InterviewSolution
Saved Bookmarks
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 fly wheel rotating about a fixed axis has a kinetic energy of `360 J`. When its angular speed is `30 rad s^(-1)`. The moment of inertia of the wheel about the axis of rotation isA. `0.6kgm^(-2)`B. `0.15kgm^(-2)`C. `0.8kgm^(-2)`D. `0.75kgm^(-2)` |
|
Answer» Correct Answer - C Rotational kinetic energy `=1/2Iomega^(2)` `360=1/2Ixx30xx30` or `I=(360xx2)/(30xx30)kgm^(2)=0.8mm^(2)` |
|
| 2. |
A constant external torque `tau` acts for a very brief period `/_ ` on a rotating system having moment of inertia `I` thenA. the angular velocity of the system will change by `tau/_ //I`B. the angular velocity of the system will change by `tau/_ //I`C. if the system was initially at rest, it will acquire rotational kinetic enegy `(tau/_ )^(2)//2I`D. the kinetic energy of the system will change by `(tau/_ )^(2)//(2I)` |
|
Answer» Correct Answer - A::B::C::D Let `L=`angular momentum `tau=(dL)/(dt)` or `dL=taudt` For constant torque `/_L=tau/_ =I/_omega` or `Iomega` (if `omega_(I)=0)` Rotational kinetic energy `=Iomega^(2)//2=(/_L)^(2)//2l` |
|
| 3. |
A disc of mass `m` and radius `R` moves in the `x-y` plane as shown in Fig. The angular momentum of the disc about the origin `O` at the instant shown is A. `-5/2mR^(2)omegahatk`B. `7/3mR^(2)omegahatk`C. `-9/2mR^(2)omegahatk`D. `5/2mR^(2)omegahatk` |
|
Answer» Correct Answer - A `vecL_(O)=I_(CM)vecomega+m(vecrxxvecv)=1/2mR^(2)omegahatk+m(4Rhati+3Rhatk)xx(omegaRhati)` `=1/2mR^(2)omegahatk-3mRomegahatk=-5/2mR^(2)omegahatk` |
|
| 4. |
If `vec(tau)xxvec(L)=0` for a rigid body, where `vec(tau)=` resultant torque & `vec(L)` =angular momentum about a point and both are non-zero. ThenA. `vec(L)=`constantB. `|vec(L)|`=constantC. `|vec(L)|` will increasesD. `|vec(L)|` may increase |
|
Answer» Correct Answer - D `sin theta=0` so `theta=0, pi` In case of `theta=0, vec(L)` will increases |
|
| 5. |
A particle has a linear momentum p and position vector r. the angular momentum of this particle about the origin will not be zero under the conditionsA. p=0B. p is perpendicular to r and does not cross rC. p is anti-parallel to rD. the particle passes through the origin. |
|
Answer» Correct Answer - B `vec(L)=vec(r)xxvec(p)=r p sin theta` If `p=0 rArr L=0` `in` is `0, pi rArr L=0` if `r=0 rArr L=0` |
|
| 6. |
The end B of the rod AB which makes angle `theta` with the floor is being pulled with a constant velocity `v_(v)` as shown. The length of the rod is `l`.A. `3/5v_(0)`B. `4/5v_(0)`C. `5/3v_(0)`D. `5/4v_(0)` |
|
Answer» Correct Answer - B Let `OA=x, OB=y` `x^(2)+y^(2)=l^(2)` and `x=lcostheta` Different `x^(2)+y^(2)=l^(2)` with respect to time. `2x(dx)/(dt)+2y(dy)/(dt)=0` `V_(B)=(dy)/(dt)=x/y (dx)/(dt)=-V_(A)cottheta=4/5v_(0)darr` |
|
| 7. |
End `A` of a rod `AB` is being pulled on the floor with a constant velocity `v_(0)` as shown. Taking the length of the rod as `l`, at an instant when the rod makes an angle `37^@` with the horizontal, calculate the velocity of the `CM` of the rodA. `5/7v_(0)` at `tan^(-1)4/3` below horizontalB. `5/7v_(0)` at `tan^(-1)3/4` below horizontalC. `5/6v_(0)` at `tan^(-1)3/4` below horizontalD. `5/6v_(0)` at `tan^(-1) 4/3` below horizontal |
|
Answer» Correct Answer - D `x_(CM)=x/2` `impliesV_(CM)=1/2(dx)/(dt)0=(v_(0))/2(rarr)` `implies y_(CM)=y/2` `y_(CM)=1/2 (dy)/(dt)=(v_(B))/2=2/3v_(0)darr` `V_(CM)=sqrt((v_(0)^(2))/4+(4v_(0)^(2))/9)=5/6v_(0)` at `tan^(-1)(4/3)` below horizontal |
|
| 8. |
A sphere of mass `M` and radius r shown in the figure on a rough horizontal plane. At some instant it has translational velocity `v_(0)`, and rotational velocity about centre `v_(0)//2r`. The percentage change of translational velocity after the sphere start pure rolling: A. `14.28%`B. `7.14%`C. `21.42%`D. none |
|
Answer» Correct Answer - A Initial angular momentum `L=2/5Mr^(2)((v_(0))/(2r))+Mrv_(0)=6/5Mrv_(0)` …………i The angular mometum about `A` is `=2/5Mr^(v/r)+Mrv=2/5Mr^(2)(v/r)+Mrv`……ii `v=6/7v_(0)`. `:. %` change of translation velocity `=[(v_(0)-6/7v_(0))/(v_(0))]xx100=1/7xx100=14.28%` |
|
| 9. |
A circular platform is mounted on a vertical frictionless axle. Its radius is `r = 2 m` and its moment of inertia`I = 200 kg m^2`. It is initially at rest. A `70 kg` man stands on the edge of the platform and begins to walk along the edge at speed `v_0 = 1 m s^-1` relative to the ground. The angular velocity of the platform is.A. `1.2 rads^(-1)`B. `0.4rads^(-1)`C. `2.0rads^(-1)`D. `0.7rads^(-1)` |
|
Answer» Correct Answer - D Net external torque is zero. Therefore angular mometnumof system will remain conserved i.e, `L_(i)=L_(f)` Initial angular momentum `L_(i)=L_(f)` Initial angular momentum `L_(i)=0` `:.` final angular mometum should be zero. or angular momentum of man `=` angular momentum of platform is opposite direction or `mv_(0)r=Iomega` `omega=(mv_(0)r)/I=((70)(1.0)(2))/200` `omega=0.7rad//s` |
|
| 10. |
In the given figure `F=10N, R=1m`, mass of the body is `2kg` and moment of inertia of the body about an axis passing through `O` and perpendicular to the plane of the body is `4kgm^(2)`. `O` is the centre of mass of the body. If the ground is smooth, what is the total kinetic energy of the body after `2s`?A. `25J`B. `16.67J`C. `50J`D. `37.5J` |
|
Answer» Correct Answer - C When frictional force is absent, the rigid body does not translate `SigmavecF_("external")=0` therefore `veca_(CM)=0` Taking torque about `CM` `Fxx2R-FxxR=I_(CM)alpha` `10xx1=4alpha` `alpha=5/2rad//s^(2)` `omega=alphat=5rad//s` `KE_("total")=1/2I_(CM)omega^(2)=1/2xx4xx5xx5=50J` Taking torque about `IC` (bottom most point) `Fxx4R-Fxx3R=[I_(CM)+M(2R)^(2)]alpha` `=10xx1=[4+2xx(2xx1)^(2)]alpha` `alpha=10/2rad//s^(2)` `omega=alphat=10/12xx2=5/2rad//s` `KE_("total")=1/2[I_CM+M(2R)^(2)]omega^(2)` `=1/2[4+2xx(2xx2)^(2)]xx(5/3)^(2)` `=16.67J` |
|
| 11. |
Calculate the moment of inertia of a. a ring of mass `M` and radius `R` about an axis coinciding with the diameter of the ring. b. as thin disc about an axis coinciding with the diameter. |
|
Answer» Let `X` and `Y` axis be along two perpendicular diameters of the ring. BY symmetry`I_(x)=I_(y)` and by perpendicular axis theorem `I_(z)=I_(x)+I_(y)`. But we know that `I_(z)=MR^(2)` `MR^(2)=I_(x)+I_(y)^(z)=2I_(x)` `implies I_(x)=I_(y)=(MR^(2))/2` Situation of a thin disc (i.e. a circular plate) the moment of inertia about a diameter is `I=1/2(1/2MR^(2))=1/4MR^(2)` |
|
| 12. |
In the given figure `F=10N, R=1m`, mass of the body is `2kg` and moment of inertia of the body about an axis passing through `O` and perpendicular to the plane of the body is `4kgm^(2)`. `O` is the centre of mass of the body. If the ground is sufficiently rough to ensure rolling, what is the kinetic energy of the body now in the given time interval of `2s`?A. `18.75J`B. `25.67J`C. `16.67J`D. none of these |
|
Answer» Correct Answer - C When frictional force is absent, the rigid body does not translate `SigmavecF_("external")=0` therefore `veca_(CM)=0` Taking torque about `CM` `Fxx2R-FxxR=I_(CM)alpha` `10xx1=4alpha` `alpha=5/2rad//s^(2)` `omega=alphat=5rad//s` `KE_("total")=1/2I_(CM)omega^(2)=1/2xx4xx5xx5=50J` Taking torque about `IC` (bottom most point) `Fxx4R-Fxx3R=[I_(CM)+M(2R)^(2)]alpha` `=10xx1=[4+2xx(2xx1)^(2)]alpha` `alpha=10/2rad//s^(2)` `omega=alphat=10/12xx2=5/2rad//s` `KE_("total")=1/2[I_CM+M(2R)^(2)]omega^(2)` `=1/2[4+2xx(2xx2)^(2)]xx(5/3)^(2)` `=16.67J` |
|
| 13. |
From a complete ring of mass `M` and radius `R`, a `30^@`sector is removed. The moment of inertia of the incomplete ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ,A. `9/12MR^(2)`B. `11/12MR^(2)`C. `11.3/12MR^(2)`D. `MR^(2)` |
|
Answer» Correct Answer - B Mass of incomplete ring `=m-M/(2pi)xxpi/6=M-M/12=(11M)/12` moment of inertia of incomplete ring `((11M)/12)R^(2)=11/12MR^(2)` |
|
| 14. |
A uniform rod of mass `m` and length is acted upon by the forces `F_(1)` and `F_(2)` Find that: a. linear-and angular acceleration of the rod. b. value of `x` for which the point `P` does not accelerate. |
|
Answer» a. Equation of motion `F_(1)+F_(2)=ma` or `2F-F=maimpliesa=(3F)/m`…….i torque equation about centre of mass. `2F.l/2-F.x=((ml^(2))/12)alpha` `implies alpha(12F(l-x))/(ml^(2))` ………ii b. If point `P` does not acceleration `veca_(P)=veca_(P,C)+veca_(C)` As `a_(P)=0=-ax+aimpliesax=a` `(12F(l-x))/(ml^(2))=(3F)/m` which given `x=l/2` |
|
| 15. |
A thread is passing through a hole at the centre of frictionless table. At the upper end a block of mass `0.5 kg` is tied and a block of mass `8.0 kg` is tied at the lower end which is freely hanging. The smaller mass is rotate, on the table with a constant angular velocity about the axis passing through the hole so as to balance the heavier mass. If the mass of the hanging block is changed from `8.0 kg` to `1.0 kg`, what is the fractional change in the radius and the angular velocity of the smaller mass so that it balances the hanging mass again? |
|
Answer» For circular motion of a body tied to a string on a horizontal plane `((mv^(2))/r)=T` Here as tension is provided by the hanging mass `M`. i.e., `T=g`, therefore So,`((mv^(2))/r)=Mg` According to the given problem, `((mv_(1)^(2)//r_(1)))/((mv_(2)^(2)//r_(2)))=(M_(1)g)/(M_(2)g)=8/1` ...........i Now as force `T` is central, so angular momentum is also conserved i.e., `mv_(1)r_(1)=mv_(2)r_(2)`..........ii So substituting the value of `v_(1)//v_(2)` from Eq. i Eq. i `[(r_(2))/(r_(1))]xx(r_(2))/(r_(1))=8/1`.............iii i.e., `(r_(2))/(r_(1))=2` so that `(/_)/(r_(1))=(r_(2)-r_(1))/(r_(1))=(r_(2))/(r_(1))-1=2=1` Furthermore as in circular motion `v=romega` so, `(omega_(2))/(omega_(1))=(v_(2))/(v_(1))xx(r_(2))/(r_(1))=[(r_(1))/(r_(2))]^(2)` [as from Eq. ii `(v_(2))/(v_(1))=[(r_(1))/(r_(2))]` `implies(omega_(1))/(omega_(2))=[1/2]^(2)=1/4` [as from eq. iii `(r_(2))/(r_(1))=2`] So, `(/_omega)/(omega)=(omega_(2)-omega_(1))/(omega_(1))=(omega_(2))/(omega_(1))=1=1/4-l=-3/4` |
|
| 16. |
A thin bar of mass `M` and length `L` is free to rotate about a fixed horizontal axis through a point at its end. The bar is brought to a horizontal position and then released. The axis is perpendicular to the rod. The angular velocity when it reaches the lowest point isA. directly proportional to its length and inversely proportional to its massB. independent of mass and inversely proportional to the square root of its lengthC. dependent only upon the acceleration due to gravity and the length of the barD. directly proportional to its length and inversely proportional to the acceleration due to gravity |
|
Answer» Correct Answer - B::C::D Gain in `KE =`Loss in `PE` `implies 1/2Iomega^(2)=mgl/2implies1/2 (ml^(2))/3omega^(2)=mgl/2` `implies omega=sqrt((3g)/l)` |
|
| 17. |
A cubical block of side a is moving with velocity V on a horizontal smooth plane as shown in Figure. It hits a ridge at point O. The angular speed of the block after it hits O is A. `(3v)/(4a)`B. `(3v)/(2a)`C. `sqrt(3/2)a`D. zero |
|
Answer» Correct Answer - A Conserving angular momentum about `O` (just before and just after ) is `mva/2=[(ma^(2))/6 (ma^(2))/2]omegaimplies omega=(3v)/(4a)` |
|
| 18. |
Illustrated is a uniform cubical block of mass `M` and side `a` Mark the correct statement (s) A. The moment of inertia about axis `A`, passing through the centre of mass is `IA=1/6Ma^(2)`B. The moment of inertia about axis `B`, which bisects one of the cube faces is `lB=5/12Ma^(2)`C. The moment of inertia about axis `C`, along one of the cube edge is `IC=2/3Ma^(2)`D. The moment of inertia about axis `D`, whch bhisects one of the horizontal cube face is `7/12` |
|
Answer» Correct Answer - A::B::C Parallel axis therorem, check the distasnce carefully. `I_(D)=I_(B)("symmetric")` `I_(B)=I_(A)+M(a/2)^(2)=5/12Ma^(2)` `I_(C)=I_(A)+M(a/(sqrt(2)))^(2)=2/3Ma^(2)` |
|
| 19. |
The radius of gyration of a body depends uponA. mass of the bodyB. nature of distribution of massC. axis of rotationD. none of these |
|
Answer» Correct Answer - B::C We know that `K=sqrt(I/M),IpropM` hence radius of gyration does not depends upon mass of the body. Further `I` depennds upon distribution of mass of body and also on the axis of rotation. Hence `K` also depends upon these factors. |
|
| 20. |
There are four solid balls with their centres at the four comers of a square of side `a`. the mass of each sphere is `m` and radius is `r`. Find the moment of inertia of the system about (i) one of the sides of the square (ii) one of the diagonals of the square. |
|
Answer» a. Moment of inertia of the arrangement about the diagonal `AC`: The moment of inertia of each of spheres `A` and `C` about their common diameter `AC=(2/5)ma^(2).` The moment of inerta of each of spheres `B` and `D` about an axis passing through their centres and parallel to `AC` is `(2/5)ma^(2)`. The distance between the axis and `AC` is `b/(sqrt(2))`, `b` being side of the square. therefore the moment of inertia of each spheres `B` and `D` about `AC` by the theorem of parallel axis is `2/5 ma^(2)+m(b/sqrt(2))^(2)=2/5 ma^(2)-(mb^(2))/2` Therefore, the moment of inertias of all the four spheres about diagonal `AC` is `I=2(2/5 ma^(2))+2[2/5ma^(2)+(mb^(2))/2]` b. The moment of inertia of each of spheres `A` and `D` about side `AD` is `(2/5) ma^(2)`. The moment of inertia of each of sphere `B` and `C` about `AD` is `(2/5)ma^(2)+mb^(2)`. Therefore the moment of inertia of the whole arrangement about side `AD` is `2[2/5ma^(2)+(2/5ma^(2)+mb^(2))]=(2m)/5[4^(2)+5b^(2)]` |
|
| 21. |
A bullet of mass `m` moving with a velocity of `u` just grazes the top of a solid cylinder of mass `M` and radius `R` resting on a rough horizontal surface as shown and is embedded in the cylinder after impact. Assuming that the cylinder rolls without slipping, find the angular velocity of the cylinder and the final velocity of the bullet. |
|
Answer» Let `v` and `omega` be the velocity of the bullet and angular velocity of the cylinder, respectively. Applying onservation of angular momentum about the point of contact of the cylinder with the floor we get `mu2R=mv2R+(I_(CM)+MR^(2))omega` here, `v=2Romega` and `I_(CM)=1//2MR^(2)` Substituting and solving, we get `v=(8mu)/(8m+3M)` and `omega=(4mu)/((8m+3M)R)` |
|
| 22. |
A cylinder of mass M and radius R is resting on a horizontal platform (which is parallel to the x-y plane) with its axis fixed along the y-axis and free to rotate about its axis. The platform is given a motion in the x-direction given by `x =A cos (omega t).` There is no slipping between the cylinder and platform. The maximum torque acting on the cylinder during its motion is .................. |
|
Answer» `x=A cosomegat` `implies(dx)/(dt)=-Aomegacosomegat` `implies (d^(2)x)/(dt)=-Aomegacosomegat` `:. |"max acceleration"|=Aomega^(2)` `:. alpha_("max")=(Aomega^(2))/R` Max torque`= Ialpha_("max")` `=1/2MR^(2)xx(Aomega^(2))/R=1/2MRAomega^(2)` |
|
| 23. |
A uniform solid cylinder of mass `2 kg` and radius `0.2 m` is released from rest at the top of a semicircular track of radius `0.7 m` cut in a block of mass `M = 3 kg` as shown in Fig. The block is resting on a smooth horizontal surface and the cylinder rolls down without slipping. Based on the above information, answer the following questions: The speed of the cylinder when it reaches the bottom of the track isA. `2/(sqrt(11))m//s`B. `8/(sqrt(11))m//s`C. `8/(sqrt(11))m//s`D. `6/(sqrt(11))m//s` |
|
Answer» Correct Answer - D `v` and `v_(0)` are the velocities w.r.t ground `omega=(v+v_(0))/r` |
|
| 24. |
A disc of mass `m` and radius `R` is free to rotate in a horizontal plane about a vertical smooth fixed axis passing through its centre. There is a smooth groove along the diameter of the disc and two small balls of mass `m//2` each are placed in it on either side of the centre of the disc as shown in Fig. The disc is given an initial angular velocity `omega_(0)` and released. The angular speed of the disc when the balls reach the end of the disc isA. `(omega_(0))/2`B. `(omega_(0))/3`C. `(2omega_(0))/3`D. `(omega_(0))/4` |
|
Answer» Correct Answer - B Let the angular speed of the disc when the balls reach the end be `omega`. From conservation of angular momentum, `1/2mR^(2)omega_(0)=1/2mR^(2)omega+m/2R^(2)omega+m/2R^(2)omega` or `omega=(omega_(0))/3` |
|
| 25. |
In the pulley system shown, if radii of the bigger and smaller pulley are `2 m` and `1 m` respectively and the acceleration of block `A` is`5 m//s^2` in the downward direction, then the acceleration of block `B` will be : .A. `0 m//s^(2)`B. `5 m//s^(2)`C. `10 m//S^(2)`D. `5//2 m//s^(2)` |
|
Answer» Correct Answer - D Given `a_(A)=2alpha=5 m//s^(2)` `rArr alpha=5//2 rad//s^(2)` `rArr a_(B)=1.(alpha)=5//2 m//s^(2)` |
|
| 26. |
A sphere is rolling down an inclined plane without slipping. The ratio of rotational kinetic energy to total kinetic energy isA. `5/7`B. `2/54`C. `2/7`D. none of the above |
|
Answer» Correct Answer - C `(KE_(rot))/(KE_(rot))=(1/2Iomega^(2))/(1/2mv^(2)+1/2Iomega^(2))=(2/5mv^(2))/(7/5mv^(2))=2/7` |
|
| 27. |
A string is wrapped several times on a cylinder of mass `M` and radius `R`. the cylinder is pivoted about its adxis of block symmetry. A block of mass `m` tied to the string rest on a support positioned so that the string has no slack. The block is carefully lifted vertically a distance `h`, and the support is removed as shown figure. a. just before the string becomes taut evalute the angular velocity `omega_(0)` of the cylinder ,the speed `v_(0)` of the falling body, `m` and the kinetic energy `K_(0)` of the system. b. Evaluate the corresponding quanitities `omega_(1), v_(1)` and `K_(1)` for the instant just after the string becomes taut. c. Why is `K_(1)` less than `K_(0)`? Where does the energy go? d. If `M=m`, what fraction of the kinetic energy is lost when the string becomes taut? |
|
Answer» a. Just before the stirng becomes taut, the block falls freely, so `v_(0)=sqrt(2gh),` There is no tension in the string so nothing causes the cylider to spin so `omega_(0)=0`. The kinetic energy of the system is `K_(0)=1/2mv_(0)^(2)+1/2lomega_(0)^(2)=mgh`. b. When the string experiences a jerk, the large impulse developed of very short duration so that the contribution of weight `mg` can be neglected during this time interval. The angular momentum of the system is conserved, as the tension is internal force for the system. Thus we have `vecL_(i)=vecL_(f)` `mv_(1)R+1/2MR^(2)omega_(1)=mv_(0)R=msqrt(2gh)R` The string is inextensible so `v_(1)=Romega_(1).` On solving for `omega_(1)`, we get ltbr `omega_(1)=sqrt((2gh))/(R[1+(M/2m)]),v_(1)=Romega_(1)=(sqrt(2gh)/([1+(M/2m)])` The final kinetic energy `K_(1)` is given by `K_(1)=1/2mv_(1)^(2)+omega_(1)^(2)` `=1/2mv_(1)^(2)+1/2(1/2MR^(2))((v_(1)^(2))/R^(2))=1/2(m+M/2)v_(1)^(2)` `=1/2[(mv_(0)^(2))/(1+(M/2m))]=(K_(0))/(1+(M/2m))` c. The situation in this case is analogous to the energy loss is complete inelastic two body collisions. The lost kinetic energy is converted to heat energy or elastic potential energy of the string or in the two objects. d.For `M=m, K_(1)=(2K_(0))/3,` so the fraction lost is `((K_(0)-K_(1))/K_(0))=1/3` |
|
| 28. |
Four forces of the same magnitude act on a square as shown in figure. The square can rotate about point `O`, mid point of one of the edges. The force which can produce greatest torque is A. `F_(1)`B. `F_(2)`C. `F3`D. `F4` |
|
Answer» Correct Answer - C Perpendicular distance of `F_(3)` is greatest from `O` hence torque of `F_(3)` is greatest. |
|
| 29. |
Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J = MV is imparted to the body at one of its ends what would be it angular velocity? A. `V/L`B. `(2V)/L`C. `V/(3L)`D. `V/(4L)` |
|
Answer» Correct Answer - A Change an angular momentum of the system` =`angular impulse given to the system `("angular momentum")_f-("angular momenum")_I` `=MVxxL/2` …..i Let the system starts rotating with the angular velocity `omega`. Momentun of inertia of the system about its axis of rotation. (centre of mass of system) `=M(L/2)^(2)+M(L/2)^(2)=(2ML^(2))/4` `=(ML^(2))/2` (from i) `Iomega-0=MVxL/2` `impliesomega=(MV)/IxxL/2=(MV)/((ML^(2))//2)xxL/2=V/L` |
|
| 30. |
A small bead of mass `m` moving with velocity `v` gets threaded on a stationary semicircular ring of mass `m` and radius `R` kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system? A. `v/R`B. `(2v)/R`C. `v/(2R)`D. `(3v)/R` |
|
Answer» Correct Answer - C Using conservation of angular momentum about `O` we get `mvR=(mR^(2)+mR^(2))omega, omega=v/(2R)` |
|
| 31. |
As shown in the figure, a disc of mass m is rolling without slipping with angular velocity `omega`. When it crosses point B disc will be in A. translational motion onlyB. pure rolling motionC. rotational motion onlyD. none of these |
|
Answer» Correct Answer - B In BC section no force or torque is applied on disc so pure rolling motion will continue. |
|
| 32. |
A disc having radius `R` is rolling without slipping on a horizontal (`x-z`) plane. Centre of the disc has a velocity `v` and acceleration `a` as shown. Speed of point `P` having coordinates `(x,y)` isA. `(vsqrt(x^(2)+y^(2)))/(R`B. `(vsqrt(x^(2)+(y+R)^(2)))/R`C. `(vsqrt(v^(2)+(y-R)^(2)))/R`D. none of these |
|
Answer» Correct Answer - B `omega=v/R`, Distance of `P` from origin `r=sqrt(x^(2)+y^(2))` origin is instantaneous centre of rotatio. So, `v_(P)=omegar=(vsqrt(x^(2)+y^(2)))/R` |
|
| 33. |
A uniform disc of mass `1 kg` and radius `20 cm` is rolling purely on a flat horizontal surface. Its centre `C` is moving with acceleration `a = 20 ms^(-2)` and velocity `v =4 m//s` at a certain instant. At this instant. points `A` and `B` are located on the disc as shown in the diagram, with `AC = BC = 10 cm`. What is the acceleration magnitude of point `A`?A. `10sqrt(17)ms^(-2)`B. `10sqrt(37)ms^(-2)`C. `10sqrt(35)ms^(-2)`D. `60sqrt(17)ms^(-2)` |
|
Answer» Correct Answer - B `KE=1/2mv^(2)=1/2Iomega^(2)=3/4mv^(2)=12J` `veca_(A//C)=veca_(A)-veca_(C)` ` implies veca_(A)=veca_(A//C)+veca_(C)=omega^(2)rhati+alpharhatj+ahati=60hati+10hatj` `veca_(A)=sqrt(3600+10)=10sqrt(37)m//s^(2)` `veca_(B)=omega^(2)rhatj+alphar(-hati)+ahati=10hati+40hatj` `a_(B)=10sqrt(17)m//s^(2)` |
|
| 34. |
A uniform disc of mass `1 kg` and radius `20 cm` is rolling purely on a flat horizontal surface. Its centre `C` is moving with acceleration `a = 20 ms^(-2)` and velocity `v =4 m//s` at a certain instant. At this instant. points `A` and `B` are located on the disc as shown in the diagram, with `AC = BC = 10 cm`. What is the kinetic energy of the disc?A. `12J`B. `8J`C. `20J`D. `10J` |
|
Answer» Correct Answer - A `KE=1/2mv^(2)=1/2Iomega^(2)=3/4mv^(2)=12J` `veca_(A//C)=veca_(A)-veca_(C)` ` implies veca_(A)=veca_(A//C)+veca_(C)=omega^(2)rhati+alpharhatj+ahati=60hati+10hatj` `veca_(A)=sqrt(3600+10)=10sqrt(37)m//s^(2)` `veca_(B)=omega^(2)rhatj+alphar(-hati)+ahati=10hati+40hatj` `a_(B)=10sqrt(17)m//s^(2)` |
|
| 35. |
The moment of inertia of a solid sphere about an axis passing through the centre radius is `1/2MR^(2)` , then its radius of gyration about a parallel axis at a distance `2R` from first axis isA. `5R`B. `sqrt(22/5)R`C. `5/2R`D. `sqrt(12/5)R` |
|
Answer» Correct Answer - b According to the theroem of parallel axis, `I=I_(CG)+M(2R)^(2)` where `I=I_(CG)=MI` about an axis through the centre of gravity. `I=2/5 MR^(2) +4MR^(2)=22/5 MR^(2)` or `MK^(2)=22/5MR^(2)impliesK=sqrt(22/5)R` |
|
| 36. |
The moment of inertia of a solid sphere about an axis passing through the centre radius is `1/2MR^(2)` , then its radius of gyration about a parallel axis t a distance `2R` from first axis isA. `5R`B. `sqrt(22/5)R`C. `5/2R`D. `sqrt(12/5)R` |
|
Answer» Correct Answer - B According to the theroem of parallel axis, `I=I_(CG)+M(2R)^(2)` where `I=I_(CG)=MI` about an axis through the centre of gravity. `I=2/5 MR^(2) +4MR^(2)=22/5 MR^(2)` or `MK^(2)=22/5MR^(2)impliesK=sqrt(22/5)R` |
|
| 37. |
Find the radius of gyration of a hollow uniform sphere of radius `R` about its tangent. |
|
Answer» Moment of inertia of hollow sphere about its tangent `I=2/3MR^(2)=5/3MR^(2)` Let radius of gyration of `K`. Hence `MK^(2)=5/3MR^(2)implies sqrt(5/3)R` |
|
| 38. |
Find the moment of inertia of a solid sphere of mass `M` and radias `R` about an axis XX shown in figure. Also find radius of gyration about the given axis. |
|
Answer» `I=2/5mR^(2)+mR^(2)=7/5mR^(2)=mK^(2)` `implies sqrt(7/5)R` |
|
| 39. |
A copper ball of mass `m = 1 kg` with a radius of `r = 10 cm` rotates with angular velocity `omega = 2 rad//s` about an axis passing through its centre. The work should be performed to increase the angular velocity of rotation of the ball two fold is. |
|
Answer» `KE=1/2(2/5)mv^(2)implies KE=2/10 mv^(2)` If `omega` is to be doubled `v` is also doubled. Therefore `/_KE-KE_(2)-KE_(1)=2/10m[v_(2)^(2)-v_(1)^(2)]` `implies /_KE=2/10m[r^(2)omega_(2)^(2)-r^(2)omega_(1)^(2)]` `=2/10xx1[(1/10)^(2)][4^(2)-2^(2)]=2.4xx10^-(2)J` |
|
| 40. |
Calculate the moment of inertia of a rectangular frame formed by uniform rods having mass `m` each as shown in about an axis passing through its centre and perpendicular to the plane of frame. Also find moment of inertia about an axis passing through `PQ` ? |
|
Answer» a. For calculation of the moment of inertia about centre, use parallel axis theorem `I=2[(Ml^(2))/12+m(b/2)^(2)]+2[(mb^(2))/12+m(l/2)^(2)]` b. Hence, we can also use parallel axis theorem `I_(PQ)=2((mb)/3)^(2)+mb^(2)=5/3mb^(2)` |
|
| 41. |
Four identical rods, each of mass `m` and length `l`, make a square frame in the `xy` plane as shown in Fig. a. Calculate its moment of inertia about the `x`-and y-axes. b. Also, calculate its moment of inertia about the `z`-axis. |
|
Answer» a. Due to symmetry the moment of inertia of the square frame about the `x`- and `y`-axes are equal i.e., `I_(x)=I_(y)` Now `I_(x)=0+2[(ml^(2))/3]+ml^(2)=ml^(2)` `I_(y)=I_(x)=5/3ml^(2)` b. Due to perpendicular axis theorem `I_(z)=I_(x)+I_(y)=5/3ml^(2)+5/3ml^(2)=10/3ml^(2)` Note that the `z`-axis does not pass through its centre of mass. |
|
| 42. |
A man of mass `m` stands on a horizontal platform in the shape of a disc of mass `m` and radius `R`, pivoted on a vertical axis thorugh its centre about which it can freely rotate. The man starts to move aroung the centre of the disc in a circle of radius `r` with a velocity `v` relative to the disc. Calculate the angular velocity of the disc. |
|
Answer» Since there is no torque acting about the axis of rotation of the disc, so the angular momentum of the system (disc`+`man) remains constant. Initially it is zero. Suppose `omega` is the angular velocity of the disc (take positive in the sense of motion of the man). The velocity of the man with respect to the ground observer will be: `vecv_("man, disc")=vecv_("man")-vecv("disc")` `impliesvecv_("man")=vecv("man,disc")-vecv_("disc")` or `vecv_("man")=v-omegar` Thus, angular momentum of the man `=m(v-omegar)r` And angular momentum of the disc is `I_("disc") omega=(MR^(2))/2omega` By conservation of angular momentum, we have `m(v-omegar)=(MR^(2))/2omega` After solving, we get `omega=(mvr)/((mvr^(2)+(MR^(2))/2)` |
|
| 43. |
A uniform disc of mass `m` is fitted (pivoted smoothly) with a rod of mass `m//2`. If the bottom of the rod os pulled with a velocity `v`, it moves without changing its orientation and the disc rolls without sliding. The kinetic energy of the system (rod + disc) is. . |
|
Answer» Rolling can be considered as pure rotation about point of contact `K_("rolling")=1/2I_(P)Iomega^(2)` `=1/2((mR^(2))/2+mR^(2))omega^(2)=3/4mR^(2)omega^(2)`……..i The rod translates with the velocity `v` hence velocity of centre of disc will also be `v` `V=omegaR` ............ii Form i and ii `K_("rolling")=3/4mv^(2)` Kinetic energy of the rod `K_("rod")=1/2(m/2)v^(2)=(mv^(2))/4`.........iii `K_("total")=k_("rolling")+k_("rod")=mv^(2)` |
|
| 44. |
Two rings of same radius and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass the ring `= m`, radius `= r`)A. `1/2mr^(2)`B. `mr^(2)`C. `3/2mr^(2)`D. `2mr^(2)` |
|
Answer» Correct Answer - c Because the plane of two ringns are mutually perpendicular and centres are coincident, hence `n` axis, which is passing through the centre of one of the rings and to its plane, will be along the diameter of other ring. Hence, moment of inertia of the system `=I_(CM)+I_("diameter")=mr^(2)+(mr^(2))/2` `=3/2mr^(2)` |
|
| 45. |
When a bicycle is in motio, the force of friction exerted by the ground on the two wheels is such that it actsA. in the backward direction on the front wheel and in the forward direction on the rear wheelB. in the forward direction on the front wheel and in the backward direction on the rear wheelC. in the backward direction on both the front and the rear wheelsD. in the forward direction on both the front and the rear wheels |
|
Answer» Correct Answer - C When the cycle is not pedaled but it is in motion (due to previous effort) the wheels move in the direction such that the centre of mass of the wheels moves forward. Rolling friction will act in the opposite direction to the relative motion of the centre of mass of the body with respect to the ground. Therefore the rolling friction will act in the backward directioin in both the wheels. The sliding friction will act in the forward direction of the rear wheel during pedaling. |
|
| 46. |
A bicycle has pedal rods of length `16 cm` connected to sprocketed disc of radius `10 cm`. The bicycle wheels are `70 cm` in diameter and the chain runs over a gear of radius `4cm`. The speed of the cycle is constant and the cyclist applies `100 N` for, that is always perpendicular to the pedal rod, as shown in figure. Assume tension in the lower part of chain is negligible. The cyclist is peddling at a constant rate of two revolutions per second. Assume that the force applied by other foot is zero when one foot is exerting `100 N` force. Neglect friction within cycle parts and the rolling friction. The power delivered by the cyclist is equal toA. `280W`B. `100W`C. `64piW`D. `32W` |
|
Answer» Correct Answer - C Power delivered `=vecF.vecv` where `vecv` is velocity of the point of application of the force. `v=16cmxx2pi.2(=Romega)` `=0.64pims^(-1)` `P=100xx0.64pi=64piW` |
|
| 47. |
A bicycle has pedal rods of length `16 cm` connected to sprocketed disc of radius `10 cm`. The bicycle wheels are `70 cm` in diameter and the chain runs over a gear of radius `4cm`. The speed of the cycle is constant and the cyclist applies `100 N` for, that is always perpendicular to the pedal rod, as shown in figure. Assume tension in the lower part of chain is negligible. The cyclist is peddling at a constant rate of two revolutions per second. Assume that the force applied by other foot is zero when one foot is exerting `100 N` force. Neglect friction within cycle parts and the rolling friction. The speed of the bicycle isA. `6.4pims^(-1)`B. `3.5pims^(-1)`C. `2.8pims^(-1)`D. `6.5pims^(-1)` |
|
Answer» Correct Answer - B `RN=rnimpliesn=(10cmxx2)/(4cm)=5cy//s` so rear wheel rotates `5 "cycles"//s` Hence `V=35/100xx2pixx5=3.5pims^(-1)` |
|
| 48. |
Two identical particles `B` and `C` each of mass `50 g` are connected by a light rod of length `30 CM`. Another particle `A` of same mass moving With a speed `u = 60 CM//s` strikes `B`, in a direction perpendicular to `AB`, and sticks to it. The whole process takes place on a smooth horizontal plane. Find the angular velocity `omega` of the system about its centre of mass, immediately after the impact. |
|
Answer» Clearly, the centreof mass of the system `(A+B+C)` is at a distance of `10 cm` from along the rod Initial angualr momentum `L_(i)=m_(A) v_(A)r_(A)` `=50xx10^(-3)xx60xx10^(-2)xx10xx10^(-2)` `=3xx10^(-3)` S.I. unit Moment of inertia about new centre of mass `I=(2xx50xx10^(-3))x(10xx10^(-2))^(2)+(50+10^(-3))xx(20xx10^(-2))^(2)` `=3xx10^(-3)` S.I. unit From `L_(i)=L_(f)impliesL_(i)=I.omega` `implies omega=L/i=1rad//s` |
|
| 49. |
A uniform rod of mass M and length `a` lies on a smooth horizontal plane. A particle of mass m moving at a speed v perpendicular to the length of the rod strikes it at a distance `a/4` from the centre and stops after the collision. Find (a). the velocity of the centre of the rod and (b). the angular velocity of the rod abut its centre just after the collision. |
|
Answer» a. Conservation linear momentum `mv=MVimplies=(mv)/M` b. coserving angular momentum of rod `A` about centre of mas, `mva/4=Iomega=(Ma^(2))/12 omegaimplies omega=(3mv)/(Ma)` |
|
| 50. |
A skater rotating about a vertical axis pulls her arms inward. Ignoring all frictional effects, which of the following statements are true? Denote the magnitude of her angular velocity by `omega`, the magnitude of her angular momentum by `L`, and her kinetic energy by `E_(k)`.A. `L` is constant, `E_(k)` increasesB. `L` is constant, `omega` increasesC. `L` and `E_(k)` are both constantD. `L` and `omega` are both constant |
|
Answer» Correct Answer - A::B `L=Iomega, E_(k)=1/2Iomega^(2)=1/2L^(2)/I` `L=` constant `Idarr, E_(k) uarr, omegauarr` |
|