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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
In `Q.137`, if we hang a body of mass `m` from the cord, the tangential acceleration of the disc is :A. `(mg)/(M + m)`B. `(mg)/(M + 2m)`C. `(2 mg)/(M + 2m)`D. `(M + 2m)/(2 mg)` |
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Answer» Correct Answer - C ( c) From above problem, `T = (Ma)/(2)` Now `mg - T ma` or `mg - M (a)/(2) = ma` or `mg = (m + (M)/(2)) a` or `a = (2 mg)/(M + 2 m)`. |
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| 52. |
A fruit vendor in Indra Vihar is selling frits by a faulty balance. He uses correct weight, but he has tied a thread of mass `100 gm` at a distance of `4 cm` from the hinge as shown. Your buy `250 gm` of grapes. How many `gm` grapes do you actually get ? . |
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Answer» Correct Answer - `0210` Balancing moment of force about hindge `250 xx g xx 10 = 100 xx g xx 4 + m xx g xx 10` `m = 210 gm`. |
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| 53. |
A mass `m` is rest on an inclined plane of mass `M` which is further resting on a smooth horizontal plane. Now if the mass starts moving the position of `C.M`. Of mass of system will : .A. remain unchangedB. change along the horizontalC. move up in the vertical directionD. move down in the vertical direction the vertical along the horizontal. |
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Answer» Correct Answer - C ( c) The mass move under the influence of gravitational pull which acts along the vertical. Thus `CM` changes along vertical while it remains unchanged in the horizontal direction. |
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| 54. |
An insulated particle of mass `m` is moving in a horizontal plane (x - y) along X-axis. At a certain height above the ground, it suddenly explodes into two fragments of masses `m//4 and 3 m//4`. An instant later, the smaller fragment is at `Y = + 15`. The larger fragment at this instant is at :A. `Y = -5 cm`B. `Y = + 20 cm`C. `Y = + 5 cm`D. `Y = -20 cm` |
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Answer» Correct Answer - A (a) Initially Y-coordinate of centre of mass is zero. After explosion Y coordinate of centre of mass of the particles should be zero. `(m)/(4)xx(15) -(3)/(4)m(Y) = 0 rArr Y = -5 cm`. |
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| 55. |
A uniform square plate has a small piece `Q` of an irregular shape removed and guled to the centre of the plate leaving a hole behind in figure. The moment of inertia about the z-axis is then, .A. increasedB. decreasedC. the sameD. changed in unpredicted manner |
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Answer» Correct Answer - B (b) In the given diagram, when the small piece `Q` removed and glued to the centre of the plate, the mass comes closer to the z-axis, hence moment of interia decreases. |
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| 56. |
For inelastic collsion between two spherical rigid bodiesA. the total kinetic energy is conservedB. the total mechanical energy is not conservedC. the linear momentum is not conservedD. the linear momentum is conserved |
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Answer» Correct Answer - D (d) In an inelastic collision, the particles do not region their shape and size completely after collision. Thus, the kinetic energy of particles no longer remains conserved. However, in the absence of external forces, law of conservation of linear momentum still holds good. |
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| 57. |
Two circular discs `A` and `B` of equal masses and thicknesses. But are made of metals with densities `d_A and d_B (d_A gt d_B)`. If their moments of inertia about an axis passing through the centre and normal to the circular faces be `I_A and I_B`, then.A. `I_A = I_B`B. `I_A gt I_B`C. `I_A lt I_B`D. `I_A ge I_B` |
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Answer» Correct Answer - C ( c) mass = volume `xx` density `rArr M = (pi R^2 t)d`…(1) where `t` is thickness `I = (MR^2)/(2)` …(2) From (1) and (2), we get : `I = (M^2)/(2 pi dt) rArr I prop (1)/(d) rArr I_A lt I_B`. |
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| 58. |
Analogue of mass in rotational motion is.A. moment of inertiaB. torqueC. radius of gyrationD. angular momentum |
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Answer» Correct Answer - A (a) Analogue of mass in rotational motion is moment of inertia. |
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| 59. |
Assertion: In an elastic collision of two bodies , the momentum and energy of each body is conserved. Reason: If two bodies stick to each other , after colliding , the collision is said to be perfectly elastic.A. If both assertion and reason are true and reason is the true explanation of assertion.B. If both assertion and reason are reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - D (d) In an elastic collision both the momentum and kinetic energy remains conserved. But this rule is not for individual bodies, but for the system of bodies before and after the collision. While collision in which there occurs some loss of kinetic energy is called inelastic collision. Collision in daily life is generally inelastic. The collision is said to be perfectly inelastic, if two bodies stick to each other. |
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| 60. |
The moment of inertia of a disc of mass `M` and radius `R` about an axis. Which is tangential to sircumference of disc and parallel to its diameter is.A. `(5)/(4)M R^2`B. `(2)/(3) MR^2`C. `(3)/(2) MR^2`D. `MR^2` |
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Answer» Correct Answer - A (a) `I = I_(cm) = Mh^2 + ((1)/(4)) MR^2 + MR^2 = ((5)/(4)) MR^2`. |
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| 61. |
A horizontal platform is rotating with uniform angular velcity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass `m` is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period :A. decreases continouslyB. remains unlteredC. decreases initially and increases againD. increases continuously |
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Answer» Correct Answer - C ( c) From law of conservation of angular momentum if no external torque is acting upon a body rotating about an axis, then the angular momentum of the body remains constant that is `J = I omega = constant` Hence, if decreases, `omega` increases and vice-versa. When liquid is dropped, mass increase hence `I` increases `(I = mr^2)` So, `omega` decreases, but as soon as the liquid starts falling `omega` increases again. |
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| 62. |
A rigid object is rotating in a counterclock wise sense around a fixed axis. If the rigid object rotates through more than `180^@` but less than `360^@`, which of the following pairs of quantities can represent and initial angular position and a final angular position of the rigid object -A. `3 rad, 6 rad`B. `-1 rad, 1 rad`C. `1 rad, 5 rad`D. `- 1 rad, 2.5 rad` |
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Answer» Correct Answer - C::D `pi lt Delta theta lt 2 pi` `3.14 rad lt Delta theta lt 6.26 rad`. |
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| 63. |
A hollow sphere of radius `R` and mass `m` fully filled with water of mass `m`. It is rolled down a horizontal plane such that its centre of mass moves with a velocity `v`. If it purely rolls.A. Kinetic energy of the sphere is `(5)/(6) mv^(2)`B. Kinetic energy of the sphere is `(4)/(5) mv^(2)`C. Angular momentum of the sphere about a fixed point on ground is `(8)/(2) mvR`D. Angular momentum of the sphere about a fixed point ground is `(14)/(5) mvR`. |
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Answer» Correct Answer - C `L = I omega + 2 mvR` `L = ((2)/(3) MR^(2)) xx (V)/(R) + 2mvR rArr L = (8)/(3) mvR`. |
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| 64. |
What will be the position of centre of mass of a half disc shown ? .A. `(2 a)/(pi)`B. `(4 a)/(3 pi)`C. `(a)/(pi)`D. `(2 a)/(3 pi)` |
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Answer» Correct Answer - B (b) Fact that `C.O.M.` of half disc is at distance of `(4 a)/(3 pi)` from centre. |
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| 65. |
An cylinder of mass `m` is rotated about its axis by an angular velocity `omega` and lowered gently on an inclined plane as shown in figure. Then : .A. it will start going upwardB. it will first going upward and then downwardC. it will go downward just after it is loweredD. it can never go upward. |
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Answer» Correct Answer - D (d) Since net force along the incline is zero, so cylinder will remain in position till it stops rotating. After that it will start moving downwards. |
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| 66. |
A playground merry-go-round of radius `R = 2.00` has a moment of inertia `I = 250 kg.m^2` and is rotating `10.0 rev//min` about a frictionless, vertical axle. Facing axle, a `25.0-kg` child hops onto the merry-go-round and manages to sit down on the edge. The new angular speed of the merry-go-round is.A. `5.25 rev//min`B. `8.45 rev//min`C. `7.14 rev//min`D. `3.14 rev//min` |
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Answer» Correct Answer - C ( c) From conservation of angular momentum, `I_i omega_i = I_f omega_f : (250 kg.m^2)(10.0 rev//min)` =`[250 kg.m^2 + (25.0 kg)(2.0 m)^2] omega_2` `omega_2 = 7.14 rev//min`. |
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| 67. |
The moment of inertia of a hollow cubical box of mass `M` and side `a` about an axis passing through the centres of two opposite faces is equal to.A. `(5 M a^2)/(3)`B. `(5 M a^2)/(6)`C. `(5 M a^2)/(12)`D. `(5 M a^2)/(18)` |
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Answer» Correct Answer - D (d) Taking mass of plate `m = (ma^2)/(6) xx 2 = (ma^2)/(3)` Then `MI` of two plates through which the axis is passing. `M.I` of `4` plates having symmetrical position from the axis. =`4 xx[(m a^2)/(12) + m((a)/(2))^2] = 4 xx[(ma^2)/(3)]` Total `MI = (4 m a^2)/(3) + ( ma^2)/(3) = (5m a^2)/(3)` using `(M)/(6) = m = MI = (5 M a^2)/(18)`. |
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| 68. |
An solid cylinder of mass `20 kg` and radius `20 cm` rotates about its axis with a angular speed `100 rad s^-1`. The angular momentum of the cylinder about its axis is.A. 40 J sB. 400 J sC. 20 J sD. 200 J s |
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Answer» Correct Answer - A (a) Here, `M = 20 kg` `R = 20 cm = 20 xx 10^-2 m, omega = 100 rad s^-1` Moment of inertia of the solid cylinder about its axis is. `I = (MR^2)/(2) = ((20 kg)(20 xx 10^-2 m)^2)/(2) = 0.4 kg m^2` Angular momentum of the cylinder about axis is. `L = I omega = (0.4 kg m^2)(100 rad s^-1) = 40 J s`. |
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| 69. |
A uniform solid disk of mass `m = 3.0 kg` and radius `r = 0.20 m` rotates about a fixed axis perpendicular to its face with angular frequecy `0.6 rad/s`. The magnitide of the angular momentum of the disk when the axis of rotation momentum of the disk when the axis of rotation passes through a point midway between the centre and the rim is.A. `0.72 kg.m^2//s`B. `0.54 kg.m^2//s`C. `0.36 kg.m^2//s`D. `1.08 kg.m^2//s` |
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Answer» Correct Answer - B (b) For a point midway between the centre and the rim, we use the parallel-axis theorem to find the moment of inertia about this point. Then, `L = I omega = [(1)/(2) MR^2 + M((R)/(2))^2] omega` =`(3)/(4) (3.00 kg)(0.200 m)^2 (6.00 rad//s)` =`0.540 kg . m^2//s`. |
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| 70. |
A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the centre of the disk. The platform has a mass of `150 kg`, a radius of `2.0 m`, and a rotational inertia of`300 kg m^2` about the axis of rotation. A `60 kg` student walks slowly from the rim of the platform toward the centre. If the angular speed of the system is `1.5 rad//s` when the student starts at the rim, what is the angular speed when she is `0.50 m` from the centre ?A. `1.2 rad//s`B. `2.6 rad//s`C. `1.5 rad//s`D. `3.6 rad//s` |
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Answer» Correct Answer - B (b) The initial rotational inertia of the system is `I_i = I_(disk) + I_(student)`, where `I_(disk) = 300 kg m^2` (which, incidentally does agree with Table `10-2`(c )) and `I_(student) = mR^2` where `m = 60 kg` and `R = 2.0 m` The rotational inertia when the student reaches `r = 0.5 m` is `I_f = I_(disk) + mr^2`. Angular momentum conservation leads to `I_i omega_i = I_f omega_f rArr omega_f = omega_i ((I_(disk) + mR^2)/(I_(disk) + mr^2))` Which yields, for `omega_i = 1.5 rad//s`, a final angular velocity of `omega_f = 2.6 rad//s`. |
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| 71. |
A body A of mass M while falling wertically downwards under gravity brakes into two parts, a body B of mass `(1)/(3)` M and a body C of mass `(2)/(3)` M. The center of mass of bodies B and C taken together shifts compared to that of body A towardsA. Body CB. Body BC. Depends on height of breakingD. Does not shift |
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Answer» Correct Answer - D (d) Since there is no external force, so centre of mass does not shift compared to `A`. |
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| 72. |
In previous problem, what is the reaction force on the hinge ?A. `W - P` upwardB. `W - P` downwardC. `W - P` to the leftD. None of these |
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Answer» Correct Answer - B (b) `R + P = W rArr R = W - P` This is the reaction on rod upward direction, so from third law, reaction on hinge will be `R` in downward direction. |
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| 73. |
A particle of mass 0.1 kg is subjected to a force which varies with distance as shown in figure. If it starts its journey from rest at `x=0`, its velocity at `x=12m` isA. `0 m//s`B. `20 sqrt(2) m//s`C. `20 sqrt(3) m//s`D. `40 m//s` |
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Answer» Correct Answer - D (d) Area between curve and displacement axis =`(1)/(2) xx (12 + 4) xx 10 = 80 J` In this time the body acquired kinetic energy `= (1)/(2) mv^2` by the law of conservation of energy `(1)/(2) mv^2 = 80 J` `rArr (1)/(2) xx 0.1 xx v^2 = 80` `rArr v^2 = 1600` `rArr v = 40 m//s`. |
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| 74. |
In a bicycle the radius of rear wheel is twice the radius of front wheel. If `v_F` and `v_r` are the speeds of top most points of front and rear wheels respectively, then :A. `v_r = 2 v_F`B. `v_F = 2 v_r`C. `v_F = v_r`D. `v_F gt v_r` |
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Answer» Correct Answer - C ( c) Speed of centres of both the wheels is same as that of cycle. We know that speed of top point is double that of centre. So speed of top points of both the wheels should be same. Hence `v_F = v_r`. |
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| 75. |
A solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first ?A. solid cylinderB. hollow cylinderC. both will take the same timeD. it cannot be predicted |
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Answer» Correct Answer - A (a) `a = (g sin theta)/(1 + (I)/(MR^2))`, For solid cylinder `I` is less, so `a` is more, hence time taken will be less. |
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| 76. |
In first figure a meter stick, half of which is wood and the other half steel is pivoted at the wooden end at `A` and a force is applied at the steel and at `O`. On second figure the stick is pivoted at the steel end at `O` and the same force is applied at the wooden end at `A`. The angular acceleration. .A. in first is greater than in secondB. equal in both first and secondC. in second is greater than in firstD. None of the above. |
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Answer» Correct Answer - C ( c) Torque will be same in both cases, but moment of inertia is less in second case, hence angular acceleration is greater in this case. |
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| 77. |
A ladders is more likely to slip when a person is near the top than when he is near the bottom. The friction between the ladder and floor decreases as he climbs up.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - C ( c) As the person climbs up, normal reaction and friction between the ladder and the wall both increase. This decrease normal reaction form the floor, decreasing limiting value of friction there. This increases the possibility of the ladder slipping. |
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| 78. |
A ladders is more likely to slip when a person is near the top than when he is near the bottom. The friction between the ladder and floor decreases as he climbs up.A. Statement-1 is True, Statement-2 is Ture , Statement -2 is a correct explanation for statement -1.B. Statement-1 is True, Statement-2 is True , Statement-2 is NOT a correct explanation fro Statement -1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - C As the person climbs up, normal reaction and friciton between the ladder and the wall both increase. This decreases normal reacction from the floor. Decreasing limiting value of friction there. This increases the possibility of the ladder slippling. |
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| 79. |
A couple produces.A. purely translational motionB. purely rotational motionC. bth translational and rotational motionD. no motion |
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Answer» Correct Answer - B (b) A couple produces ratation without translation. |
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| 80. |
An autmobile moves on road with a speed of `54 km//h`. The radius of its wheel is `0.45 m` and the moment of inertia of the wheel about its axis of rotation is `3 kg m^(2)`. If the vehicle is brought to rest in `15 s`, the magnitude of average torque tansmitted by its brakes to the wheel is :A. `2.86` kg `m^2 s^-2`B. `6.66` kg `m^2 s^-2`C. `8.58` kg `m^2 s^-2`D. `10.86` kg `m^(2) s^-2` |
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Answer» Correct Answer - B (b) Velocity of the automobile `v = 54 xx (5)/(18) = 15 m//s` `omega_0 = (v)/(R) =(15)/(0.45) = (100)/(3) rad//s` So angular acceleration `prop = (Delta omega)/(f) =(omega_f - omega_0)/(t) = -(100)/(45) rad//s^2` So, torque `I prop = 3 xx (100)/(45) = 6.66 kg - m^2s^-2`. |
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| 81. |
Figure shows two identical particles `1` and `2`, each of mass `m`, moving in opposite directions with same speed `vec V` along parallel lines. At a particular instant, `vec r_1` and `vec r_2` are their respective position vectors drawn from point `A` which is in the plane of the parallel lines. Which of the following is the correct statement ? .A. Angular momentum `vec L_1` of particle `1` about `A` is `vec L_1 = mv vec r_1 odot`B. Angular momentum `vec L_2` of particle `2` about `A` is `vec L_2 = mv vec r_2 odot`C. Total angular momentum of the system about is `vec L = mv (vec r_1 + vec r_2) odot`D. Total angular momentum of the system about `A` is `vec L = mv(d_2 - d_1) otimes` [Here, `otimes` represents a unit vector going into the page and `odot` represents a unit vector coming out of the page]. |
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Answer» Correct Answer - D (d) Angular momentum of particle `1` about `A` is `vec L_1 = mvd_1 odot` Angular momentum of particle `2` about `A` is `vec L_2 = mvd_2 otimes` :. Total angular momentum of the system about `A` is `vec L = vec L_2 - vec L` and `(because L_1 and L_2` are in opposite directions) `vec L = mv(d_2 - d_1) otimes`. |
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| 82. |
A sphere starting from rest rolls for 5.3s without slipping along a plane which is 1m in length. The upper end of the plane is 1 cm higher than the lower end. Find the accelerationi due to gravity. |
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Answer» Correct Answer - 9.97 m`s^(2)` |
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| 83. |
The figure shows an isosceles triangle plate of mass `M` and base `L`. The angle at the apex is `90^@`. The apex lies at the origin and the base is parallel to `X-`axis. The moment of inertia of the plate about the `y-`axis isA. `(ML^(2))/(6)`B. `(ML^(2))/(8)`C. `(ML^(2))/(24)`D. none of these |
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Answer» Correct Answer - C From `_|_` axis theorem `I_(z) = I_(x) + I_(y)` `I_(y) = I_(z) - I_(x) = (mL^(2))/(6) - (mL^(2))/(8) rArr I_(y) = (mL^(2))/(24)`. |
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| 84. |
If `vec F` is the force acting in a particle having position vector `vec r` and `vec tau` be the torque of this force about the origin, thenA. `vec r. vec tau ne 0 and vec F. vec tau = 0`B. `vec r. vec tau gt 0 and vec F. vec tau lt 0`C. `vec r. vec tau = 0 and vec F. vec tau = 0`D. `vec r. vec tau = 0 and vec F. vec tau ne 0` |
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Answer» Correct Answer - C ( c) Torque is an axial vector i.e., its direction is always perpendicular to the plane containing vectors `vec r` and `vec F`. `vec tau = vec r xx vec F` Torque is perpendicular to both `vec r` and `vec F` :. `vec tau . vec r = 0` `vec F . vec tau = 0`. |
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| 85. |
If there are no external forces, the centre of mass of a double star moves like a free particle. If we go to the centre of mass frame, then we find that the two starts are moving in a circle about the centre of mass, which is at rest.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - B (b) In our frame of reference, the trajectories of the starts are a combination of (i) Uniform motion in a straight line of the centre of mass and. (ii) circular orbits of the starts about the centre of mass. |
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| 86. |
The velocity of the `CM` of a system changes from `vec v_1 = 4 hat im//s to vecv_2 = 3 hatj m//s` during time `Deltat = 2 s`. If the mass of the system is `m = 10 kg`, the constant force acting on the system is :A. 25 NB. 20 NC. 50 ND. 5 N |
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Answer» Correct Answer - A (a) `vec F = m(Delta vec v_(cm))/(Deltat) =(10(3 hat j - 4 hat i))/(2) = 5(3 hat j - 4 hati)` `|vec F| = 5 sqrt(3^2 + 4^2) = 25 N`. |
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| 87. |
The torque `vec tau` on a body about a given point is found to be equal to `vec A xx vec L` where `vec A` is a constant vector and `vec L` is the angular momentum of the body about the point. From this its follows that -A. `(d vec(L))/(dt)` is perpendicular to `vec L` at all instants of time.B. the components of `vec L` in the direction of `vec A` does not change with time.C. the magnitude of `vec L` does not change with timeD. `vec L` does not change with time. |
| Answer» Correct Answer - A::B::C | |
| 88. |
The force `7 hat i+ 3 hat j - 5 hat k` acts on a particle whose position vector is `hat i- hat j + hat k`. What is the torque of a given force about the origin ?A. `2 hat i+ 12 hat j + 10 hat k`B. `2 hat i+ 10 hat j + 12 hat k`C. `2 hat i+10 hat j + 10 hat k`D. `10 hat i+ 2 hat j + hat k` |
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Answer» Correct Answer - A (a) Here, `vec r = hat i - hat j + hat k` `vec F = 7 hat i + 3 hat j - 5 hat k` Torque, `vec tau = vec r xx vec F` `vec tau = |(hat i)/(1/7) (hat j)/(-1/3) (hatk)/(1/(-5))| = hat i (5 - 3) +_hat j (7 -(-5)) + hat k (3-(-7))` or `vec tau = 2 hat i + 12 hat j + 10 hat k`. |
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| 89. |
Given that, `vecr = 2 hat i + 2 hatj` and `vec F = 2 hat i + 6 hat k`. Find the magnitude of torque. |
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Answer» We know that, `vec tau = vecr xx vec F` `rArr vec r(2hat i + 3 hat j) xx (2 hat i + 6 hatk)` =`12(-hatj) + 6 (-hatk) + 18 hat i` =`-12 hat j - 6 hat k + 18 hat i` [Note : `hat ixx hat i= 0, hat ixx hat j = k, hat j xx hat i= -K` etc] Now, `|vectau| = sqrt((-12)^(2) +(-6)^(2) + (18)^(2))` =`sqrt(144 + 36 +324) = sqrt(504)`. |
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| 90. |
A constant torque acting on a uniform circular wheel changes its angluar momentum from `A_0` to `4 A_0` in `4 s`. The magnitude of this torque is equal to.A. `3 A_0//4`B. `4 A_0`C. `A_0`D. `12 A_0` |
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Answer» Correct Answer - A (a) `tau = (Delta L)/(Delta t) = (4 A_0 - A_0)/(4) = (3 A_0)/(4)` |
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| 91. |
A wheel of moment of inertia `2.0 xx 10^3 kgm^2` is rotating at uniform angular speed of `4 rads^-1`. What is the torque required to stop it in one second.A. `0.5 xx 10^3 Nm`B. `8.0 xx 10^3 Nm`C. `2.0 xx 10^3 Nm`D. none of these |
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Answer» Correct Answer - B (b) `prop = (omega)/(t) = (4)/(1) = 4 rad//s^2` `tau = I prop = 2 xx 10^3 xx 4 = 8 xx 10^3 N`. |
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| 92. |
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. |
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Answer» `Delta J = 4 A_(0) - A_(0) = 3A_(0)` and `Delta T = 4` :. `tau = (Delta J)/(Delta T) = (3 A_(0))/(4)`. |
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| 93. |
When a torque acting upon a system is zero, which of the following will be constant ?A. ForceB. Linear impulseC. Linear momentumD. Angular momentum |
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Answer» Correct Answer - D (d) When the total external torque acting on the system zero, then the total angular momentum of the system is conserved. |
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| 94. |
Torque on a body can be zero even if there is a net force on it. Torque and force on a body are always perpendicular.A. If both assertion and reason are true and reason is the true explanation of assertion.B. If both assertion and reason are reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - B (b) |
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| 95. |
A man stands on a rotating platform with his arms stretched holding a `5 kg` weight in each hand. The angular speed of the platform is `1.2 rev s^-1`. The moment of inertia of the man together with the platform may be taken to be constant and equal to `6 kg m^2`. If the man brings his arms close to his chest with the distance `n` each weight from the axis changing from `100 cm` to`20 cm`. The new angular speed of the platform is.A. `2 rev s^-1`B. `3 rev s^-1`C. `5 rev s^-1`D. `6 rev s^-1` |
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Answer» Correct Answer - B (b) Initial moment of inertia, `I_1 = 6 + 2 xx 5 xx (1)^2 = 16 kg m^2` Initial angular velocity, `omega_1 = 1.2 rev s^-1` Initial angular momentum is `L_2 = I_2 omega_2` According to law of conservation of angular momentum, `L_1 = L_2` or `I_1 omega_1 = I_2 omega_2` `omega_2 = (I_1 omega_1)/(I_1) = ((16 kg m^2)(1.2 rev s^-1))/((6.4 kg m^2)) = 3 rev s^-1`. |
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| 96. |
A pulley systemis attached to a massless board as shown below. The board pivots only at the pivot point. A `10 g` mass `M` sits exaxtly in the middle of the board. If the angle `theta` is `60^@`, what is the force `F(i n N)` necessary to hold the board in the position shown. . |
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Answer» Correct Answer - `25 N` Balancing the torque about pivot `2F l = Mg(l)/(2)` `F = (Mg)/(4) rArr F = 25 N`. |
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| 97. |
A straight rod of length `L` is released on a frictionless horizontal floor in a vertical position. As it falls + slips, the distance of a point on the rod the lower end, which follows a quarter circular locus is.A. `L//2`B. `L//4`C. `L//8`D. None |
| Answer» Correct Answer - B | |
| 98. |
A hoop of radius `r` mass `m` rotating with an angular velocity `omega_(0)` is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it cases to slip ?A. `(r omega_(0))/(2)`B. `r omega_(0)`C. `(r omega_(0))/(4)`D. `(r omega_(0))/(3)` |
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Answer» Correct Answer - A `C.O.A.M` (about bottom) `mr^(2) omega_(0) = 2mr^(2) (v_(cm))/( r)` `v_(cm) = (omega_(0) r)/(2)`. |
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| 99. |
Initial angular velocity of a circular disc of mass `M` is `omega_(1)`. Then two small spheres of mass `m` are attached gently to two diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc -A. `((M +m)/(M)) omega_(1)`B. `((M +m)/(m)) omega_(1)`C. `((M )/(M + 4m)) omega_(1)`D. `((M )/(M + 2m)) omega_(1)` |
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Answer» Correct Answer - C Conservation of angular momentum gives `(1)/(2) MR^(2) omega_(1) = ((1)/(2) MR^(2) + 2 MR^(2)) omega_(2)` `rArr (1)/(2) MR^(2) omega_(1) = (1)/(2) R^(2) (M + 4m) omega_(2)` :. `omega` `= ((M)/(M + 4m)) omega_(1)`. |
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| 100. |
A horizonral circular plate is rotating about a vertical axis passing through its centre with an angular velocity `omega_(0)`. A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is `K` intially, its final kinetic energy will beA. 2 KB. `K//2`C. KD. `K//4` |
| Answer» Correct Answer - B | |