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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.
| 201. |
Radius of gyration of a ring about a transverse axis passing through its centre isA. `0.5xx` diameter of ringB. diameter of ringC. `2xx` diameter of ringD. `"(diameter of ring)"^(2)` |
| Answer» Correct Answer - A | |
| 202. |
The term moment of momentum is calledA. angular momentumB. torqueC. forceD. couple |
| Answer» Correct Answer - A | |
| 203. |
Angular momentum is vector product ofA. radius vector and linear momentumB. linear momentum and angular velocityC. momentum of inertia and angular accelerationD. linear velocity and radius vector |
| Answer» Correct Answer - A | |
| 204. |
The centre of mass of a symmetrical and uniform distribution of mass of a rigid body isA. at the centre of the surfaceB. outside the bodyC. inside the bodyD. at the geometric centre of the body |
| Answer» Correct Answer - D | |
| 205. |
A ring and a disc have same mass and same radius. Then ratio of moment of inertia of ring to the moment of inertia of disc isA. 4B. 2C. `0.5`D. 1 |
| Answer» Correct Answer - B | |
| 206. |
Moment of inertia in rotatory motion is comparable to the quantity in translatory motionA. momentumB. massC. weightD. velocity |
| Answer» Correct Answer - B | |
| 207. |
A unit mass at position vector `vecr=(3hati+hatj)` is moving with velocity `vecv=(5hati-6hatj)`. What is the angular momentum of the body about the origin?A. `2` units along the `z`-axisB. `38` units along the `x`-axisC. `38` units along the `y`-axisD. `38` units along the `z`-axis |
| Answer» `vecl=vecrxxvecp=` `|[hati,hatj,hatk],[3,4,0],[5,-6,0]|` `=-38hatk` | |
| 208. |
If a body starts rotating from rest because of a torque of 2 Nm, then its kinetic energy after 20 revolutions will beA. `60 pi J`B. `80 pi J`C. `70 pi J`D. `40 pi J` |
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Answer» Correct Answer - B Change in K.E. `=(1)/(2)I(omega_(2)^(2)-omega_(1)^(2))` `=(1)/(2)I(omega_(2)+omega_(1))((omega_(2)-omega_(1)))/(t)t` `=(1)/(2)I alpha (omega_(2)+omega_(1))t = tau theta` In one rev …….. `2 pi` rad `therefore` 20 rev …… ? `theta = 20xx2pi = 40 pi` |
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| 209. |
A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved?A. centre of the circleB. on the circumference of the circle.C. inside the circleD. outside the circle. |
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Answer» Correct Answer - A (a) The net force acting on a particle undergoing uniform circular motion is centripetal force which always passes through the centre of the circle. The torque due to this force about the centre si zero, therfore, `vecL` is conserved about O. |
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| 210. |
A body is rolling down an inclined plane. If `K.E.` of rotation is `40%` of `K.E.` in translatory state, then the body is aA. ringB. cylinderC. hollow ballD. solid ball |
| Answer» `(K_(R))/(K_(l))=((1)/(2)mv^(2)(k^(2)//R^(2)))/((1)/(2)mv^(2))=(k^(2))/(R^(2))=(40)/(100)=(2)/(5)`, solid sphere, i.e. solid ball | |
| 211. |
Rotational motion can beA. one or two dimensionalB. one or three dimensionalC. two or three dimensionalD. one dimensional |
| Answer» Correct Answer - C | |
| 212. |
Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most its mass concentrated near the axis. Which statement(s) is (are) correct?A. Both cylinders `P` and `Q` reach the ground at the same time.B. Cylinder `P` has larger linear acceleration than cylinder `Q`.C. Both cylinders `P` and `Q` reach the ground with the same translational kinetic energy.D. Cylinder `Q` reaches the ground with larger angular speed. |
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Answer» `P:` Hollow cylinder `(k^(2)//R^(2)=1)` `Q:` Solid cylinder `(k^(2)//R^(2)=1//2)` `v_(P)=sqrt((2gh)/(1+1))=sqrt(gh)` `v_(Q)=sqrt((2gh)/(1+1//2))= sqrt((4gh)/(3))` `v_(Q)gtv_(P)` `omega_(Q)gt omega_(P)` |
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| 213. |
A body is set into rotational motion due toA. a force acting at a point on the bodyB. equal and opposite forces acting at two different points on the same bodyC. equal forces acting at two different points on the bodyD. a force acting at the centre of the body |
| Answer» Correct Answer - B | |
| 214. |
The radius of a wheel is R and its radius of gyration about its axis passing through its center and perpendicualr to its plane is K. If the wheel is roling without slipping. Then the ratio of tis rotational kinetic energy to its translational kinetic energy isA. `(k^(2))/(R^(2))`B. `(R^(2))/(k^(2))`C. `(k^(2))/(R^(2)+k^(2))`D. `(R^(2))/(R^(2)+k^(2))` |
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Answer» Correct Answer - A `("K.E. rot")/("K.E. roll")=(1)/(2)MK^(2)(v^(2))/(R^(2))=(1)/(2)Mv^(2)(1+(K^(2))/(R^(2)))` |
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| 215. |
The rotational motion of a body can be produced due to applyingA. torqueB. momentumC. inertiaD. force |
| Answer» Correct Answer - A | |
| 216. |
The radius of gyration of a body depends uponA. mass of the bodyB. distribution of mass of the bodyC. axis of rotation and distribution of mass of the bodyD. all of the above |
| Answer» Correct Answer - C | |
| 217. |
If a solid sphere of mass 1 kg rolls with linear speed of 1m/s, then the rolling kinetic energy will beA. 0.7 JB. 1.4 JC. 2.1 JD. 2.8 J |
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Answer» Correct Answer - A `KE_("roll")=(1)/(2)mv^(2)[1+(K^(2))/(R^(2))]` `=(1)/(2)xx1xx1xx[1+(2)/(5)]=(1)/(2)xx(7)/(5)` = 0.7 J |
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| 218. |
In the arrangement shown, the double pulley has a mass M and the two mass less threads have been tightly wound on the inner (radius `= r`) and outer circumference (radius `R = 2r`). The block shown has a mass 4 M. The moment of inertia of the double pulley system about a horizontal axis passing through its centre and perpendicular to the plane of the figure is `I = (Mr^(2))/(2)`. (a) Find the acceleration of the center of the pulley after the system is released. (b) Two seconds after the start of the motion the string holding the block breaks. How long after this the pulley will stop ascending? |
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Answer» Correct Answer - (a) `(6g)/(11)` (b) `(18)/(11)s` |
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| 219. |
Three identical metal balls each of radius `r` are placed touching each other on a horizontal surface such that an equilateral triangle is formed, when the center of three balls are joined. The center of mass of system is located at theA. horizontal surfaceB. centre of one of the ballsC. line joining the centres of any two ballsD. point of intersection of the medians |
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Answer» Correct Answer - d |
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| 220. |
A particle of mass `M` is revolving along a circule of radius `R` and nother particle of mass `m` is recolving in a circle of radius `r`. If time periods of both particles are same, then the ratio of their angular velocities isA. `1`B. `R/r`C. `r/R`D. `sqrt(R/r)` |
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Answer» Correct Answer - a |
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| 221. |
A fan is making `600` revolutions per minute. If after some time it makes `1200` revolutions per minute, then the increase in its angular velocity isA. `10 pi rad//s`B. `20 pi rad//s`C. `40 pi rad//s`D. `60 pi rad//s` |
| Answer» `Delta omega=600 rev//min =(2pixx600)/(60)=20 pi rad//s` | |
| 222. |
A uniform rod AB has mass M and length L. It is in equilibrium supported in vertical plane by three identical springs as shown in figure. The springs are connected at A, C and D such that `AC = CD = (L)/(3)`. Assume that the springs are very stiff and the angle a made by the rod with the horizontal in equilibrium position is very small. (All springs are nearly vertical). Calculate the tension in the three springs. |
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Answer» Correct Answer - `T_(1) = (Mg)/(12); T_(2) = (Mg)/(3); T_(3) = (7 Mg)/(12)` |
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| 223. |
A 100 g weight is tied at the end of a string and whirled around in horizontal circle of radius 15 cm at the rate of 3 revolutions per second. What is the tension in the string? |
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Answer» Correct Answer - 5.334 N |
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| 224. |
A gramophone disc rotates at 60 rpm. A coin of mass 18 g is placed at a distance of 8 cm from the centre. Calculate the centrifugal force on the coin. Take `pi^(2) = 9.87`. |
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Answer» Correct Answer - 4105.92 dyne |
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| 225. |
The moment of inertia of a uniform circular ring, having a mass M and a radius R, about an axis tangential to the ring and perpendicular to its plane, isA. `MR^(2)//2`B. `3//2 MR^(2)`C. `MR^(2)`D. `2 MR^(2)` |
| Answer» Correct Answer - B | |
| 226. |
Moment of inertia of a disc about its own axis is I. Its moment of inertia about a tangential axis in its plane isA. `(mR^(2))/(4)`B. `(3MR^(2))/(2)`C. `(5)/(4)MR^(2)`D. `(7MR^(2))/(4)` |
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Answer» Correct Answer - C `I_(0)=I_(c )+m h^(2)` `= (MR^(2))/(2)+MR^(2)=(3)/(2) MR^(2)` |
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| 227. |
A trains has to negotiate a curve of radius `400 m`. By how much should the outer rail be raised with respect to the inner rail for a for a speed of `48 km h^(-1)` ? The distance between the rails is 1m. |
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Answer» Correct Answer - `0.0454 m` |
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| 228. |
Radius of gyration of a body about an axis is 12 cm. Radius of gyration of the same body about a parallel axis passing through its centre of gravity is 13 cm. T?hen perpendicular distance between the two axes isA. 5 cmB. 1 cmC. 15 cmD. 10 cm |
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Answer» Correct Answer - A `K_(0)^(2)=K_(c )^(2)+h^(2)` `therefore h^(2)= K_(0)^(2)-K_(c )^(2)` `= 13^(2)-12^(2)=169-144` `h^(2)=25` `h=5 cm`. |
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| 229. |
The radius of gyration of a disc of mass 100 g and radius 5 cm about an axis passing through its centre of gravity and perpendicular to the plane isA. 0.5 cmB. 2.5 cmC. 3.54 cmD. 6.54 cm |
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Answer» Correct Answer - C `K = (R )/(sqrt(2))= R xx 0.707 = 3.535 cm`. |
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| 230. |
Radius of gyration of disc rotating about an axis perpendicular to its plane passing through through its centre is (If R is the radius of disc)A. `( R)/(2)`B. `( R)/(sqrt2)`C. `( R)/(sqrt3)`D. `( R)/(3)` |
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Answer» Correct Answer - B `(MR^(2))/(2)=MK^(2) therefore K^(2)=(R^(2))/(2) therefore K = (R )/(sqrt(2))`. |
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| 231. |
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and a circular ring of the same radius about a tengential axis in the plane of the ring isA. `2 : 3`B. `2 : 1`C. `sqrt(5) : sqrt(6)`D. `1 : sqrt(2)` |
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Answer» Correct Answer - c |
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| 232. |
In the above probllem, the normal force between the ball and the shell in position B is (m=mass of ball)A. `(12)/(7)mg`B. `(7)/(9)mg`C. `(17)/(7)mg`D. `(10)/(7)mg` |
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Answer» Correct Answer - C |
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| 233. |
Two particles of mass 1 kg and 3 kg move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is 2m//s, their centre of mass has a velocity of 0.5 m/s. When the relative velocity of approach becomes 3 m/s. When the relative velocity of approach becomes 3m/s, the velocity of the centre of mass is 0.75 m/s. |
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Answer» Since no external force is acting on the two particle system `:. a_(cm) = 0` `rArr V_(cm) = constant.` |
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| 234. |
The mass of a thin circular plate is `M` and its radius is `R`. About an axis in the plane of plate at a perpendicular distance `R//2` from centre of plate, its moment of inertia isA. `MR^(2)//2`B. `3//2 MR^(2)`C. `MR^(2)`D. `2 MR^(2)` |
| Answer» Correct Answer - D | |
| 235. |
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and a circular ring of the same radius about a tengential axis in the plane of the ring isA. `2:3`B. `2:1`C. `sqrt5:sqrt6`D. `1:sqrt2` |
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Answer» Disc: `I_(1)=(1)/(4)mr^(2)+mr^(2)=(5)/(4)mr^(2)=mk_(1)^(2)` Ring: `I_(2)=(1)/(2)mr^(2)+mr^(2)=(3)/(2)mr^(2)=mk_(2)^(2)` `(k_(1)^(2))/(k_(2)^(2))=(5//4)/(3//2)=(5)/(6)implies (k_(1))/(k_(2))=(sqrt5)/(sqrt6)` |
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| 236. |
If momentum of an object is increased by 10%, then is kinetic energy will increase byA. `40%`B. `20%`C. `10%`D. `21%` |
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Answer» Correct Answer - D `(K.E._(2))/(K.E._(1))=(L_(2)^(2))/(L_(1)^(2))` `= (R_(2)^(2)omega_(2)^(2))/(R_(1)^(2)omega_(1)^(2))=(1.1)^(2)=1.21` `therefore (K.E._(2))/(K_(.E._(1))-1=1.21-1 therefore K.E._(1)=21%` `therefore K.E._(2)` increased by 21%. |
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| 237. |
The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous exis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless, stick as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed `omega` the motion at any instant can be taken as a combination of (i) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed `omega` in this case Now consider two similar system as shown in the figure: Case (a) the disc with its face vertical and parallel to x-z plane, Case (b) the disc with its face making an angle of `45^@` with x-y plane and its horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed `omega` about the z-axis. .` Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct?A. It is vertical for both thecases (a) and (b)B. It is vertical for case (a), and it at `45^@` to the x-z plane and lies in the plane of the disc for case (b).C. It is vertical for case (a), and is `45^@` to the x-z plane and is normal to the plane of the disc for case (b)D. It is vertical for case (a), and is `45^@` to the x-z plane and is normal to the plane of the disc for case (b). |
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Answer» Correct Answer - A (a) Axis of rotation is parallel to z-axis. |
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| 238. |
The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous exis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless, stick as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed `omega` the motion at any instant can be taken as a combination of (i) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed `omega` in this case Now consider two similar system as shown in the figure: Case (a) the disc with its face vertical and parallel to x-z plane, Case (b) the disc with its face making an angle of `45^@` with x-y plane and its horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed `omega` about the z-axis. Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct?A. it is `sqrt2omega` for both the casesB. it is `omega` for cases (a), and `omega//sqrt2` for case (b)C. it is `omega` for case (a), and `sqrt2omega` for case (b)D. it is `omega` for both the cases. |
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Answer» Correct Answer - B (d) Since the body is rigid, `omega` is same for any point of the body. |
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| 239. |
A uniform disc of mass 2kg is rotated about an axis perpendicular to the plane of the disc . If radius of gyration is 50cm, then the M.I. of disc about same axis isA. `0.25 kg m^(2)`B. `0.5kg m^(2)`C. `2kg m^(2)`D. `1kg m^(2)` |
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Answer» Correct Answer - B `I = MK_(c )^(2)=2xx(0.5)^(2)=2xx0.25=0.5`. |
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| 240. |
A rigid spherical body is spinning around an axis without any external torque. Due to temperature its volume increases by `3%`. Then percentage change in its angular speed is:A. `-2%`B. `-1%`C. `-3%`D. `1%` |
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Answer» Correct Answer - A |
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| 241. |
A uniform thin rod AB of mass M and length l attached to a string OA of length `=1/(2)` is supported by a smooth horizontal plane and rotates with angular velocity `omega` around a vertical axis through O. A peg P is inserted in the plane in order that on striking it the rod will come to rest. A. magnitude of angular momentum of rod about O is `(4)/(3) l^(2)omega`B. Magnitude of tension in string is `M//omega^(2)`C. Location of peg for rod coming to rest is x`=(13)/(12)l`D. magnitude of angular impulse by peg on the rod is `(4)/(3) l^(2)omega` |
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Answer» Correct Answer - B::C |
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| 242. |
A rod of mass m and length l is hinged at one of its end A as shown in figure. A force F is applied at a distance x from A. The acceleration of centre of mass a varies with x as A. B. C. D. |
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Answer» Correct Answer - B |
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| 243. |
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along.A. a line perpendicular to the plane of rotationB. the line making an angle of `45^(@)` to the plane of rotationC. the radiusD. the tangent to the orbit |
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Answer» Correct Answer - a |
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| 244. |
The kinetic energy of a lamina moving in its planeisA. `M(v_(cm)^(2) + K^(2)omega^(2))`B. `1/2M(v_(CM)^(2) + K^(2)omega^(2))`C. `1/2komegap^(2)`D. None of these |
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Answer» b) if a lamina is moving with `v_(CM)` having angualr velocity `omega`. Then, KE = Angular kinetic energy + Translatory kinetic energy `=1/2_(CM)omega^(2) + 1/2 mv_(CM)^(2)` `=1/2MK^(2)omega^(2) + 1/2 Mv_(CM)^(2) = `1/2 M(K^(2)omega^(2)+ v_(CM)^(2))` |
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| 245. |
Angular momentum of the particle rotating with a central force is constant due toA. constant torqueB. constatn forceC. constant linear momentumD. zero torque |
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Answer» Correct Answer - D (d) We know that `vectau_c = (dvecL_c)/(dt)` `where `vectau_c` torque about the center of mass of the body and `vecL_c`= Angular momentum about the center of mass of the body. Central forces act along the center of mass Therefore torque about center of mass is zero. `when vectau_c =0 then vecL_c = constt.` |
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| 246. |
A playground merry-go-round is at rest, pivoted about a frictionless axis. A child of mass `m` runs along a path tangential to the rim with speed `v` and jumps on to the merry-go-round. If `R` is the radius of the merry-go-round and `I` is its moment of inertia, then the angular velocity of the merry-go-round isA. `(mvR)/(mR^(2)+I)`B. `(mvR)/(I)`C. `(mR^(2))/(mvR)`D. `(I)/(mvR)` |
| Answer» `mvR=(I+mR^(2))omegaimpliesomega=(mvR)/(I+mR^(2))` | |
| 247. |
A particle of mass `2 kg` located at the position `(hati + hatj)m` has velocity `2(hati - hatj + hatk) m//s` . Its angular momentum about Z-axis in `kg m^(2)//s` isA. zeroB. `+8`C. 12D. `-8` |
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Answer» Correct Answer - D |
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| 248. |
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along.A. radiusB. tangent to the orbitC. axis of rotationD. circumference of the circle |
| Answer» Correct Answer - C | |
| 249. |
Two persons of masses `55 kg` and `65 kg` respectively are at the opposite ends of a boat. The length of the boat is `3.0 m` and weights `100 kg`. The `55 kg` man walks up to the `65 kg` man and sits with him. If the boat is in still water the centre of mass of the system shifts by.A. `3 m`B. `2.3 m`C. zeroD. `0.75 m` |
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Answer» Correct Answer - c |
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| 250. |
A body rolling on a horizontal plane is an example ofA. rotational motionB. oscillatory motionC. translation motionD. rotational and translational motion |
| Answer» Correct Answer - D | |