553 + Interview Questions in ROTATIONAL-MOTION IN GENERAL AWARENESS Page 7 InterviewSolution

301.	A string can withstand a tension of 25 N. What is the greatest speed at which a body of mass 0.5 kg can be whirled in a circle using 0.5 m length of the string ? Neglect the force of gravity on the body.
Answer» Correct Answer - `5ms^(-1)`

Discussion

302.	Two cylinders A and B have been placed in contact on an incline. They remain in equilibrium. The dimensions of the two cylinders are same. Which cylinder has larger mass?
Answer» Correct Answer - `M_(A) lt M_(B)`

Discussion

303.	Consider the object shown in the figure. It consist of a solid hemisphere of mass M and radius R. There is a solid rod welded at its centre. The object is placed on a flat surface so that the rod is vertical. Mass of the rod per unit length is `(M)/(2R)`. What is the maximum length of the rod that can be welded so that the object can performoscillations about the position shown in diagram? Note : Centre of mass of a solid hemisphere is at a distance of `(3R)/(8)` from its base.
Answer» Correct Answer - Less than `sqrt((3)/(2))R`

Discussion

304.	A flywheel of moment of inertia `10^(7) g cm^(2)` is rotating at a speed of 120 rotations per minute. Find the constant breaking torque required to stop the wheel in 5 rotations.
Answer» Correct Answer - `2.513xx10^(7)` dyne cm

Discussion

305.	A disc of mass 100 g is rolling without slipping on a horizontal surface with a velocity of `10 cm s^(-1)`. Calculate its total energy.
Answer» Correct Answer - `7.5xx10^(-4) J`

Discussion

306.	A rod revlving 60 times in a minute about an axis passing through an end at right angles to the length, has kinetic energy of 400 J. find the moment of inertia of the rod.
Answer» Correct Answer - `20.26 kg m^(2)`

Discussion

307.	The centre of mass of three particles of masses `1kg, 2 kg` and `3kg` lies at the point (3m, 3m, 3m). Where should a fourth particle of mass 4 kg be positioned so that the centre of mass of the four particle system lies at the point (1m, 1m, 1m)?
Answer» Correct Answer - `(-2m, -2m, -2m)`

Discussion

308.	The mass per unit length `(lambda)` of a non-uniform rod varies linearly with distance x from its one end accrding to the relation, `lambda = alpha x`, where `alpha` is a constant. Find the centre of mass as a fraction of its length L.
Answer» Correct Answer - `2L//3`

Discussion

309.	Two bodies of masses 10 kg and 2 kg are moving with velocities `2 hat(i)-7 hat(j)+3 hat(k)` and `-10 hat(i)+35 hat(j)-3hat(j) ms^(-1)` respectively. Find the velocity of the centre of mass of the system.
Answer» Correct Answer - `2hat(k)ms^(-1)`

Discussion

310.	Two bodies of masses 2 kg and 1 kg are moving along the same line with speeds `2ms^(-1)` and `5 ms^(-1)` respectively. What is the speed of the centre of mass of the system if the two bodies are moving (i) in same direction and (ii) in opposite directions?
Answer» Correct Answer - (i) `3ms^(-1)` (ii) `(1)/(3) ms^(-1)`, in the direction of motion of 1kg mass

Discussion

311.	Two bodies of mass `1 kg` and `3 kg` have position vectors `hat i+ 2 hat j + hat k` and `- 3 hat i- 2 hat j+ hat k`, respectively. The centre of mass of this system has a position vector.A. `-2 hat(i)+2hat(k)`B. `-2hat(i)-hat(j)+hat(k)`C. `2hat(i)-hat(j)-2hat(k)`D. `-hat(i)+hat(j)+hat(k)`
Answer» Correct Answer - b

Discussion

312.	Four identical thin rods each of mass `M` and length `l`, from a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane isA. `4/3 MI^(2)`B. `2/ MI^(2)`C. `13/3 MI^(2)`D. `1/3 MI^(2)`
Answer» Correct Answer - a

Discussion

313.	Three thin rods each of length Land mass M are placed along x, y and z-axes such that one of each rod is at origin. The moment of inertia of this system about z-axis isA. `2.3ML^(2)`B. `(4ML^(2))/3C. `(5ML^(2))/3D. `(ML^(2))/3
Answer» a) Moment of inertia of the rod lying along z-axis will be zero. MI of the rods along x and y-axes will be `(ML^(2))/3` each. Hence, total moment of inertia is `2/3` M`L^(2)`.

Discussion

314.	AOB is a frictionless parabolic track in vertical plane. The equation of parabolic track can be expressed as `y = (3)/(2H) x^(2)` for co-ordinate system shown in the figure. The end B of the track lies at `y = (H)/(2)`. When a uniform small ring is released on the track at A it was found to attain a maximum height of `h_(1)`, above the ground after leaving the track at B. There is another track DEF which is in form of an arc of a circle of radius H subtending an angle an angle of `150^(@)` at the centre. The radius of the track at D is horizontal. The same ring is released on this track at point D and it rolls without sliding. The ring leaves the track at F and attains a maximum height of `h_(2)` above the ground. Find the ratio `(h_(1))/(h_(2))`.
Answer» Correct Answer - `(h_(1))/(h_(2)) = (14)/(11)`

Discussion

315.	A body is said to be rigid, if distance between any two position of the particleA. changes with forceB. remains constant with the forceC. changes with the force initially and maximum force changes laterallyD. erratical changes with force
Answer» Correct Answer - B

Discussion

316.	Calculate the M.I. of a thin uniforn ring about an axis tangent to the ring and in a plane of the ring, if its M.I. about an axis passing through the centre and perpendicular to plane is `4kg m^(2)`.A. `12kg m^(2)`B. `3kg m^(2)`C. `6kg m^(2)`D. `9 kg m^(2)`
Answer» Correct Answer - C `(I_(T))/(I_(e ))=((3)/(2)MR^(2))/(MR^(2))=(3)/(2)` `I_(T)=(3)/(2)I_(c )=(3)/(2)xx4=6`

Discussion

317.	Two men each of mass m stand on the rim of a horizontal circular disc, diametrically opposite to each other. The disc has a mass M and is free to rotate about a vertical axis passing through its centre of mass. Each mass start simultaneously along the rim clockwise and reaches their original starting positions on the disc. The angle turned through by disc with respect to the ground (in radian) isA. `(8mpi)/(4m+M)`B. `(2mpi)/(4m+M)`C. `(mpi)/(M+m)`D. `(4mpi)/(2M+m)`
Answer» Correct Answer - A

Discussion

318.	If radius of solid sphere is doubled by keeping its mass constant, thenA. `(I_(1))/(I_(2))=(1)/(4)`B. `(I_(1))/(I_(2))=(4)/(1)`C. `(I_(1))/(I_(2))=(3)/(2)`D. `(I_(1))/(I_(2))=(2)/(3)`
Answer» Correct Answer - A `(I_(2))/(I_(1))=((2)/(5)MR_(2)^(2))/((2)/(5)MR_(1)^(2))=((R_(2))/(R_(1)))^(2)=(4)/(1)` `therefore (I_(1))/(I_(2))=(1)/(4)`.

Discussion

319.	The moment of inertia of a disc of mass `M` and radius `R` about a tangent to its rim in its plane isA. `2/3 MR^(2)`B. `3/2 MR^(2)`C. `4/5 MR^(2)`D. `5/4 MR^(2)`
Answer» Correct Answer - d

Discussion

320.	Moment of inertia of a disc about an axis which is tangent and parallel to its plane is I . Then the moment of inertia of disc about a tangent, but perpendicular to its plane will beA. `(3I)/(4)`B. `(3I)/(2)`C. `(5I)/(6)`D. `(6I)/(5)`
Answer» Correct Answer - D `I_(t)=I, I_(T)=?` `(I_(T))/(I_(t))=((3)/(2)MR^(2))/((5)/(4)MR^(2))` `I_(T)=(6)/(5)I_(t)=(6)/(5)I`

Discussion

321.	A rod of length 2 m is kept vertical inside a smooth spherical shell of radius 2m. The rod starts slipping inside the shell . Mass of the rod is 4 kg. Angular speed of the rod(in rad/s) in the position when it becomes horizontal is A. 4.6B. 6.8C. 3.2D. 7.2
Answer» Correct Answer - C

Discussion

322.	If a uniform solid sphere of diameter 0.2 m and mass 10 kg is rotated about its diameter with an angular velocity of 2 rad/s, then the its angular momentum in kg `m^(2)//s` will beA. 0.01B. 0.02C. 0.08D. 0.04
Answer» Correct Answer - C `L = I omega = (2)/(5)MR^(2)omega =(2)/(5)xx10xx10^(-2)xx2` `= 0.08 kgm^(2)//s`.

Discussion

323.	A uniform rod of mass `m` and length L is fixed to an axis, making an angle `theta` with it as shown in the figure. The rod is rotated about this axis so that the free end of the rod moves with a uniform speed ‘v’. Find the angular momentum of the rod about the axis. Is the angular momentum of the rod about point A constant?
Answer» Correct Answer - `(1)/(3) mv L sin theta`, No

Discussion

324.	ABCFED is a uniform plate (shown in figure). ABC and DEF are circular arcs with common centre at O and having radii `a` and `2a` respectively. This plate is lying on a smooth horizontal table. A particle of mass half the mass of the plate strikes the plate at point A while travelling horizontally along the `x` direction with velocity `u`. The particle hits the plate and rebounds along negative `x` with velocity `(u)/(2)` . Find the velocity of point D of the plate immediately after the impact. [Take `(28)/(9 pi) cong 1` ]
Answer» Correct Answer - `(3u)/(4)`

Discussion

325.	The radius of gyration of a solid hemisphere of mass M and radius Rn about an axis parallel to the diameter at a distance `(3)/(4)`R is given by (centre of mass of the hemisphere lies at a height `3R//8` from the base.) A. `(3R)/(sqrt10)`B. `(5R)/(4)`C. `(5R)/(8)`D. `sqrt((2)/(5))R`
Answer» Correct Answer - D

Discussion

326.	The position of a particle is given by `vecr=(hati+2hatj-hatk)` and momentum `vecp=(3hati+4hatj-2hatk)`. The angular momentum is perpendicular to theA. `x`-axisB. `y`-axisC. `z`-axisD. line at equal angles to all the three axes
Answer» `vecl=vecrxxvecp=\|[hati,hatj,hatk],[1,2,-1],[3,4,-2]\|=-hatj-2hatk`

Discussion

327.	The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height `h` from rest without slipping will be.A. `sqrt(10/7 gh)`B. `sqrt(gh)`C. `sqrt(6/5 gh)`D. `sqrt(4/3 gh)`
Answer» Correct Answer - a

Discussion

328.	A force `F` is applied on the top of a cube as shown in the figure. The coefficient of friction between the cube and the ground is `mu`. If `F` is gradually increased, find the value of `mu` for which the cube will topple before sliding. A. `mugt(1)/(4)`B. `mult(1)/(2)`C. `mugt(1)/(2)`D. `mult1`
Answer» Correct Answer - C

Discussion

329.	The wheel of radius R rolls wihtout slipping on horizontal rough surface and its centre O has an acceleration `a_(0)` in forward direction. A point P on the wheel is a distance r from O and angular position ` theta` from horizontal. Find the angle ` theta` for which point P can have zero acceleration in this position . A. `cos^(-1) ( r)/(R )`B. `tan^(-1) ( r)/(R )`C. `sin^(-1) ( r)/(R )`D. `cos^(-1) ( r)/(2R )`
Answer» Correct Answer - C

Discussion

330.	A uniform metal sheet of mass M has been folded to give it L shape and it is placed on a rough floor as shown in figure. Wind is blowing horizontally and hits the vertical face of the sheet as shown. The speed of air varies linearly from zero at floor level to `v_(0)` at height L from the floor. Density of air is `rho`. Find maximum value of `v_(0)` for which the sheet will not topple. Assume that air particles striking the sheet come to rest after collision, and that the friction is large enough to prevent the sheet from sliding
Answer» Correct Answer - `(v_(0))_("max") = (1)/(L) sqrt((3Mg)/(rho))`

Discussion

331.	A disc of moment of inertia `9.8//pi^(2)kg m^(2)` is rotating at 600 rpm . If the frequency of rotation changes from 600 rpm to 300 rpm, then what is the work done?A. 1470 JB. 1452 JC. 1567JD. 1632J
Answer» Correct Answer - A Work done `= Delta K.E. = (1)/(2)I(omega_(2)^(2)-omega_(1)^(2))` `= (9.8)/(2xx pi^(2))xx 4pi^(2)` `(n_(1)^(2)-n_(2)^(2))=19.6xx75` = 1470 J.

Discussion

332.	A disc of moment of inertia `9.8//pi^(2)kg m^(2)` is rotating at 600 rpm. If the frequency of rotation changes from 600 rpm to 300 rpm, then what is the work done ?A. 1567 JB. 1452 JC. 1467 JD. 1632 J
Answer» Correct Answer - C `omega = KE_(2)-KE_(1)=(1)/(2)I (omega_(2)^(2)-omega_(1)^(2))` `= (1)/(2)xx(9.8)/(pi^(2))xx4 pi^(2)(n_(2)^(2)-n_(1)^(2))` `= 4.9xx4(25-100)` `=4.9xx4xx75` `= 4.9xx300` `= 1470.0` `= 1467 J`

Discussion

333.	The angular position of a point on the rim of a rotating wheel is given by `theta=4t^(3)-2t^(2)+5t+3` rad. Find (a) the angular velocity at `t=1 s`, (b) the angular acceleration at `t=2 s`. (c ) the average angular velocity in time interval `t=0` to `t=2 s` and (d) the average angular acceleration in time interval `t=1` to `t=3 s`.
Answer» `theta=4t^(3)-2t^(2)+5t+3` `omega=(d theta)/(dt)=12t^(2)-4t+5` `alpha=(domega)/(dt)=24t-4` (a) At `t=1 s`, `omega=12(1)^(2)-4(1)+5=13 rad//s` (b) At `t=2 s`, `alpha=24(2)-4=44 rad//s^(2)` (c ) `t_(1)=0, theta_(1)=3 rad` `t_(2)=2 s, theta_(2)=4(2)^(3)-2(2)^(2)+5(2)+3=37 rad` `undersetomega(-)=(theta_(2)-theta_(1))/(t_(2)-t_(1))=(37-3)/(2-0)=17 rad//s` (d) At `t_(1)=1 s`, ` omega_(1)=12(1)^(2)-4(1)+5=13 rad//s` At `t_(2)=3 s`, `omega_(2)=12(3)^(2)-4(3)+5=101 rad//s` `undersetalpha(-)=(omega_(2)-omega_(1))/(t_(2)-t_(1))=(101-13)/(3-1)=44 rad//s^(2)`

Discussion

334.	A dumb – bell has a rigid mass less stick and two point masses at its ends. Each mass is m and length of the stick is L. The dumb- bell leans against a frictionless wall, standing on a frictionless ground. It is initially held motionless, with its bottom end an infinitesimal distance from the wall. It is released from this position and its bottom end slides away from the wall where as the top end slides down along the wall. (a) Show that centre of mass of the dumb-bell moves along a circle. (b) When the dumb-bell loses contact with the wall what is speed of the centre of mass?
Answer» Correct Answer - (b) `sqrt((gL)/(6))`

Discussion

335.	A uniform thin bar of mass `6 m` and length `12 L` is bend to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the hexagon is :A. `20 mL^(2)`B. `6 mL^(2)`C. `(12)/(5)mL^(2)`D. `30 mL^(2)`
Answer» Correct Answer - A

Discussion

336.	A uniform rod `AB` of mass `m` and length `l` is at rest on a smooth horizontal surface. An impulse `J` is applied to the end `B`, perpendicular to the rod in the horizontal direction. Speed of particlem `P` at a distance `(l)/(6)` from the centre towards `A` of the rod after time `t = (pi m l)/(12 J)` is.A. `2J/M`B. `(J)/(sqrt2M)`C. `J/M`D. `sqrt2J/M`
Answer» Correct Answer - D

Discussion

337.	Two particles connected by a rigid light rod AB, lying on a smooth horizontal table. An impulse J is applied at A in the plane of thetable and perpendicular at AB. Then the velocity of particle at A isA. `(J)/(2m)`B. `(J)/(m)`C. `(2J)/(m)`D. zero
Answer» Correct Answer - B

Discussion

338.	A uniform rod of mass m and length l is lying on a smooth table. An impulse J acts on the rod momentarily as shown in figure at point R. Just after that:
Answer» Correct Answer - (A)R,(B)S,(C)P,(D)Q

Discussion

339.	A fly wheel of M.I. `0.32 kg m^(2)` is rotated steadily at 120 rad/s by 50 w electric motor. Then the value of the frictional couple opposing rotation is,A. 4.2 NmB. 0.42 NmC. 0.042 NmD. 42 Nm
Answer» Correct Answer - B `P = tau omega`

Discussion

340.	If a constant torque of 500 Nm turns a wheel of moment of inertia `100 kg m^(2)` about an axis through its centre, find the gain in angular velocity in 2s.
Answer» Correct Answer - `10 rad s^(-1)`

Discussion

341.	A disc of moment of inertia `2 kg m^(2)` is acted upon by a constant torque of 40 Nm. If it is initially at rest, then the time taken by it to acquire an angular velocity 100 rad/s will beA. 20 sB. 10 sC. 5 sD. 4 s
Answer» Correct Answer - C `t=(I(Delta omega))/(lambda)=(2xx100)/(40)=5` sec.

Discussion

342.	A constant torque acting on a uniform circular wheel changes its angular momentum from `A_(0)` to `4 A_(0)` in `4` seconds. Find the magnitude of this torque.A. 0.75AB. 4AC. AD. 12A
Answer» Correct Answer - A `tau =(dL)/(dt)=(4A-A)/(4)=0.75 A`

Discussion

343.	A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x-y plane with centre at O and constant angular speed `omega`. If the angular momentum of the system. calculated about O and P are denoted. by `vecL_O and vecL_P` respectively, then. A. `vecL_(0)` and `vecL_(P)` do not vary with timeB. `vecL_(0)` varies with time while `vecL_(P)` remains constantC. `vecL_(0)` remains constant while `vecL_(P)` varies with timeD. `vecL_(0)` and `vecL_(P)` both vary with time
Answer» With respect to `P`, the position vector of a particle changes with time.

Discussion

344.	A wheel of moment of inertia `5xx10^(-3) kg m^(2)` is making 20 revolutions per second. If it is stopped in 20 s, then its angular retardation would beA. `pi rad //s^(2)`B. `4pi rad//s^(2)`C. `2pi rad//s^(2)`D. `8pi rad //s^(2)`
Answer» Correct Answer - C `alpha = (omega_(2)-omega_(1))/(t)=(2pi(n_(2)-n_(1)))/(t)=(2pi(0-20))/(t)=2pi`

Discussion

345.	A wheel of mass 40 kg and radius of gyrating 0.5 m comes to rest from a speed of 180 revolutions per minute in 3 s. Assuming that the retardation is uniform, then the value of retarding torque `tau` in Nm isA. `10 pi`B. `20 pi`C. `30 pi`D. `40 pi`
Answer» Correct Answer - B `I=m K^(2)=40xx0.25=10 kg m^(2)` `alpha = (omega_(2)-omega_(1))/(t)=(0-2pi xx 3)/(3)= - 2pi` `tau = I alpha =-2pi xx 10=-20 pi`

Discussion

346.	A wheel of mass 40 kg and radius of gyration 0.5 m comes to rest from a speed of 180 rpm in 30 s assuming that the retardation is uniform then the value of the retarding torque, in Nm will beA. `1 pi`B. `3 pi`C. `2 pi`D. `4 pi`
Answer» Correct Answer - C `tau = I alpha = M k^(2) alpha`.

Discussion

347.	A flywheel is in the form of solid circular wheel of mass 72 kg and radius of 0.5 m and it takes 70 rpm, then the energy of revolution isA. 24 JB. 2.4 JC. 240 JD. 2400 J
Answer» Correct Answer - C `K.E. = (1)/(2) I omega^(2)` `=(1)/(2)(MR^(2))/(2)xx4 pi^(2)n^(2)`

Discussion

348.	A fly wheel of mass 60 kg and radius 40 cm is revolving at 300 rpm, then its rotational K.E. isA. `(48)/(pi^(2))J`B. `480 pi J`C. `(48)/(pi)J`D. `240 pi^(2)J`
Answer» Correct Answer - D K.E. of rotation `=(1)/(2) I omega^(2)` `=(1)/(2)((1)/(2)mR^(2))omega^(2)` `=(1)/(4)mR^(2)(2pi n)^(2)=pi^(2)n^(2)m R^(2)` `= pi^(2)((300)/(60))^(2)xx60xx(0.4)^(2)=240 pi^(2)J`

Discussion

349.	A child stands at the turn table center of a turnable with his two arms outstretched. The turnable is set rotating with an angular speed of `10 rad//s`. If the child folds his hands back so that the moment of inertia reduces to `3//5` times the initial value, find the new angular speed. Why does the kinetic energy of child increase?
Answer» `I_(1)=I,omega_(1)=10 rad//s` `I_(2)=(3)/(5)I, omega_(2)=?` `I_(1)omega_(1)=I_(2)omega_(2)` (conservation of angular momentum) `Ixx10=(3I)/(5)xxomega_(2)implies omega_(2)=(50)/(3) rad//s` Initial `K.E., K_(i)=(1)/(2)I_(1)omega_(1)^(2)=(1)/(2)I(10)^(2)=50I` Final `K.E., K_(f)=(1)/(2) I_(2) omega_(2)^(2)=(1)/(2) xx(3I)/(5)((50)/(3))^(2)=(250)/(3)I` `K_(f)gtK_(i)` The child uses his inernal muscular energy (used in folding his hands ) to increase the rotational kinetic energy.

Discussion

350.	A man stands holding a weight in each hand and with his arms outstretched on a frictionless platform which is rotating at a speed of 1 revolution per sec. In ths position the total rotational intertia of the man, the weights and the platform is `6kgm^(2)`. If by drawing in his arms, the man decreases the rotational inertia of the system by `2kg m^(2)`, calcualte the resulting speed of the platform and the increase in kinetic energy. How do you account for the increase of kinetic energy?A. 2 rpsB. 4 rpsC. 3 rpsD. 6 rps
Answer» Correct Answer - C `I_(1)omega_(1)=I_(2)omega_(2)`

Discussion

Explore topic-wise InterviewSolutions in .

A string can withstand a tension of 25 N. What is the greatest speed at which a body of mass 0.5 kg can be whirled in a circle using 0.5 m length of the string ? Neglect the force of gravity on the body.

Two cylinders A and B have been placed in contact on an incline. They remain in equilibrium. The dimensions of the two cylinders are same. Which cylinder has larger mass?

A flywheel of moment of inertia `10^(7) g cm^(2)` is rotating at a speed of 120 rotations per minute. Find the constant breaking torque required to stop the wheel in 5 rotations.

A disc of mass 100 g is rolling without slipping on a horizontal surface with a velocity of `10 cm s^(-1)`. Calculate its total energy.

A rod revlving 60 times in a minute about an axis passing through an end at right angles to the length, has kinetic energy of 400 J. find the moment of inertia of the rod.

The centre of mass of three particles of masses `1kg, 2 kg` and `3kg` lies at the point (3m, 3m, 3m). Where should a fourth particle of mass 4 kg be positioned so that the centre of mass of the four particle system lies at the point (1m, 1m, 1m)?

The mass per unit length `(lambda)` of a non-uniform rod varies linearly with distance x from its one end accrding to the relation, `lambda = alpha x`, where `alpha` is a constant. Find the centre of mass as a fraction of its length L.

Two bodies of masses 10 kg and 2 kg are moving with velocities `2 hat(i)-7 hat(j)+3 hat(k)` and `-10 hat(i)+35 hat(j)-3hat(j) ms^(-1)` respectively. Find the velocity of the centre of mass of the system.

Two bodies of masses 2 kg and 1 kg are moving along the same line with speeds `2ms^(-1)` and `5 ms^(-1)` respectively. What is the speed of the centre of mass of the system if the two bodies are moving (i) in same direction and (ii) in opposite directions?

Two bodies of mass `1 kg` and `3 kg` have position vectors `hat i+ 2 hat j + hat k` and `- 3 hat i- 2 hat j+ hat k`, respectively. The centre of mass of this system has a position vector.A. `-2 hat(i)+2hat(k)`B. `-2hat(i)-hat(j)+hat(k)`C. `2hat(i)-hat(j)-2hat(k)`D. `-hat(i)+hat(j)+hat(k)`

Four identical thin rods each of mass `M` and length `l`, from a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane isA. `4/3 MI^(2)`B. `2/ MI^(2)`C. `13/3 MI^(2)`D. `1/3 MI^(2)`

Three thin rods each of length Land mass M are placed along x, y and z-axes such that one of each rod is at origin. The moment of inertia of this system about z-axis isA. `2.3ML^(2)`B. `(4ML^(2))/3C. `(5ML^(2))/3D. `(ML^(2))/3

A body is said to be rigid, if distance between any two position of the particleA. changes with forceB. remains constant with the forceC. changes with the force initially and maximum force changes laterallyD. erratical changes with force

Calculate the M.I. of a thin uniforn ring about an axis tangent to the ring and in a plane of the ring, if its M.I. about an axis passing through the centre and perpendicular to plane is `4kg m^(2)`.A. `12kg m^(2)`B. `3kg m^(2)`C. `6kg m^(2)`D. `9 kg m^(2)`

If radius of solid sphere is doubled by keeping its mass constant, thenA. `(I_(1))/(I_(2))=(1)/(4)`B. `(I_(1))/(I_(2))=(4)/(1)`C. `(I_(1))/(I_(2))=(3)/(2)`D. `(I_(1))/(I_(2))=(2)/(3)`

The moment of inertia of a disc of mass `M` and radius `R` about a tangent to its rim in its plane isA. `2/3 MR^(2)`B. `3/2 MR^(2)`C. `4/5 MR^(2)`D. `5/4 MR^(2)`

Moment of inertia of a disc about an axis which is tangent and parallel to its plane is I . Then the moment of inertia of disc about a tangent, but perpendicular to its plane will beA. `(3I)/(4)`B. `(3I)/(2)`C. `(5I)/(6)`D. `(6I)/(5)`

A rod of length 2 m is kept vertical inside a smooth spherical shell of radius 2m. The rod starts slipping inside the shell . Mass of the rod is 4 kg. Angular speed of the rod(in rad/s) in the position when it becomes horizontal is A. 4.6B. 6.8C. 3.2D. 7.2

If a uniform solid sphere of diameter 0.2 m and mass 10 kg is rotated about its diameter with an angular velocity of 2 rad/s, then the its angular momentum in kg `m^(2)//s` will beA. 0.01B. 0.02C. 0.08D. 0.04

The radius of gyration of a solid hemisphere of mass M and radius Rn about an axis parallel to the diameter at a distance `(3)/(4)`R is given by (centre of mass of the hemisphere lies at a height `3R//8` from the base.) A. `(3R)/(sqrt10)`B. `(5R)/(4)`C. `(5R)/(8)`D. `sqrt((2)/(5))R`

The position of a particle is given by `vecr=(hati+2hatj-hatk)` and momentum `vecp=(3hati+4hatj-2hatk)`. The angular momentum is perpendicular to theA. `x`-axisB. `y`-axisC. `z`-axisD. line at equal angles to all the three axes

The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height `h` from rest without slipping will be.A. `sqrt(10/7 gh)`B. `sqrt(gh)`C. `sqrt(6/5 gh)`D. `sqrt(4/3 gh)`

A force `F` is applied on the top of a cube as shown in the figure. The coefficient of friction between the cube and the ground is `mu`. If `F` is gradually increased, find the value of `mu` for which the cube will topple before sliding. A. `mugt(1)/(4)`B. `mult(1)/(2)`C. `mugt(1)/(2)`D. `mult1`

A disc of moment of inertia `9.8//pi^(2)kg m^(2)` is rotating at 600 rpm . If the frequency of rotation changes from 600 rpm to 300 rpm, then what is the work done?A. 1470 JB. 1452 JC. 1567JD. 1632J

A disc of moment of inertia `9.8//pi^(2)kg m^(2)` is rotating at 600 rpm. If the frequency of rotation changes from 600 rpm to 300 rpm, then what is the work done ?A. 1567 JB. 1452 JC. 1467 JD. 1632 J

A uniform thin bar of mass `6 m` and length `12 L` is bend to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the hexagon is :A. `20 mL^(2)`B. `6 mL^(2)`C. `(12)/(5)mL^(2)`D. `30 mL^(2)`

Two particles connected by a rigid light rod AB, lying on a smooth horizontal table. An impulse J is applied at A in the plane of thetable and perpendicular at AB. Then the velocity of particle at A isA. `(J)/(2m)`B. `(J)/(m)`C. `(2J)/(m)`D. zero

A uniform rod of mass m and length l is lying on a smooth table. An impulse J acts on the rod momentarily as shown in figure at point R. Just after that:

A fly wheel of M.I. `0.32 kg m^(2)` is rotated steadily at 120 rad/s by 50 w electric motor. Then the value of the frictional couple opposing rotation is,A. 4.2 NmB. 0.42 NmC. 0.042 NmD. 42 Nm

If a constant torque of 500 Nm turns a wheel of moment of inertia `100 kg m^(2)` about an axis through its centre, find the gain in angular velocity in 2s.

A disc of moment of inertia `2 kg m^(2)` is acted upon by a constant torque of 40 Nm. If it is initially at rest, then the time taken by it to acquire an angular velocity 100 rad/s will beA. 20 sB. 10 sC. 5 sD. 4 s

A constant torque acting on a uniform circular wheel changes its angular momentum from `A_(0)` to `4 A_(0)` in `4` seconds. Find the magnitude of this torque.A. 0.75AB. 4AC. AD. 12A

A wheel of moment of inertia `5xx10^(-3) kg m^(2)` is making 20 revolutions per second. If it is stopped in 20 s, then its angular retardation would beA. `pi rad //s^(2)`B. `4pi rad//s^(2)`C. `2pi rad//s^(2)`D. `8pi rad //s^(2)`

A wheel of mass 40 kg and radius of gyrating 0.5 m comes to rest from a speed of 180 revolutions per minute in 3 s. Assuming that the retardation is uniform, then the value of retarding torque `tau` in Nm isA. `10 pi`B. `20 pi`C. `30 pi`D. `40 pi`

A wheel of mass 40 kg and radius of gyration 0.5 m comes to rest from a speed of 180 rpm in 30 s assuming that the retardation is uniform then the value of the retarding torque, in Nm will beA. `1 pi`B. `3 pi`C. `2 pi`D. `4 pi`

A flywheel is in the form of solid circular wheel of mass 72 kg and radius of 0.5 m and it takes 70 rpm, then the energy of revolution isA. 24 JB. 2.4 JC. 240 JD. 2400 J

A fly wheel of mass 60 kg and radius 40 cm is revolving at 300 rpm, then its rotational K.E. isA. `(48)/(pi^(2))J`B. `480 pi J`C. `(48)/(pi)J`D. `240 pi^(2)J`