(2xx9.8)/(0.1/0.0025+0.2/0.01+2)` `=0.316m/s^2` `rarr T_2=I_2 a/r_2^2` ltbr.gt `=(0.20xx0.316)/0.01=6.32N` |
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| 57. |
A string is wrapped on a wheel of moment of inertia 0.20 kg-m^2 and radius 10 cm and goes through a light pulley to support a block of mas 2.0 kg as shown in figure. Find the acceleration of the block. |
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Answer» Correct Answer - B `I=0.20kg-m^2` (bigger pulley) `r=10cm=0.1m, Smaller pulley is light `Mass of the block m=2kg` Therefore `mg-T=ma`……….1 `Txxr=Ixxalpha` `rarr T=(Ia)/r^2`………….2 `rarr mg=(m+I/r^2)a` [Using 2 and 1] `rarr a=(mg)/((m+I/r^2))` `=(2xx9.8)/{(2+(0.2/0.01)}` `=19.6/22=0.89m/s^2` therefore acceleration of the block `=0.89m/s^2` |
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| 58. |
A shaft initially rotating at 1725 rpm is brought to rest uniformly in 20s. The number of revolutions that the shaft will make during this time isA. 1680B. 575C. 287D. 627 |
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Answer» `0=omega_(0)-alphat` `thereforealpha=(omega_(0))/(t)=((2pi)(1725//60))/(20)` `=9rad//s^(2)` `theta=(1)/(2)alphat^(2)=(1)/(2)xx9xx(20)^(2)=1800rad` `N=(theta)/(2pi)=287` |
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| 59. |
A wheel of moment of inertia `0.10 kg-m^2` is rotating about a shaft at an angular speed o 160 rev/minute. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both the wheels finally rotate with as common angular speed of 200 rev/minute. Find the moment of inertia o the second wheel. |
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Answer» Correct Answer - B::D Wheel 1 has : `I_1=10kg-m^2` `omega_1=160 rev/min Wheel 2 has `I_2=?` `omega_2=300 rev/min` Given that after they are coupled `=omega =200 rev/min ` Theredore if we take the two wheels to be an isolated system. Totla exterN/Al torque =0 Therefore `I_1omega_1+I_2omega_2=(I_1+I_2)omega` `gt 0.10x160+I_2xx300=(0.10+I_2)xx200` `rarr 16+300I-2=20+200I_2` ltbr. `rarr 100I_2=4` `rarr I_2=4/100=0.04kg-m^2` |
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| 60. |
A flywheel rotating at a speed of `600` rpm about its axis is brought to rest by applying a constant torque for `10` seconds. Find the angular deceleration and angular velocity `5` second after the application of the torque. |
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Answer» Correct Answer - A::B A wheel rotating at a speed of 600 rpm `omega_0=600 rpm` =10 revolutions per second r=10sec (in 10 sec it comes to rest) omega=0` therefore `omega_0=-at` `rarr a=-10/10=-`rev/s^2` rarr omega=omega_0+at` `=10-1xx5=5rev/s` Therefore angular deceleration `=1rev/s^2 and Angular velocity after 5 sec is 5 rev/s. |
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| 61. |
A wheel of mass 10 kg and radius 0.2 m is rotating at an angular speed of 100 rpm, when the motion is turned off. Neglecting the friction at the axis. Calculate the force that must be applied tangentially to the wheel to bring it to rest in 10 rev. Assumed wheel to be a disc. |
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Answer» `omega^(2)=omega_(0)^(2)-2alphatheta` `0=((100)/(60)xx2pi)^(2)(10xx2pi)` `alpha=0.87N=(tau)/(I)` `=(F.R)/((1)/(2)mR^(2))` `thereforeF=(0.87)(0.5mR)` `=(0.87)(0.5)(10)(0.2)` `=0.87N` |
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| 62. |
Find the angular velocity of a body rotating wilth an acceleration of `2rev/s^2` as it completes the 5th revolution after the start |
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Answer» Correct Answer - B `theta=5. alpha=2rev/s^2, omega_0=0, omega=?` `omega^2=(2alphatheta)` `rarr omega=sqrt((2xx2xx5))` `=2sqrt5 rev/s` `or theta=10pirad` `alpha=4pirad/s` `omega_0=0` `omega=?` `omega=sqrt(2alphatheta)` `=(2xx4pixx10pi)` `=4pisqrt5rad/s` `=2sqrt5rev/s` |
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| 63. |
The wheel of a motor, accelerated uniformly from rest, rotates through 2.5 radian during the first second. Find the angle rotated during the next second. |
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Answer» As the angular acceleration is constant, we have `theta=omega_0t+1/2alphat^2=1/2alphat^2`. `Thus `2.5rad=1/2alpha(1s)^2` `alpha=5rad/s^2` or `alpha=5rad/s^2` the angle rotated during the first two seconds is `=1/2xx(5rad/s^2)(2s)62=10rad`. Thus the angle rotated during the 2nd second is `10rad-2.5rad=7.5rad.` |
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| 64. |
A angular positio of a point on the rim of a rotating wheel is given by `theta=4t-3t^(2)+t^(3)` where `theta` is in radiuans and `t` is in seconds. What are the angualr velocities at (a).`t=2.0` and (b). `t=4.0s` (c). What is the average angular acceleration for the time interval that begins at `t=2.0s` and ends at `t=4.0s`? (d). What are the instantaneous angular acceleration at the biginning and the end of this time interval? |
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Answer» Angular velocity `omega=(dtheta)/(dt)=(d)/(dt)(4t-3t^(2)+t^(3))` or `omega=4-6t+3t^(2)` (a). At `t=2.0s,omega=4-6xx2+3(2)^(2)` or `omega=4rad//s` (b). At `t=4.0s,omega=4-6xx5+3(4)^(2)` or `omega=28rad//s` (c). Average angular acceleration `alpha_(av)=(omega_(f)-omega_(i))/(t_(f)-t_(i))=(28-4)/(4-2)` or `alpha_(av)=12rad//s^(2)` (d). Instantaneous angular acceleartion is , `alpha=(domega)/(dt)=(d)/(dt)(4-6t+3t^(2))` or `alpha=-6+6t` at `t=2.0s` `alpha=-6+6xx2=6rad//s^(2)` at `t=4.0s` `alpha=-6+6xx4=18rad//s` |
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| 65. |
Starting from rest, a fan takes five seconds to attain the maximum speed of 40 rpm(revolutions per minute). Asuming constant acceleration find the time taken by the fan in attaining half the maximum speed. |
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Answer» Let the agular acceleratio be `alpha`. According to the question `400 rev/min=0+alpha5s`………i Let t be the time taken in attaining the speed of 200 rev/m whilch is half the maximum Then, `200 rev/min=0+alphat` ………ii Dividing i by ii we get, `2=5s/t or t=2.5s` |
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| 66. |
A wheel having moment of inertia 2 kig m^2 about its xis, rotates at 50 rpm about this axis. Find the torque that can stop the wheel in one minute. |
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Answer» The initial angular velocity `=50 rpm=(5pi)/3rad/s` Using `omega=omega_0+alphat` alpha=(omega-omega_0)/t=(0-(5pi)/3)/60 rad/s^I2=-pi/36rad/s^2`. The torque that can produce this deceleratioin is `Gamma =I|alpha|=(2kg-m^2)(pi/36rad/s^2)=pi/18N-m.` |
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| 67. |
A wheel rotating wilth unifrom angular acceleration covers 50 revolutions in the first five seconds after the start. Find the angular acceleration and the angular velocity at the end of five seconds. |
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Answer» Correct Answer - B::D `theta=100 pi,T=5sec ` `theta=1/2alphat^2` `rarr 100 pi=1/2alopha 25` `rarr alpha=8pi rad/s^2` `=omega=omega_0+2t` `=0+8pixx5=40pirad/s ` `=520 rev/s` |
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| 68. |
The angular velocity of a gar is controlled according to `omega=12-3t^(2)` where `omega` in radian per second, is positive in the clockwise sense and `t` is the time in seconds. Find the net angular displacement `triangletheta` from the time `t=0` to `t=3`s. Also, find the number of revolutions `N` through which the gear turns during the 3s. |
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Answer» `theta=int_(0)^(3)omegadt=int_(0)^(3)(12-3t^(2))dt` `=36-27=9rad` `N=(theta)/(2pi)=(9)/(2xx3.14)=1.43` |
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| 69. |
A horizontal disc rotates freely with angular velocity `omega` about a vertical axes through its centre. A rig having the same mass and radius as the disc, is now gently placed coaxially on the disc. After some time, the two rotate with a common angular velocity. then.A. no friction exists between te disc and the ringB. the angular momentum of the system is conservedC. the final common angular velocity is `(1)/(2)omega`D. all of the above |
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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)+mR^(2)))omega=(omega)/(3)` |
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| 70. |
A circular table rotates about a vertical axis with a constant angular speed `omega`. A circular pan rests on the turn table (with the centre coinciding with centre of table) and rotates with the table. The bottom of the pan is covered with a uniform small thick layer of ice placed at centre of pan. The ice starts melting. The angular speed of the turn table.A. remains the sameB. decreaseC. increaseD. may increase or decrease dependingon the thickness of ice layer |
| Answer» After melting, ice will distribute in whole pan. So moment of inertia I will increase Hence angular speed `omega` will decrease (as `Iomega=` constant) | |
| 71. |
A wheel is making revolutions about its ais with unifrom angular acceleration. Starting from rest, ilt reaches 100 rev/sec in 4 seconds. Find the angular acceleration. Find the angle rotated during these four seconds. |
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Answer» Correct Answer - A::B::D Given that `omega_0=0` `omega=100 rev/s` `omega=omega_0+alphat` `: alpha=omega/t` `alpha=100/4rev/s^2=25 rev/s^2` `thet=theta_0t+1/2alphat^2` `=1/2x25xx16` `=200 degre `200xx2pi radians `=400 pi radians`. |
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| 72. |
A thin spherical shell lying on a rough horizontal surface is hit by as cue in such a way that the line of action passes through the centre of the shell. As a result the shell starts moving with a linear speed v without any initial angualr velocity. Find the linear speed of the shell after it starts pure rolling on the surface. |
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Answer» Correct Answer - C The shell will move a velocity nearly equal to v due to this motion a frictioN/Al force wil act. In the background direction, for which after some time the shell attains a pure roling. If we consider moment about A, then it wil be zero. Therefore Net angular momentum about A before pure rolling. =net angular moment after pure rolling `=(2/3)MR^2xx(V_0/R)+(MxV_0xxR)=MVR` `rarr (5/3)V_0=V` `rarr V_0=(3V)/5` |
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| 73. |
A rod of length 1 m rotates in the xy plane about the fixed point O in the anticlockwise sense, as shown in figure with velocity `omega=a+bt` where `a=10rads^(-1)` and `b=5rads^(-2)`. The velocity and acceleration of the point A at `t=0` isA. `+10hatims^(-1)` and `+5hatims^(-2)`B. `+10hatjms^(-1)` and `(-100hati+5hatj)ms^(-2)`C. `-10hatjms^(-1)` and `(100hati+5hatj)ms^(-2)`D. `-10hatjms^(-1)` and `-5hatjms^(-1)` |
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Answer» `omega=10+5t` `alpha=(domega)/(dt)=5` At `t=0,omega=10rad//s` and `alpha=5rad//s^(2)impliesr=OA=1m` `v=(romega)hatj=(10hatj)m//s` `a=(ralpha)hatj-(romega^(2))hati` `=(-100hati+5hatj)m//s^(2)` |
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| 74. |
A uniform rod of mass 300 g and length 50 cm rotates at a uniform asngulr speed of 2 rad/s about an axis perpendicular to the rod through an end. Calculte a. the angular momentum of the rod about the axis of rotation b. the speed of thecentre of the rod and c. its kinetic energy. |
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Answer» Correct Answer - A::B::C A uniform of rod mass 300 gm and length =50 cm rotates wilth a uniform angular velocity =2 rad/s abut an axis `_|_` to te rod through an end. a. `L=Iomega` `I at the end =(mL^2)/3=({0.3xx(0.5)^2})/3` `=((.3xx0.25))/3` `=0.075/3=0.025kg-m^2` Therefore angular momentum about that point `=0.025xx2` `-=0.05kg-m^2/s` b. Sped ofthe centre of rod `V=omegar=2xx(50/2)` `=50 cm/s=0.5m/s` c. It kinetic energy c. `1/2Iomega^2=(1/2)xx0.025xx2^2` `=0.25joule |
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| 75. |
One end of a uniform rod of mas m and length l is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a unifrom angular velocity `omega`. Theforce exerted by the clamp on the rod has a horizontal componentA. `momega^2l`B. `zero`C. `mg`D. `1/2momega^2l` |
| Answer» Correct Answer - D | |
| 76. |
Two small balls A and B each of mass m, are joined rigidly to the ends of a light rod of length L figure. The system translates on a frictionless horizontal surface with a velocity `v_0` in a direction perpendicular to the rod. A particle P of mass kept at rest on the surface sticks to the ball A as the ball collides with it . Find a. the linear speeds of the balls A and B after the collision, b. the velocity of the centre of mass C of the system A+B+P and c. the angular speed of the system about C after the collision. |
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Answer» Correct Answer - A::B::C::D Two ball A and B each of mass m are joined rigidly to the ends of a light rod of length L. The system moves with a velocity `V_0` in a direction perpendicular to the rod. A particle P of mass m kept at rest rod. A particle OP of masss m kept at rest on teh surface sticks to the ball A as the ball collides with it. a. The light rod wil exert a force on the bal B only along its length. So collision will not affect its velocity. B has a velocity `=V_0` if we consider the three bodies to be a system Applying L.C.L.M. Therefore `mv_0=2mxxv` `rarr v=v_0/2` Therefore A has velocity `=v_0/2` b. If we consider the three bodies tobe a system net exterN/Asl force =0` Therefore `V_(VCM)=(mxxv_0+2mxx(v_0/2))/(m+2m)` ltbr.gt `=(mv_0+mv_0)/(3m)` `=(2v_0)/3` (along the iniltial velocity as befoe collision) c. The velocity of (A+P) w.r.t `the center of pass ={((2v_0)/3)}-(v_0-2)` `=v_0/6` and the velocity of B w.r.t the centre of mass `=v_0-(2v-0)/3=v_0/3` (only magnitude has been taken) Distance of the (A+P) from centre of mass `=l/3 and for B it is ((2l)/3)` therfore `P_(cm)=L_(cm)xxomega` `rarr 2mxxV_0/6xxl/3+mxxV_0/3xx(2l)/3 `{2m(l/3)^2+((2l)/3)^2m}xxomega` rarr 6m(V_0l)/18=18((6ml)/9)omega` `rarr omega=(v_0/(2l))` |
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| 77. |
A sphere starts roling down can incline of inclination theta. Find the speed of its centre whenit has covered a distance l. |
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Answer» Correct Answer - A A sphere is rolling in an inclined plane with incliN/Ation `theta` therfore according to the principle ` _gt mglsintheta=1/2Iomega^2+1/2mmv^2` `rarr 1/2xx2/5mv^2+1/2mv^2=7/10` `rarr gl sintheta=7/10v^2` `rarr v=sqrt((10/7 glsintheta))` |
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| 78. |
In the shown figure, accelearated pure rolling with takes place, if `a=Ralpha`, find the case if. (a). `agtRalpha` (b). `alphaltRalpha` |
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Answer» (a). If `agtRalpha`, then at any instant `vgtRomega`, so, it is a case of forward slipping. (b). If `altRalpha`, then at any instant `vltRomega`. So, it is a case of backward slipping. |
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| 79. |
A disc is rolling without slipping with linear velocity `v` as shown in figure. With the concept of instantaneous axis of rotation, find velocities of point A, B, C and D. |
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Answer» `v=romega`, where `r` is the distance from A (position of instantaneous axis of rotation) For A, `r=0impliesv_(A)=0` For B and D, `r=sqrt(2)R` `impliesv_(B)=v_(D)=sqrt(2)Romega=sqrt(2)v` for C, `r=2Rimpliesv_(C)=2Romega=2v` Direction of velocity of a general point `P` perpendicular to AP in the sense of rotation. |
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| 80. |
A solid sphere is rolling without slipping as shown in figure. Prove that `(1)/(2)mv^(2)+(1)/(2)l_(C)omega^(2)=(1)/(2)l_(0)omega^(2)` |
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Answer» `I_(C)=(2)/(5)mR^(2),I_(0)=(7)/(5)mR^(2)` and `omega=(v)/(R)` substituting these values in the given equation we get the result. |
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| 81. |
A solid disc is rolling without slipping on a horizontal ground as shown in figure. Its total kinetic energy is 100 J. what is its translational and rotational kinetic energy? |
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Answer» In case of pure rooilling ratio of rational kinetic energy and tranlational kinetic energy is `(1)/(2)` or `(K_(R))/(K_(T))=(1)/(2)impliesthereforeK_(R)=((1)/(1+2))` (total kinetic energy) `=(1)/(3)(100J)=(100)/(3)J` Similarly `K_(T)=((2)/(1+2))` (total kintic energy ) `=(2)/(3)(100J)=(200)/(3)J` |
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| 82. |
A solid disk is rolling without slipping on a level surface at a constant speed of `2.00m//s`. How far can it roll up a `30^(@)` ramp before it stops? (take `g=9.8m//s^(2))` |
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Answer» `mgh=K_(R)+K_(T)=(3)/(4)mv^(2)` Here `h=ssintheta` `thereforegssintheta=(3)/(4)v^(2)` or `s=(3v^(2))/(4gsintheta)` `=(3xx(2.0)^(2))/(4xx9.8xx(1)/(2))=0.612m` |
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| 83. |
Assertion: A ring is rolling without slipping on a rough ground. It strikes elastically with a smooth wall as shown in figure. Ring will stop after some time while travelling in opposite direction. Reason: After impact net angular momentum about an axis passing through bottommost point and perpendicular to plane of paper 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 second figure. Net angular momentum about bottommost axis `L=mvR-I_(C)omega` `=mvR-(mr^(2))((v)/(R))` `=0` |
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| 84. |
A closed cylindrica tube containing some water (not filling the entire tube) lies in a horizontal plane. If the tueb is rotated about a perpendicular bisector, the moment of inertia of water about the axisA. increasesB. decreasesC. remains constantD. increases if the rotatinis clockwise and decreases if it is anticlockwises |
| Answer» Correct Answer - A | |
| 85. |
The moment of inertia of hollow sphee (mass M) of inner radius R and outer radius 2R, having material of uniform density, about a diametric axis isA. `31MR^(2)//70`B. `43MR^(2)//90`C. `19MR^(2)//80`D. None of these |
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Answer» `rho=(M)/({(4)/(3)pi(2R)^(3)}-{(4)/(3)piR^(3)})` `=(3M)/(28piR^(3))` `M_(2R)=rho((4)/(3))(pi)(2R)^(3)` `=((3)/(28))((4)/(3))(8)M=(8)/(7)M` `M_(R)=rho((4)/(3)piR^(3))` `=((3)/(28))((4)/(3))M=(1)/(7)M` Now, `I=I_(2R)-I_(R)` `=(1)/(2)M_(2R)(2R)^(2)-(1)/(2)M_(R)R^(2)` `=(1)/(2)((8)/(7)M)(4R^(2))-(1)/(2)((M)/(7))R^(2)` `=(31)/(14)MR^(2)` |
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| 86. |
A solid sphere, a ring and a disc all having same mass and radius are placed at the top of an incline and released. The friction coefficient between the objects and the incline are same but not sufficient to allow pure rolling. Least time will be taken in reaching the bottom byA. the solid sphereB. the hollow sphereC. the discD. all will take same time |
| Answer» Correct Answer - D | |
| 87. |
A disc and a solid sphere of same mass and radius roll down an inclined plane. The ratio of thhe friction force acting on the disc and sphere isA. `(7)/(6)`B. `(5)/(4)`C. `(3)/(2)`D. depends on angle of inclination |
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Answer» `f=(mgsintheta)/(1+mR^(2)//I)` `d_("disc")=(mgsintheta)/(1+2)=(1)/(3)mgsintheta` `f_("sphere")=(mgsintheta)/(1+5//2)=(2)/(7)mgsintheta` `therefore` the ratio is `(7)/(6)` |
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| 88. |
A sphere cannot roll onA. a smooth horizontal surfaceB. a smooth inclined surfaceC. a rough horizontal surfaceD. a rough inclined surface |
| Answer» Correct Answer - B | |
| 89. |
A thin hollow sphere of mass `m` is completely filled with non viscous liquid of mass `m`. When the sphere roll-on horizontal ground such that centre moves with velocity `v`, kinetic enerrgy of the system is equal toA. `mv^(2)`B. `(4)/(3)mv^(2)`C. `(4)/(5)mv^(2)`D. none of these |
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Answer» Viscous liquid has only translational kinetic energy `thereforeKE=` translational `+` rotaional kinetic energy of hollow sphere `+` translational kinetic energy of liquid `=(1)/(2)mv^(2)+(1)/(2)xx(2)/(3)mR^(2)((v)/(R))^(2)+(1)/(2)mv^(2)` `=(4)/(3)mv^(2)` |
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| 90. |
A uniform sphere of mass 200 g rols withiout slipping on a plane surface so that its centre mioves at a speed of 2.00 cm/s. Find its kinetic energy. |
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Answer» As the sphere rolls without slipping on the plane surface, it sangular speed about the center is `omega=(v_(cm))/r`. The kinetic energy is `K=1/2l_(cm)omega^2+1/2Mv^2_(cm)` `=1/2.2/5mr^2omega^2+1/2Mv^2_(cm)` `=1/5Mv^2_(cm)+1/2Mv^2_(cm)=7/10v_(cm)^2` `7/10(0.200kg)(0.02m/s)62=5.6xx10^-5J`. |
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| 91. |
A solid uniform disc of mass `m` rols without slipping down a fixed inclined plank with an acceleration a. The frictional force on the disc due to surface of the plane isA. `(1)/(4)ma`B. `(3)/(2)ma`C. `ma`D. `(1)/(2)ma` |
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Answer» `a=(gsintheta)/(1+I//mR^(2))=(gsintheta)/(1+(1//2))` `=(2gsintheta)/(3)` `thereforegsintheta=(3)/(2)a` Now, `f=(mgsintheta)/(1+mR^(2)//I)=(m(3//2a))/(1+2)` `=(1)/(2)ma` |
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| 92. |
Moment of inertia of a uniform rod of length `L` and mass `M`, about an axis passing through `L//4` from one end and perpendicular to its length isA. `(7)/(36)ML^(2)`B. `(7)/(48)ML^(2)`C. `(11)/(48)ML^(2)`D. `(ML^(2))/(12)` |
| Answer» `I=(ML^(2))/(12)+M((L)/(4))^(2)=(7)/(48)ML^(2)` | |
| 93. |
A kid of mass M stands at the edge of a platform of radius R which can be freely rotated about its axis. The moment of inerti of the platform is I . The system is at rest when a friend throws as ball of mas mand the kid cathces it. If the velocity of the ball is v horiontally along the tangent to the edge of the platform when it was caught by the kid find the angular speed of the platform after the event. |
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Answer» Correct Answer - B A kid of mass M stands at tehedge of the platform of a raduis R which has a moment of inertia I. S ball thrown to him and with horizontal velocity of the ball -v when we catches it. Therefoe if we take the total bodies as a system `mvR={1+(M+m)R^2}omega` The moment of inerti aofkid adn ball about the axis=`(M+m)R^2` `rarr omega=(mvR)/(1+(M+m)R^2)` |
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| 94. |
A particle of mass `m=3kg` moves along a straight line `4y-3x=2` where `x` and `y` are in metre, with constant velocity `v=5ms^(-1)` the magnitude of angular momentum about the origin isA. `12kgm^(2)s^(-1)`B. `6.0kgm^(2)s^(-1)`C. `4.5mgm^(2)s^(-1)`D. `8.0kgm^(2)s^(-1)` |
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Answer» `y=(3)/(4)x+0.5` slope `=tantheta=(3)/(4)` `thereforetheta=37^(@)` `r_(bot)=0.5sin53^(@)=0.4m` `L=mvr_(bot)=(3)(5)(0.4)` `=6kg-m^(2)//s` |
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