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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. |
Three particles, each of mass m are situated at the vertices of an equilateral triangle ABC of side L figure. Find thee moment of inertia of the system about the line AX perpendicular to AB in the plane of ABC |
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Answer» Correct Answer - `(5 mL^(2))/4` |
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| 52. |
From a circular disc of mass `M` and radius `R`, a part of `60^(@)` is removed. The `M.I.` of the remaining portion of disc about an axis passing through the center and perpendicular to plane of disc is A. `(5)/(6)MR^(2)`B. `(5)/(3)MR^(2)`C. `(5)/(12)MR^(2)`D. `(5)/(24)MR^(2)` |
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Answer» Mass of the removed portion `=Mxx(60^(@))/(360^(@))=(M)/(6)` Its `M.I. I_(1)=(1)/(2)(M)/(6)R^(2)` `M.I.` of the complete disc `I_(2)=(1)/(2)MR^(2)` `M.I.` of the remaining portion `=I_(2)-I_(1)=(5)/(12)MR^(2)` |
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| 53. |
A point mass m collides with a disc of mass m and radius R resting on a rough horizontal surface as shown . Its collision is perfectly elastic. Find angular velocity of the disc after pure rolling starts A. `((2u)/(3R))`B. `((3u)/(3R))`C. `((5u)/(3R))`D. `((2u)/(5R))` |
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Answer» Correct Answer - A |
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| 54. |
A solid sphere is kept over a smooth surface as shown in figure. It is hit by a cute at height h above the centre C. If the surface is rough, then after hitting the sphere, in which case the force of friction is in forward direction.A. In case 1B. In case 2C. In both the casesD. In none of the cases |
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Answer» Correct Answer - B |
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| 55. |
A solid sphere is kept over a smooth surface as shown in figure. It is hit by a cute at height h above the centre C. In case 1, `h=(R)/(2)` and in case 1 the sphere acquires a total kinetic energy `k_(1)` and in case 2 total kinetic energy is `k_(2)` . Then,A. `k_(1)=k_(2)`B. `k_(1) gt k_(2)`C. `k_(1)lt k_(2)`D. Data is sufficient |
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Answer» Correct Answer - C |
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| 56. |
A ring of mass 10 kg and diameter 0.4m is rotated about an axis passing through its centre and perpendicular to its plane moment of inertia of the ring isA. `1.4kg m^(2)`B. `2.4kg m^(2)`C. `0.4kg m^(2)`D. `2kg m^(2)` |
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Answer» Correct Answer - C `I = mR^(2)` |
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| 57. |
A circular disc is to be made by using iron and aluminium, so that it acquires maximum moment of inertia about its geometrical axis. It is possible withA. Iron and aluminium layers in alternate orderB. Aluminium at interior and iron surrounding itC. Iron at interior and aluminium surrounding itD. Either (`1`) or (`3`) |
| Answer» Density of iron `gt` density of aluminium, mass of iron `gt` mass of aluminium | |
| 58. |
The moment of inertia of a ring about its geometrical axis is I, then its moment of inertia about its diameter will beA. 1 IB. I/2C. ID. I/4 |
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Answer» Correct Answer - B `I_(d)= (I_(c ))/(2)=(I)/(2)`. |
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| 59. |
An iron disc and a wooden disc have the same radius. Then which will have larger moment of inertia about the diameter ?A. Wooden discB. Iron discC. same for bothD. can not be predicted |
| Answer» Correct Answer - B | |
| 60. |
A rigid large uniform square platform is resting on a flat horizontal ground supported at its vertices by four identical spring. At vertex l a wooden block, 6 cm high, is inserted below the spring. Calculate the change in height of the centre of the platform. Assume change in height to be small compared to dimension of the platform. |
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Answer» Correct Answer - `1.5 cm` |
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| 61. |
A circular disc is to be made by using iron and aluminium, so that it acquires maximum moment of inertia about its geometrical axis. It is possible withA. aluminimum is at the interior and iron surrounds itB. iron is at the interior and aluminimum surrrounds itC. aluminium and iron layers are in alternate orderD. sheet of iron is used at both external surfaces and aluminium sheet as inner material |
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Answer» Correct Answer - a |
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| 62. |
A solid sphere of radius `R` is placed on a smooth horizontal surface. A horizontal force `F` is applied at height `h` from the lowest point. For the maximum acceleration of the centre of massA. `h=R`B. `h=2R`C. `h=0`D. the acceleration will be same whatever `h` may be |
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Answer» Correct Answer - d |
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| 63. |
A uniform solid cylinder of mass 5kg and radius 0.1m is resting on a horizontal platform (parallel to the x-y plane) and is free to rotate about its axis along the y-axis the platform is given a motion in the x direction given by x=0.2 cos (10t) m if there is no slipping then maximum torque acting on the cylinder during its motion isA. `0.2 N-m`B. `2.0 N-m`C. `5.0 N -m`D. `10.0 N-m` |
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Answer» Correct Answer - C |
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| 64. |
A solid sphere of radius R is resting on a smooth horizontal surface. A constant force F is applied at a height h from the bottom. Vhoose the correct alternative. A. Sphere will always slide whatever be the value of hB. Sphere will roll without sliding when `hge104R`C. Sphere will roll without sliding if h=1.4RD. None of the above |
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Answer» Correct Answer - C |
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| 65. |
A ball tied to a string takes us to complete revolution along a horizontal circle. If, by pulling the cord, the radius of the circle is reduceed to half of the previous value, then how much time the ball will take in one revolution? |
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Answer» Correct Answer - 1 s |
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| 66. |
A solid body rotates with angular velocity `vecomega=3thati+2t^(2) hatj rad//s`. Find (a) the magnitude of angular velocity and angular acceleration at time `t=1 s` and (b) the angle between the vectors of the angular velocity and the angular acceleration at that moment. |
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Answer» `vecomega=3thati+2t^(2)hatj` ` vecalpha=(dvecomega)/(dt)=3hati+4thatj` `vecomegacdotvecalpha=9t+8t^(3)` (a) At `t=1 s`, `vecomega=3hati+2hatj` `|vecomega|=omega=sqrt((3)^(2)+(2)^(2))=sqrt13rad//s` `vecalpha=3hati+4hatj` `|vecalpha|=alpha=sqrt((3)^(2)+(4)^(2))=5rad//s^(2)` (b) `vecomegacdotvecalpha=9(1)+8(1)=17` Angle between `vecomega` and `vecalpha` `cos theta=(vecomegacdotvecalpha)/(|vecomega||vecalpha|)=(17)/(sqrt13xx5)` `theta=cos^(-1)((17)/(5sqrt13))` |
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| 67. |
The sun rotates round itself once in 27 days. What will be period of revolution if the sun were to expand to twice its present radius? Assume the sun to be a sphere of uniform density. |
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Answer» Correct Answer - 108 days |
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| 68. |
A disc and a hoop of the same and size roll down on inclined plane starting simultaneously. Then the one which reaches bottom firstA. hoopB. discC. depends upon angle of inclineD. both |
| Answer» Correct Answer - B | |
| 69. |
An aeroplane describes a vertical circle when looping. Find the radius of the greatest possible loop if the velocity of the aeroplane at the lowest point of its path is `50 ms^(-1)`. Take `g = 10 ms^(-2)`. |
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Answer» Correct Answer - 50 m |
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| 70. |
A disc of mass m and radius R is placed over a plank of same mass m. There is sufficient friction between disc and plank to prevent slipping. A force F is applied at the centre of the disc. Acceleration of the plank isA. `(F)/(2m)`B. `(3F)/(4m)`C. `(F)/(4m)`D. `(3F)/(2m)` |
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Answer» Correct Answer - C |
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| 71. |
A disc of mass m and radius R is placed over a plank of same mass m. There is sufficient friction between disc and plank to prevent slipping. A force F is applied at the centre of the disc. Force of friction between the disc and the plank isA. `(F)/(2)`B. `(F)/(4)`C. `(F)/(3)`D. `(2F)/(3)` |
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Answer» Correct Answer - B |
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| 72. |
The acceleration a of the plank P required to keep the centre C of a cylinder in a fixed position during the motion is (no slipping take place between cylinder and plank) A. `(g)/(2)sintheta`B. `2gsintheta`C. `gsintheta`D. `g tantheta` |
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Answer» Correct Answer - B |
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| 73. |
A flywheel rotates with a uniform angular acceleration. Its angular speed increases from `2pirad//s` to `10pirad//s` in `4 s`. Find the number of revolutions in this period.A. `80`B. `100`C. `120`D. `150` |
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Answer» `theta=bar(omega)t=((omega_(0)+omega)/(2))t=((20pi+40pi)/(2))(10)=300pi` Number of rotations `n=(theta)/(2pi)=150` |
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| 74. |
A flywheel revolves at 100 rev/min, a torque is applied to the flywheel for 10 s If the torque increases the speed to 200 rev/min, then the angular acceleration of the flywheel will beA. `(pi)/(6)rad//s^(2)`B. `(pi)/(5)rad//s^(2)`C. `(pi)/(4)rad//s^(2)`D. `(pi)/(3)rad//s^(2)` |
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Answer» Correct Answer - D `alpha =(omega_(2)-omega_(1))/(t)=(2pi(n_(2)-n_(1)))/(t)=(2pi)/(6)=(pi)/(3)` |
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| 75. |
A ring of mass m and radius R is rolling on a horizontal plane. If the ring suddenly starts rotating about its diameter, then its angular velocity in this will beA. `(V)/( R)`B. `(2V)/( R)`C. `(V)/(2R)`D. `sqrt2(V)/( R)` |
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Answer» Correct Answer - B `KE_("roll")=KE_("rot")` `(1)/(2)mv^(2)(1+(k^(2))/(R^(2)))=(1)/(2)m omega^(2)` `mv^(2)(1+(k^(2))/(R_(2)))=(mR^(2))/(4)omega^(2)` `omega =(2v)/(R )` |
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| 76. |
A cyclist riding at a speed of `14sqrt(3) ms^(-1)` takes a turn around a circular road of radius `20 sqrt(3)m`. What is the inclination to the vertical? |
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Answer» Correct Answer - `60^(@)` |
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| 77. |
The mass of a bicycle rider along with the bicycle is 100 kg. he wants to cross over a circular turn of radius 100 m with a speed of `10 ms^(-1)`. If the coefficient of friction between the tyres and the road is 0.6, will the rider be able to cross the turn? Take `g = 10 ms^(-2)`. |
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Answer» Correct Answer - Yes |
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| 78. |
Two cubes A and B of same shape, size and mass are placed on a rough surface in the same manner. Equal forces are applied on the both cubes. But at the cube A, the force is applied at the top in horizontal direction. But at the cube B just above the center of mass of the cube in the same manner. Then,A. A will topple firstB. B will topple firstC. Both will topple at the same timeD. None of the above |
| Answer» a) The greater height of the point of applying force from ground, more change of troppling | |
| 79. |
The ladder shown in the figure is light and stands on a frictionless horizontal surface. Arms AB and BC are of equal length and M and N are their mid points. Length of MN is half that of AB. A man of mass M is standing at the midpoint of BM. Find the tension in the mass less rod MN. Consider the man to be a point object. |
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Answer» Correct Answer - `(sqrt3)/(4) Mg` |
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| 80. |
A regular polygon of n sides is placed on a rough surface vertically as such one of the side of regular polygaon touches the surface. A force is applied horizontally at the top. The chosen value of n are 3,5 and 8. For which value of n, the polygon first is likely to topple?A. 3B. 5C. 8D. All of these |
| Answer» c) The body has smaller base, greater chance of toppling. When the value of n increases, base size decreases. | |
| 81. |
A uniform frictionless ring of mass M and radius R, stands vertically on the ground. A wall touches the ring on the left and another wall of height R touches the ring on right (see figure).There is a small bead of mass m positioned at the top of the ring. The bead is given a gentle push and it being to slide down the ring as shown. All surfaces are frictionless. (a) As the bead slides, up to what value of angle `theta` the force applied by the ground on the ring is larger than Mg? (b) Write the torque of force applied by the bead on the ring about point A as function of `theta`. (c) What is the maximum possible value of torque calculated in (b)? Using this result tell what is the largest value of `(m)/(M)` for which the ring never rises off the ground ? |
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Answer» Correct Answer - (a) `theta = cos^(-1) ((2)/(3))` (b) `mgR (2 cos theta - 3 cos^(2) theta)` (c) `tau_("max") = (mgR)/(3); ((m)/(M))_("max") = 3` |
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| 82. |
The masses of two uniform discs are in the ratio `1 : 2` and their diameters in the ratio `2 : 1`. The ratio of their moment, of inertia about the axis passing through their respective centres and perpendicular to their planes isA. `1:1`B. `1:2`C. `2:1`D. `1:4` |
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Answer» Correct Answer - C `(I_(1))/(I_(2))=(M_(1))/(M_(2))((R_(1))/(R_(2)))^(2)` `= (1)/(2)xx(4)/(1)=(2)/(1)` |
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| 83. |
Two circular discs A and B have equal masses and uniform thickness but have densities `rho_(1)` and `rho_(2)` such that `rho_(1)gt rho_(2)` . Their moments of inertia isA. `I_(1)gt I_(2)`B. `I_(1)gt gt I_(2)`C. `I_(1)lt I_(2)`D. `I_(1)=I_(2)` |
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Answer» Correct Answer - C `I_(1)=(MR_(1)^(2))/(2)` and `I_(2)=(MR_(2)^(2))/(2)` `(I_(1))/(I_(2))=(R_(1)^(2))/(R_(2)^(2))` and `rho_(1)=(m)/(A_(1))` and `rho_(2)=(m)/(A_(2))` `(rho_(1))/(rho_(2))=(A_(2))/(A_(1))` `= (R_(1)^(2))/(R_(1)^(2))` |
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| 84. |
A disc of mass `m` and radius R lies flat on a smooth horizontal table. A mass less string runs halfway around it as shown in figure. One end of the string is attached to a small body of mass `m` and the other end is being pulled with a force F. The circumference of the disc is sufficiently rough so that the string does not slip over it. Find acceleration of the small body. |
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Answer» Correct Answer - `a = (F)/(4m)` |
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| 85. |
Two circular discs are of same thickness. The diameter of `A` is twice that of `B`. The moment of inertia of `A` as compared to that of `B` isA. twice as largeB. four times as largeC. sixteen times as largeD. eight times of large |
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Answer» Correct Answer - C `(I_(1))/(I_(2))=(M_(1)R_(1)^(2))/(M_(2)R_(2)^(2))` ……(i) `M_(1)=rho v_(1)= rho pi R_(1)^(2)t` `M_(2)= rho v_(2)=rho pi R_(2)^(2) t` `(M_(1))/(M_(2))=(R_(1)^(2))/(R_(2)^(2))` Therefore equation nbo. (i) becomes, `therefore (I_(1))/(I_(2))=((R_(1))/(R_(2)))^(2)xx((R_(1))/(R_(2)))^(2)=((R_(1))/(R_(2)))^(4)` `(I_(1))/(I_(2))=((2R_(2))/(R_(2)))^(4)=16` |
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| 86. |
A disc of mass m and radius R is moving on a smooth horizontal surface with the flat circular face on the surface. It is spinning about its centre with angular peed `omega` and has a velocity V (see figure). It just manages to hit a stick AB at its end A. The stick was lying free on the surface and stick to the disc. [The combined object becomes like a badminton racket]. Mass and length of the stick are m and 2R respectively. (a) Calculate the angular speed of the combined object assuming `V = R omega` (b) Calculate loss in kinetic energy. Why is energy lost ? (c) If `V = eta (R omega)`, loss in kinetic energy is minimum. Find `eta`. (Assume `omega` is given] |
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Answer» Correct Answer - (a) `omega_(0) =(9)/17omega` (b) `Delta E = (5)/(68)mV^(2) + (7)/(34)mR^(2)omega^(2) - (3)/(17)mV R omega` (c) `eta = (6)/(5)` |
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| 87. |
Two circular discs A and B have equal masses and uniform thickness but have densities `rho_(1)` and `rho_(2)` such that `rho_(1)gt rho_(2)` . Their moments of inertia isA. `I_(1)gt I_(2)`B. `I_(1)gt gt I_(2)`C. `I_(1)ltI_(2)`D. `I_(1)=I_(2)` |
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Answer» Correct Answer - C `(I_(1))/(I_(2))=(M_(1))/(M_(2))xx((R_(1))/(R_(2)))^(2)=((R_(1))/(R_(2)))^(2)` `M_(1)=M_(2)` `rho_(1)v_(1)=rho_(2)v_(2)` `rho_(1)pi R_(1)^(2)t=rho_(2)pi R_(2)^(2)t` `(rho_(1))/(rho_(2))=((R_(2))/(R_(1)))^(2)` `therefore (I_(1))/(I_(2))=(rho_(2))/(rho_(1))` As, `rho_(1) gt rho_(1)` `(I_(1))/(I_(2)) lt 1` `I_(1) lt I_(2)`. |
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| 88. |
A uniform ball of mass M and radius R can rotate freely about any axis through its centre. Its angular velocity vector is directed along positive x axis. A bullet is fired along negative Z direction and it pierces through the ball along a line that is at a perpendicular distance`r (le R)` from the centre of the ball. The bullet passes quickly and its net effect is that it applies an impulse on the ball. Mass of the bullet is m and its velocity charges from u to`v (le u)` as it passes through the ball. As a result the ball stops rotating about X axis and begins to rotate about y axis. The angular speed of the ball before and after the hit is `omega`. Find `r`. |
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Answer» Correct Answer - `(2sqrt2)/(5) (MR^(2) omega)/(m (u - v))` |
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| 89. |
Two discs A and B are moving with their flat circular surface on a smooth horizontal surface. Mass, radius and velocity of the two discs are `- m_(A) = 2M, m_(B) = M, r_(A) = R, = 2R, v_(A) = v,` and `v_(B) = 2v`. The velocities of the two discs are oppositely directed so that they just cannot avoid collision and stick to each other (see figure) (a) Find the angular speed of the composite system after collision (b) Find loss in kinetic energy due to collision |
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Answer» Correct Answer - (a) `omega = (2v)/(3R)` (b) `kE_("loss") = Mv^(2)` |
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| 90. |
A uniform square plate ABCD has mass M side length `a`. It is sliding on a horizontal smooth surface with a velocity of`vec(v) = v_(0) (4 hat(i) + 2 hat(j))`. There is no rotation. Vertex A of the plate is suddenly fixed by a nail. Calculate the velocity of centre of the plate immediately after this |
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Answer» Correct Answer - `v_(c) = (3v_(0))/(4) (hat(i) - hat(j))` |
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| 91. |
A fly wheel of mass 4 kg has moment of inertia `16 kg m^(2)`, then radius of gyration about the central axis perpendicular to its plane isA. 1mB. 2mC. 4mD. 16m |
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Answer» Correct Answer - B `I = MK^(2)` `K = sqrt((I)/(m)) = sqrt((16)/(4)) = sqrt(4) = 2`. |
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| 92. |
A torque of 100 Nm acting on a wheel at rest, rotates it through 200 radians in 10 s. The angular acceleration of the wheel, in `rad//s^(2)` isA. 2B. 4C. 1D. 8 |
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Answer» Correct Answer - B `theta = omega_(1) t+(1)/(2)alpha t^(2)` |
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| 93. |
A constant torque of `31.4 N-m` id exterted on a pivoted wheel. If the angular acceleration of the wheel is `4 pi rad//s^2`, then the moment of inertia will be.A. `2.5 kg-m^(2)`B. `3.5 kg-m^(2)`C. `4.5 kg-m^(2)`D. `5.5 kg-m^(2)` |
| Answer» `tau=I alphaimplies I=(tau)/(alpha)=(3.14)/(4pi)=2.5kg m^(2)` | |
| 94. |
Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J = MV is imparted to the body at one of its ends what would be it angular velocity? A. `V//L`B. `2V//L`C. `V//3L`D. `V//4L` |
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Answer» By the conservation of angular momentum about `c.m.` `MV(L)/(2)=[M((L)/(2))^(2)+M((L)/(2))^(2)]omegaimpliesomega=(V)/(L)` |
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| 95. |
A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then A. (`i`),(`ii`)B. (`ii`),(`iii`)C. (`i`),(`iv`)D. (`ii`), (`iv`) |
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Answer» As in problem `133`, `vecv_(C)=2vhati` , `vecv_(B)=vecvhati`, `vecv_(A)=0` (`i`) `vecv_(C)-vecv_(A)=2(vecv_(B)-vecv_(C))=2(vhati-2vhati)=-2vhati`, incorrect (`ii`) `vecv_(C)-vecv_(B)=2vhati-vhati=vhati` , `vecv_(B)-vecv_(A)=vhati-0=vhati`, correct (`iii`) `|vecv_(C)-vecv_(A)=2v`, `2|vecv_(B)-vecv_(C)|=2v`, correct (`iv`) `|vecv_(C)-vecv_(A)|=2v`, `4|vecv_(B)|=4v`, incorrect |
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| 96. |
A ring of mass M and radius R lies flat on a horizontal table. A light thread is wound around it and its free end is pulled with a constant velocity `v`. (a) Two small segment A and B (see fig.) in the ring are rough and have a coefficient of friction `mu` with the table. Rest of the ring is smooth. Find the speed with which the ring moves. (b) Find the speed of the ring if coefficient of friction is `mu` everywhere, for all points on the ring. |
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Answer» Correct Answer - (a) `(v)/(2)` (b) `(v)/(2)` |
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| 97. |
Light thread is tightly wound on a uniform solid cylinder of radius R. The cylinder is placed on a smooth horizontal table and the thread is pulled horizontally as shown, by applying a constant force F. How much length of the thread is unwound from the cylinder by the time its kinetic energy becomes equal to K. |
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Answer» Correct Answer - `(2K)/(3F)` |
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| 98. |
A solid cylinder of mass `50 kg` and radius `0.5 m` is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other end hanging freely. Tension in the string required to produce an angular acceleration of `2` revolution `s^(-2)` isA. `25 N`B. `50 N`C. `78.5 N`D. `157 N` |
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Answer» Correct Answer - d |
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| 99. |
The ratio of the accelerations for a solid sphere (mass `m, and radius R`) rolling down an incline of angle `theta` without slipping, and slipping down the incline without rolling isA. `5 : 7`B. `2 : 3`C. `2 : 5`D. `7 : 5` |
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Answer» Correct Answer - a |
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| 100. |
The ratio of the accelerations for a solid sphere (mass `m, and radius R`) rolling down an incline of angle `theta` without slipping, and slipping down the incline without rolling isA. `2 : 3`B. `2 : 5`C. `7 : 5`D. `5 : 7` |
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Answer» Rolling without slipping `a_(1)=(g sin theta)/(1+(k^(2))/(R^(2)))=(gsin theta)/(1+(2)/(5))=(5)/(7) g sin theta` Slipping without rolling `a_(2)= g sin theta` `(a_(1))/(a_(2))=(5)/(7)` |
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