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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.
| 501. |
Dimensions of angular momentum isA. `[L^(1)M^(1)T^(-2)]`B. `[L^(-2)M^(1)T^(-1)]`C. `[L^(2)M^(1)T^(-1)]`D. `[L^(0)M^(1)T^(-1)]` |
| Answer» Correct Answer - C | |
| 502. |
The orbital angular momentum and angular momentum (classical analogue) for the electron of 4s-orbital are respectively, equal to:A. `[L^(-2)M^(1)T^(-1)]`B. `[L^(2)M^(1)T^(-1)]`C. `[L^(2)M^(1)T^(1)]`D. `[L^(2)M^(2)T^(-2)]` |
| Answer» Correct Answer - B | |
| 503. |
The correct statement about angular momentum isA. directly proportional to moment of inertiaB. a scalar quantityC. inversely proportional to moment ofD. its direction always is tangential at a point |
| Answer» Correct Answer - A | |
| 504. |
If the angular momentum of any rotating body increases by `200%`, then the increase in its kinetic energyA. `400%`B. `800%`C. `200%`D. `100%` |
| Answer» Correct Answer - B | |
| 505. |
When a circular disc of radius 0.5 m is rotating about its own axis, then the direction of its angular momentum isA. along the tangent drawn at every pointB. radialC. perpendicular to the direction of angular velocityD. along the axis of rotation of the body |
| Answer» Correct Answer - D | |
| 506. |
The angular momentum of a particle isA. parallel to its linear momentumB. perpendicular to its linear momentumC. inclined to its linear momentumD. a scalar quantity |
| Answer» Correct Answer - B | |
| 507. |
A heavy disc is rotating with uniform angular velocity `omega` about its own axis. A piece of wax sticks to it. The angular velocity of the disc willA. increaseB. decreaseC. becomes zeroD. remain unchanged |
| Answer» Correct Answer - B | |
| 508. |
A wooden log of mass M and length L is hinged by a frictionless nail at O.A bullet of mass m strikes with velocity v and sticks to it. Find angular velocity of the system immediately after the collision aboutO. |
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Answer» Correct Answer - A::C We know that `tau = (dvacL)/(dt)` `tauxxdt = dvecL` when angular impulse `(tauxxd vect)` is zero, the anglar momentum is constant. In this case for the wooden log-bullet system the angular impulse about O is constant . therefore. `[angular momentum of the system]_(initial]` `=[angular momentum of the system]_(inital)` `rArr mvxxL = I_0xxomega .....(i)` where `I_0` is the moment of inertia of the wooden log - bullet system after collision aobut O `I_0 = I_(wooden log) + I_(bullet)` `=(1)/(3)ML^2 +ML^2....(ii)` From (i) and (ii) `omega = (mvxxL)/([(1)/(3)ML^2 +mL^2]` `rArr oemga = (mv)/([ML)/(3) + mL)]` = (3mv)/((M+3m))L)` |
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| 509. |
A uniform disc of radius `R = 2 sqrt3 m` is moving on a horizontal surface without slipping. At some instant its angular velocity is `omega = 1` rad/s and angular acceleration is a `alpha = sqrt3 "rad"//s^(2)`. (a) Find acceleration of the top point A. (b) Find acceleration of contact point B. (c) Find co - oridnates `(r , theta)` for a point P which has zero acceleration |
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Answer» Correct Answer - (a) `sqrt(156) ms^(-2)` (b) `2 sqrt3 ms^(-2)` (c) `(3 , pi//6)` |
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| 510. |
A particle moves in a circle of radius r with angular velocity `omega`. At some instant its velocity is v radius vector with respect to centre of the circel is r.At this particular instant centripetal acceleration `a_(c)` of the particle would beA. `omegaxxv`B. `vxxomega`C. `omegaxx(omegaxxr)`D. `vxx(rxxomega)` |
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Answer» Correct Answer - A::C |
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| 511. |
A uniform bar of length `6a` and mass `8m` lies on a smooth horizontal table. Two point masses `m` and `2m` moving in the same horizontal plane with speeds `2v` and `v`, respectively, strike the bar (as shown in the figure) and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity by `omega`, `E` and `V_(C)`, respectively, we have after collision A. `v_(c)=0`B. `omega=(3v)/(5a)`C. `omega=(v)/(5a)`D. `E=(mv^(2))/(5)` |
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Answer» Correct Answer - A::C::D |
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| 512. |
Two astronauts having mass of `75 kg` and `50 kg` are connected by a rope of length `L = 10m` and negligible mass. They are in space, orbiting their centre of mass at an angular speed of `omega_(0) = 5`rad/s. The centre of mass itself is moving uniformly in space at a velocity of `10 m//s`. By pulling on the rope, the astronauts shorten the distance between them to `(L)/(2) = 5m` . How much work is done by the astronauts in shortening the distance between them ? Assuming that the astronauts are athletic and each of them can generate a power of 500 watt, is it possible for the two astronauts to reduce the distance between them to `5 m`, within a minute? |
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Answer» Correct Answer - 112.5 KJ ; No. |
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| 513. |
The radius of gyration of a disc about an axis coinciding with a tangent in its plane isA. `(R )/(2)`B. `(R )/(sqrt(2))`C. `sqrt(2)R`D. `(sqrt(5))/(2)R` |
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Answer» Correct Answer - D `K_(t)=sqrt((5)/(4)), R = (sqrt(5))/(2) R`. |
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| 514. |
The angular speed of a body changes from `omega_1` to `omega_2` without applying a torque but due to change in its moment of inertia. The ratio of radii of gyration in the two cases is :-A. `omega_(2):omega_(1)`B. `omega_(2)^(1//2):omega_(1)^(1//2)`C. `omega_(2)^(2):omega_(1)^(2)`D. `(1)/(omega_(2)):(1)/(omega_(1))` |
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Answer» Correct Answer - B `I_(1)omega_(1) = I_(2)omega_(2)`. |
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| 515. |
The moment of inertia of a body is `2.5 kg m^(2)`. Calculate the torque required to produce an angular acceleration of `18 rad s^(-2)` in the body. |
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Answer» Correct Answer - 45 Nm |
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| 516. |
The torque is a physical quantity which causes to produceA. property of a bodyB. linear motion of the bodyC. rotational motion of the bodyD. rolling motion of the body |
| Answer» Correct Answer - C | |
| 517. |
The dimensional formula of torque is same as that ofA. powerB. angular momentumC. impulseD. kinetic energy |
| Answer» Correct Answer - D | |
| 518. |
A stone of mass m tied to the end of a string, is whirled around in a horizontal circle. (Neglect the force due to gravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by `T = Ar^n` where A is a constant, r is the instantaneous radius of the circle and n=....A. `1`B. ` -1`C. ` -2`D. ` -3` |
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Answer» `L=mvrimpliesv=(L)/(mr)` `T=(mv^(2))/(r )=(m)/(r )((L)/(mr))^(2)=(L^(2))/(mr^(3))=(L^(2))/(m)r^(-3)=Ar^(n)` `n=-3` |
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| 519. |
A stone of mass m tied to the end of a string, is whirled around in a horizontal circle. (Neglect the force due to gravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by `T = Ar^n` where A is a constant, r is the instantaneous radius of the circle and n=.... |
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Answer» Angular momentum of store is constant. `Iomega = constant =K` `mr^(2)omega=Kimplies omega=(k)/(mr^(2))` Tension provides necessary centripetal force `T=momega^(2)r=m((k)/(mr^(2)))^(2)r=(K^(2))/(m)r^(-3)=Ar^(n)` `n=-3` |
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| 520. |
In the figure shown a smooth ring is connected to rod AB, while rod CD passes through ring. At the given instant angular velocity of rod AB about hinge A is 1 `rod//s` and AC=CB. Instantaneous angular velocity of rod CD about hinge C is A. `1rad//s`B. `1//2rad//s`C. `sqrt(3//2)rad//s`D. `3//2 rad//s` |
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Answer» Correct Answer - D |
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| 521. |
A rod of length l slided down along the inclined wall as shown in figure. At the instant shown in figure, the speed of end A is v, then the speed of B will be A. `(vsinbeta)/(sinalpha)`B. `(vsinalpha)/(sinbeta)`C. `(vcosbeta)/(cosalpha)`D. `(vcosalpha)/(cosbeta)` |
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Answer» Correct Answer - C |
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| 522. |
An aeroplane travelling at a speed of `500 km h^(-1)` tilts at an angle of `30^(@)` as it makes a turn. What is the radius of the curve? |
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Answer» Correct Answer - `3.41 xx 10^(3)m` |
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| 523. |
For traffic moving at `60 kmh^(-1)`, if the radius of the curve is `0.1 km`, what is the correct angle of banking of the road? Take `g = 10ms^(-2)`. |
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Answer» Correct Answer - `15.5^(@)` |
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| 524. |
The moment of inertia of a uniform rod of length `2l` and mass `m` about an axis `xy` passing through its centre and inclined at an enable `alpha` is A. `(ml^(2))/(3)sin^(2)alpha`B. `(ml^(2))/(12)sin^(2)alpha`C. `(ml^(2))/(6)cos^(2)alpha`D. `(ml^(2))/(2)cos^(2)alpha` |
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Answer» Correct Answer - A |
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| 525. |
A thin rod of mass `m` and length `2L` is made to rotate about an axis passing through its center and perpendicular to it. If its angular velocity changes from `O` to `omega` in time `t`, the torque acting on it isA. `(mL^(2)omega)/(12t)`B. `(mL^(2)omega)/(3t)`C. `(mL^(2)omega)/(t)`D. `(4mL^(2)omega)/(6t)` |
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Answer» `I=(m(2L)^(2))/(12)=(mL^(2))/(3)` `tau Delta t=Delta L= I Delta omega` `tau t=(mL^(2))/(3)(omega-0)implies tau=(mL^(2)omega)/(3t)` |
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| 526. |
A uniform rod of mass M and length L is hinged at its end to a wall so that it can rotate freely in a horizontal plane. When the rod is perpendicular to the wall a constant force F starts acting at the centre of the rod in a horizontal direction perpendicular to the rod. The force remains parallel to its original direction and acts at the centre of the rod as the rod rotates. (Neglect gravity). (a) With what angular speed will the rod hit the wall ? (b) At what angle `theta` (see figure) the hinge force will make a `45^(@)` angle with the rod ? |
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Answer» Correct Answer - (a) `sqrt((3F)/(ML))` (b) `theta = tan^(-1) ((1)/(10))` |
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| 527. |
A uniform rod of mass M and length `l` is hinged at point O and is free to rotate on a horizontal smooth surface. Point O is at a distance of `(l)/(4)` from one end of the rod. A sharp impulse `P_(0) = 2 sqrt(130) kg m//s` is applied along the surface at one end of the rod as shown in figure `[tan theta = (9)/(7)]` (a) Find the angular speed of the rod immediately after the hit (b) Find the impulse on the rod due to the hinge. |
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Answer» Correct Answer - (a) `omega = (36P_(0) cos theta)/(7ml) = (36)/(mL)` (b) `P = P_(0) sqrt(sin^(2) theta + ((2)/(7) cos theta)^(2)) = sqrt(85 )N- s` |
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| 528. |
A uniform quarter circular thin rod of mass M and radius R is pivoted at a point B on the floor. It can rotate freely in the vertical plane about B. It is supported by a smooth vertical wall at its other free end A so that it remains at rest. Find the reaction force of wall on the rod. |
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Answer» Correct Answer - `N = (1 - (2)/(pi)) Mg` |
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| 529. |
A circular disc of moment of inertia `I_(t)` is rotating in a horizontal plane about its symmetry axis with a constant angular velocity `omega_(i)`. Another disc of moment of inertia `I_(b)` is dropped co-axially onto the rotating disc. Initially, the second disc has zero angular speed. Eventually, both the discs rotate with a constant angular speed `omega_(f)`. Calculate the energy lost by the initially rotating disc due to friction.A. `(1)/(2)(I_(b)I_(t))/(I_(t)+I_(b))omega_(1)^(2)`B. `(1)/(2)(I_(b)^(2))/(I_(t)+I_(b))omega_(1)^(2)`C. `(1)/(2)(I_(t)^(2))/(I_(t)+I_(b))omega_(1)^(2)`D. `(I_(b)-I_(t))/((I_(t)+I_(b)))omega_(1)^(2)` |
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Answer» `I_(t)omega_(1)=(I_(t)+I_(b))omegaimpliesomega=(I_(t)omega_(1))/((I_(t)+I_(b)))` `DeltaK=(1)/(2)I_(t)omega_(1)^(2)-(1)/(2)(I_(t)+I_(b))((I_(t)omega_(1))/(I_(t)+I_(b)))^(2)` `=(1)/(2)(I_(t)I_(b))/((I_(t)+I_(b)))omega_(1)^(2)` |
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| 530. |
A homogeneous rod AB of length L = 1.8 m and mass M is pivoted at the center O in such a way that it can rotate it can rotate freely position. An insect S of the same mass M falls vertically with speed V on the point C, midway between the points O and B. Immediately after falling, the insect moves towards the end B such that the rod rotates with a constant angular velocity omega. (a) Determine the angular velocity `omega` in terms of V and L. (b) If the insect reaches the end B when the rod has turned through an angle of `90^@`, determine V. |
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Answer» Let the angular velocity be `omega`, when the insect strikes the rod at `C`. By the conservation `OhatI` angular momentum about `C` `Mv(L)/(4)=I_(0)omega=[(ML^2)/(12)+M((L)/(4))^(2)]omega` `=(7)/(48)ML^(2)omega` `omega=(12v)/(7L)` |
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| 531. |
A uniform rod of mass M and length L is hinged at its lower end so as to rotate freely in the vertical plane of the fig. There is a small tight fitting the hinged end. A small mass less pin welded to the rod supports the bead. The system is released from the vertical position shown. It was observed that the bead just begins to slide on the rod when the rod becomes horizontal. (a) Find the normal contact force between the rod and the bead when the rod gets horizontal. What is the direction of this force? (b) Find the coefficient of friction between the bead and rod. |
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Answer» Correct Answer - (a) `(2Mg)/(123)` Vertically down (b) `mu = 22.5` |
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| 532. |
A uniform rod of mass m and length `l` pivoted at one of its top end is hanging freely in vertical plane. Another identical rod moving horizontally with velocity v along a line passing through its lower end hits it and sticks to it. The two rods were perpendicular during the hit and later also they remain perpendicularly connected to each other. Find the maximum angle turned by the two-rod system after collision. |
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Answer» Correct Answer - `cos^(-1){(3)/(1sqrt(10))(1-(v^(2))/(5gl))}+cos^(-1)((3)/(sqrt(10)))` |
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| 533. |
A horizontal disc of radius R and mas 20 M is pivoted to rotate freely about a vertical axis through its centre. A small insect A of mass M and another small insect B of mass `m (M)/(4)` are initially at diametrically opposite points on the periphery of the disc. The whole system is imparted an angular speed `omega_(0)` Insect A walks along the diameter with constant velocity v relative to the disc unit it reaches B which remains at rest on the disc. A then eats B and returns to its starting point along the original path with same speed v relative to the disc. (a) Find the angular speed of the disc when A reaches the centre after eating B. (b) Plot approximately, the variation of angular speed of the disc with time for the entire journey of the insect A. |
| Answer» Correct Answer - (a) `(9)/(8) omega_(0)` | |
| 534. |
A ring is made to rotate about its diameter at a constant angular speed of `omega_(0)` . A small insect of mass m walks along the ring with a uniform angular speed `omega` relative to the ring (see figure). Radius of the ring is R. (a) Find the external torque needed to keep the ring rotating at constant speed as the insect walks. Express your answer as a function of `theta`. For what value of q is this torque maximum? [given your answer for `0 le theta le 90^(@)` ] (b) Find the component of force perpendicular to the plane of the ring, that is applied by the ring on the insect. For what value of `theta` is this force maximum? Argue quantitatively to show that indeed the force should be maximum for this value of `theta`. [Give your answer for `0^(@) le theta le 90^(@)` ] |
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Answer» Correct Answer - `tau = mR^(2) omega_(0) omega sin 2 theta`; `tau` is maximum for `theta = 45^(@)` (b) `F _|_ = 2 m R omega_(0) omega cos theta; theta = 0^(@)` |
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| 535. |
A uniform rod of mass m and length `l` has been placed on a smooth table A particle of mass m, travelling perpendicular to the rod, hits it at a distance `x = (l)/(sqrt6)` from the centre C of the rod. Collision is elastic. (a) Find the speed of the centre of the rod and the particle after the collision. (b) Do you think there is a chance of second Collision? If yes, how is the system of particle and stick moving after the second collision? |
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Answer» Correct Answer - Both have speed `(u)/(2)` (b) The stick is at rest. Particle moves in original direction with speed. |
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| 536. |
A uniform thin cylindrical disk of mass M and radius R is attaached to two identical massless springs of spring constatn k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane. the unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity `vecV_0 = vacV_0hati.` The coefficinet of friction is `mu.` The centre of mass of the disk undergoes simple harmonic motion with angular frequency `omega` equal to -A. `sqer((k)/(M))`B. `sqrt((2)/(M))`C. `sqrt((2k)/(3M))`D. `sqrt((4k)/(3M))` |
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Answer» Correct Answer - D (d) As derived in ans 4. `|F_(net)|=(4k)/(3)x` `For S.H.M. |F_(net)| = Moemga^2_x` `:. Momega^2 = (4k)/(3) rArr omega= sqrt(4k)/(3M) …..(iii)` |
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| 537. |
A uniform thin cylindrical disk of mass M and radius R is attaached to two identical massless springs of spring constatn k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane. the unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity `vecV_0 = vacV_0hati.` The coefficinet of friction is `mu.` The maximum value of `V_0` for whic the disk will roll without slipping is-A. `mugsqrt((M)/(k))`B. `mugsqrt((M)/(2k))`C. `mugsqrt((3M)/(k))`D. `mugsqrt((5M)/(2k))` |
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Answer» Correct Answer - C (c ) From (i) & (ii) `rArr - 2kx + f=-2f rArr f=(2k)/(3)xx x` We see that the frictional force depends on x. As x increases, f increases. Also, the frictional force is maximum at x=A where ?A is the amplitude of S.H.M. Therefore the maximum frictional force `f_(max) = (2k)/(3)xxA` The force should be utmost equal to the limiting friction (muMg) for rolling without slipping. `:.muMg = (2k)/(3)xxA ....(iv)` For S.H.M. Velocity amplitude `= A omega :. V_0 = Aomega` `:. V_o = (3muMg)/(2k) omega from (iv)` `:. V_o = (3muMg)/(2k)xxsqrt((4k)/(3M)) from (iii)` `:. V_o = mugsqrt((3M)/(k))` |
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| 538. |
A hollow cylindrical pipe A has mass M and radius R. With the help of two identical springs (each of force constant k) it is connected to solid cylinder B having mass M and radius R. The springs are connected symmetrically to the axle of the cylinders. Moment of inertia of the two Bodies A and B about their axles are `I_(A) = MR^(2)` and `I_(B) = (1)/(2) MR^(2)` respectively. Cylinders are pulled apart so as to stretch the springs by x0 and released. During subsequent motion the cylinders do not slip. (a) Find acceleration of the centre of mass of the system immediately after it is released. (b) Find the distance travelled by cylinder A by the time it comes to rest for the first time after being released. |
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Answer» Correct Answer - (a) `(kx_(0))/(6M)` (b) (6x_(0))/(7)` |
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| 539. |
A hollow pipe of mass `M = 6 kg` rests on a plate of mass `m = 1.5 kg` . The thickness of the pipe is negligible. The coefficient of friction at all contacts is `mu = 0.2`. The system is initially at rest. A horizontal force F of magnitude 25N is applied on the plate as shown in figure. Will the cylinder slide on the plate? Find the acceleration of the centre of the cylinder. |
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Answer» Correct Answer - No, `(10)/(9) m//s^(2)` |
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| 540. |
A wire of linear mass density `lambda (kg//m)` is bent into an arc of a circle of radius R subtending an angle `2 theta_(0)` at the centre. Calculate the moment o inertia of this circular arc about an axis passing through its midpoint (M) and perpendicular to its plane. |
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Answer» Correct Answer - `I = 4 lambda R^(3) (theta_(0) - sin theta_(0))` |
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| 541. |
Four particles each of mass m are kept at the four corners of a square of edge a. Find the moment of inertia of the system about an axis perpendicular to the plane of the system and passing through the centre of the square. |
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Answer» Correct Answer - `2 ma^(2)` |
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| 542. |
Determine the moment of inertia of the shaded area about `y` axis. The mass of the shaded area is `M`. |
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Answer» Correct Answer - `(ML^(2))/(6) ((h_(1) + 3h_(2))/(h_(1) + h_(2)))` |
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| 543. |
A body of mass 2 kg and radius of gyration 0.5 m is rotating about an axis. If its angular speed is 2 rad/s, then the angular momentum of the body will beA. `1kg m^(2)//s`B. `3kgm^(2)//s`C. `2kg m^(2)//s`D. `4 kg m^(2)//s` |
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Answer» Correct Answer - A `L = I omega = mK^(2)omega` |
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| 544. |
A particle of mass M and radius of gyration K is rotating with angular acceleration `alpha`. The torque acting on the particle isA. `(1)/(2)mK^(2)alpha`B. `(1)/(4)mK^(2)alpha`C. `2 mK^(2)alpha`D. `mK^(2)alpha` |
| Answer» Correct Answer - D | |
| 545. |
Three identical solid discs, each of mass M and radius R, are arranged as shown in figure. The moment of inertia of the system about an axis AB will be A. `(11)/(4)MR^(2)`B. `(15)/(4)MR^(2)`C. `(13)/(4)MR^(2)`D. `(21)/(5)MR^(2)` |
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Answer» Correct Answer - B `I=I_(1)+I_(2)+I_(3)` `=(5)/(4)MR^(2)+(5)/(4)MR^(2)+(5)/(4)MR^(2)=(15MR^(2))/(4)`. |
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| 546. |
Two metal discs, one with radius `r` and mass `m` and the other with radius `2r` and mass `2m`, are welded together and mounted on a frictionless axis through their common center. (a) What is the total moment of inertia of the two discs? (b) A light string is wrapped around the edge of the smaller disc and a block of mass `m_(0)` is suspended from the free end of the string. If the block is released from rest at a distance `h`, what is its speed just before it strikes the floor? |
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Answer» (`a`) `I_(1)=(1)/(2)mr^(2), I_(2)=(1)/(2)xx2mxx(2r)^(2)=4mr^(2)` `I=I_(1)+I_(2)=(9mr^(2))/(2)` (`b`) Loss in the gravitational potential of block `=` gain in the kinetic energy of (block `+` disc) `m_(0)gh=(1)/(2)m_(0)v^(2)+(1)/(2)Iomega^(2)` `v=romegaimplies omega= (v)/(r)` `=(1)/(2)m_(0)v^(2)+(1)/(2)xx(9mr^(2))/(2)xx(v^(2))/(r^(2))` `((1)/(2)m_(0)+(9)/(4)m)v^(2)=((2m_(0)+9m)/(4))v^(2)` `v=sqrt((4m_(0)gh)/(9m+2m_(0)))` |
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| 547. |
A uniform rod of length `l` and mass `m` is free to rotate in a vertical plane about `A` as shown in Fig. The rod initially in horizontal position is released. The initial angular acceleration of the rod is A. `(3g)/(2 l)`B. `(2 l)/(3 g)`C. `(3g)/(2 l^(2))`D. `"mg" (I)/(2)` |
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Answer» Correct Answer - a |
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| 548. |
A uniform rod of mass 4m and length 2r can rotate in horizontal plane about a vertical axis passing through its centre O. Two small balls each of mass m are attached to its ends. A fixed gun fires identical balls with speed v in horizontal direction. The firing is being done at suitable intervals so that the fired balls either hit the ball at end A or B while moving in the direction of velocity of A or B. All collisions are elastic. (i) Initial angular velocity of the rod is zero and its angular velocity after `n^(th)` collision is `omega_(n)` Write `omega_(n + 1)` in terms of `omega_(n)` (ii) Solve the above equation to get `omega_(n) (iii) Find the limiting value of `omega`. |
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Answer» Correct Answer - (i) `omega_(n+1) = (6v)/(13r) + (7)/(13)omega_(n)` (ii) `omega_(n) = (v)/(r) [1-((7)/(13))^(n-1)]` (iii) `(v)/(r)` |
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| 549. |
A uniform rod of mass M and length 2L lies on a smooth horizontal table. There is a smooth peg O fixed on the table. One end of the rod is kept touching the peg as shown in the figure. An impulse J is imparted to the rod at its other end. The impulse is horizontal and perpendicular to the length of the rod. Find the magnitude of impulse experienced by the peg. |
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Answer» Correct Answer - `(J_(0))/(2)` |
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| 550. |
A L shaped, uniform rod has its two arms of length `l` and `2l`. It is placed on a horizontal table and a string is tied at the bend. The string is pulled horizontally so that the rod slides with constant speed. Find the angle `theta` that the longer side makes with the string. Assume that the rod exerts uniform pressure at all points on the table. |
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Answer» Correct Answer - `theta = pi - tan^(-1) ((1)/(4))` |
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