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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. |
A particle P is moving in a circle of radius r with a uniform speed u. C is the centre of the circle and AB is diameter. The angular velocity of P about A and V are in the ratio :A. `1:2`B. `2:1`C. `1:3`D. `3:1` |
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Answer» Correct Answer - a |
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| 52. |
A small body of mass m can slide without friction along a through bent which is in the form of a semi-circular arc of radius R. At what height h will the body be at rest with respect to the trough, if the trough rotates with uniform angular velocity `omega` about a vertical axis: A. RB. `R-(2g)/(omega^(2))`C. `R+(2g)/(omega^(2))`D. `R-(g)/(omega^(2))` |
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Answer» Correct Answer - d |
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| 53. |
Effect of friction between pulley and thread : In ideal cases i.e., when pulley and strings are massless and no friction exists at any contact surface, then tension in the string is constant throughout its length. But consider a, massless pulley and massless string but friction exists between pulley and string With coefficient `mu`. Then tension at the two ends of the pulley will be different. As shown in figure, consider an element of string : `dN=(T+dT) "sin" (d theta)/2+T "sin"(d theta)/2` `(T+dT)"cos"(d theta)/2 -T"cos"(d theta)/2-mu dN=dr a=0` (massless strings) `dT"cos"(d theta)/2=mu dN` `dT"cos"(d theta)/2=mu [(T+dt) "sin" (d theta)/2+T "sin" (d theta)/2]` `dt."cos" (d theta)/2=mu[T "sin" (d theta)/2+dT. "sin" (d theta)/2+T "sin" (d theta)/2]` `dT=mu[T. (d theta)/2+0+T(d theta)/2]=mu T cos theta` int_(T_(1))^(T_(2)) (d T)/T=int _(0)^(pi)mu d theta rArr ln(T_(2)/T_(1))=mu pi rArrT_(2)/T_(1)=e^(mu pi)` Suppose cofficient of friction between the string and pulley is `mu =1/pi` The tension on side of lighter mass will be:A. `m_(1)g`B. `m_(2)g`C. `(2m_(2)g)/3`D. `(4m_(1)g)/3` |
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Answer» Correct Answer - d |
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| 54. |
Effect of friction between pulley and thread : In ideal cases i.e., when pulley and strings are massless and no friction exists at any contact surface, then tension in the string is constant throughout its length. But consider a, massless pulley and massless string but friction exists between pulley and string With coefficient `mu`. Then tension at the two ends of the pulley will be different. As shown in figure, consider an element of string : `dN=(T+dT) "sin" (d theta)/2+T "sin"(d theta)/2` `(T+dT)"cos"(d theta)/2 -T"cos"(d theta)/2-mu dN=dr a=0` (massless strings) `dT"cos"(d theta)/2=mu dN` `dT"cos"(d theta)/2=mu [(T+dt) "sin" (d theta)/2+T "sin" (d theta)/2]` `dt."cos" (d theta)/2=mu[T "sin" (d theta)/2+dT. "sin" (d theta)/2+T "sin" (d theta)/2]` `dT=mu[T. (d theta)/2+0+T(d theta)/2]=mu T cos theta` int_(T_(1))^(T_(2)) (d T)/T=int _(0)^(pi)mu d theta rArr ln(T_(2)/T_(1))=mu pi rArrT_(2)/T_(1)=e^(mu pi)` Suppose cofficient of friction between the string and pulley is `mu =1/pi` The tension on side ofheavierass will be:A. `m_(1)g`B. `m_(2)g`C. `(2m_(2)g)/3`D. `(2m_(1)g)/3` |
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Answer» Correct Answer - c |
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| 55. |
In the system shown in the figure `m_(1) gt m_(2)`. System is held at rest by thread BC. Just after the thread BC is burnt : A. initial acceleration of `m_(2)` will be upwardsB. magnitude of initial acceleration of both blocks will be equal to `((m_(1)-m_(2))/(m_(1)+m_(2)))g`C. initial acceleration of `m_(1)` will be equal to zeroD. magnitude of initial acceleration of two blocks will be non-zero and unequal. |
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Answer» Correct Answer - a,c |
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| 56. |
A ball is held at rest in position A by two light cords. The horizontal cord is now cut and the ball swings to the position B. What is the ratio of the tension in the cord in position B to that in position A? A. `3//4`B. `1//2`C. 3D. 1 |
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Answer» Correct Answer - a |
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| 57. |
The force exerted by the floor of an elevator on the foot of a person standing there is more than the weight of the person if the elevator is :A. going up and speeding upB. going down and slowing downC. going up and slowing downD. going down and speeding up |
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Answer» Correct Answer - a,b |
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| 58. |
A person standing oin the floor of an elevator drops as coin. The coin reaches the floor of the elevator in a time `t_1` if the elevator is stationary and in the `t_2` if it is moving uniformly. ThenA. `t_(1) lt t_(2)`B. `t_(1) gt t_(2)`C. `t_(1)=t_(2)`D. `t_(1) lt t_(2)` or `t_(1) gt t_(2)` depending on whether the lift is going up or down. |
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Answer» Correct Answer - c |
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| 59. |
A particle moves along on a road with constant speed at all points as shown in figure. The normal reaction of the road on the particle is : A. Same at all pointsB. Maximum at point BC. Maximum at point CD. Maximum at point E |
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Answer» Correct Answer - d |
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| 60. |
A partical of mass `m` oscillates along the horizontal diameter `AB` inside a smooth spherical `AB` inside a smooth sperical shell of radius `R`. At any instate `K.E.` of the partical is `K`. Then force applied by partical on the on the shell at this instant is: A. `(K)/(R)`B. `(2K)/(R)`C. `(3K)/(R)`D. `(K)/(2R)` |
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Answer» Correct Answer - c |
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| 61. |
A rod of length L is pivoted at one end and is rotated with as uniform angular velocity in a horizontal plane. Let `T_1 and T_2` be the tensions at the points L//4 and 3L//4 away from the pivoted ends.A. `T_(1) gt T_(2)`B. `T_(2) gt T_(1)`C. `T_(1) = T_(2)`D. The relation between `T_(1)` & `T_(2)` depends on whether the rod rotates clockwise or anticlockwise |
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Answer» Correct Answer - a |
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| 62. |
The driver of a car-travelling at speed V suddenly sees a wall at a distance r directly infront of him. To avoid collision. He should :A. apply the brakesB. turn the car simply away from the wallC. do any of the above optionsD. none of these |
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Answer» Correct Answer - a |
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| 63. |
A smooth circular rod of radius r is banked for a speed `v=40 km//hr`. A car of mass `m` attempts to go on the circular road. The friction coefficient between the type and the road is negligible, The correct statements are :A. The car cannot make a turn without skiddingB. If the car turns at a speed less than 40 km/hr, it slips downC. If the car turns at the correct speed of 40 km/hr, the force by the road on the car is equal to `mv^(2)//r`D. If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as well as greater than `mv^(2)//r` |
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Answer» Correct Answer - b,d |
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| 64. |
The magnitude of difference in accelerations of block of mass `m` in both the cases shown below is A. `g`B. `(2g)/(3)`C. zeroD. `g//31` |
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Answer» Correct Answer - b,c |
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| 65. |
A car is moving with speed `v` and is taking a turn on a circular road of radius 10 m. The angle of banking is `37^(@)`. The driver wants that car does not slip on the road. The coefficient of friction is `0.4 (g = 10 m//sec^(2))` The friction force acting when `v=10 m//sec` and mass of car is 50 kg is :A. 400 NB. 100 NC. 300 ND. 200 N |
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Answer» Correct Answer - b |
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| 66. |
A car is moving with speed `v` and is taking a turn on a circular road of radius 10 m. The angle of banking is `37^(@)`. The driver wants that car does not slip on the road. The coefficient of friction is `0.4 (g = 10 m//sec^(2))` If the car were moving on a flat road and distance between the front tyres is 2 m and the height of the centre of the mass of the car is 1m from the ground, then, the minimum velocity for which car topples is :A. `5 m//sec`B. `5 sqrt(3) m//sec`C. `3 sqrt(5) m//sec`D. `10 m//sec` |
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Answer» Correct Answer - d |
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| 67. |
A particle is acted upon by constant magnitude force perpendicular to ot which is always perpendicular to velocity of particle. The motion is taking place in a plane it follows that :A. Velocity is constantB. acceleration is constantC. kinetic energy is constantD. it moves in circular path |
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Answer» Correct Answer - c,d |
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| 68. |
A body is undergoing uniform circular motion then which of the following quantity is constantA. velocityB. accelerationC. forceD. kinetic energy |
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Answer» Correct Answer - d |
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| 69. |
The position vector of a particle in a circular motion about the origin sweeps out equal area in equal time. ItsA. Its velocity remains constantB. Its speed remains constantC. Its acceleration remains constantD. Its tangential remains constant |
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Answer» Correct Answer - b,d |
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| 70. |
A block of mass `m` is placed on a smooth wedge of inclination `theta` & mass M. The whole system is slip on the wedge. Then the normal reaction on the wedge acting from the ground :A. `(M+m)g`B. `(M+m sin theta)g`C. `Mg`D. `((M+m sin theta)/(M+m))g cos theta` |
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Answer» Correct Answer - a |
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| 71. |
A block of mass m is placed on a smooth wedge of inclination. The whole system is accelerated horizontally so that block does not slip on the wedge Find the i) Acceleration of the wedge.ii) Force to be applied on the wedge.iii) Force exerted by the wedge on the block .A. `mg//cos theta`B. `mg cos theta`C. `mg`D. `mg tan theta` |
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Answer» Correct Answer - a |
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| 72. |
A wedge of mass 2m and a cube of mass m are shown in figure. Between cube and wedge, there is no friction. The minimum coefficient of friction between wedge and ground so that wedge does not move is A. `0.20`B. 0.25C. `0.10`D. `0.50` |
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Answer» Correct Answer - a |
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| 73. |
A very small cube of mass 2kg is placed on the surface of a funnel as shown in figure. The funnel is rotating about its vertical axis of symmetry with angular velocity `omega`. The wall of funnel makes an angle `37^(@)` with horizontal. The distance of cube from the axis of rotation is 20 cm and friction coefficient is `mu` (Take `g = 10 m//s^(2)`) The maximum value of angular velocity for which no relative slipping occur and also direction of frictional force is : (take `mu = 2//3`)A. `sqrt((25)/(3)` rad/sec, down the surface of funnelB. `sqrt((25)/(3))` rad/sec, up the surface of funnelC. `sqrt((25)/(9))` rad/sec, up the surface of funnelD. `sqrt((25)/(9))` rad/sec, down the surface of funnel |
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Answer» Correct Answer - a |
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| 74. |
A particle is resting on an inverted cone as shown. It is attached to cone by a thread of length 20 cm. String remains parallel to slope of cone. The cone is givenangular acceleration of `0.5 rad//sec^(2)` then at what time does mass leave contact with surface (assuming sufficient friction) : A. `20 sec`B. `10 sec`C. `40 sec`D. `5 sec` |
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Answer» Correct Answer - a |
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| 75. |
A carriage of mass M and length l is joined to the end of a slope as shown in the Fig 2E.81 (a). A block of mass m is released from the slope from height h. It slides till end of the carriage (The friction between the body and the slope and also friction between carriage and horizontal floor is negligible) Coefficient of friction between block and carriage is `mu`. Find ,minimum h in the given terms. A. `mu (1+M/m)l`B. `2mu(1+m/M)l`C. `mu(2+m/M)l`D. `mu(1+m/M)l` |
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Answer» Correct Answer - d |
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| 76. |
A cyclist moves with uniform velocity down a rough inclined plane of inclination `alpha`. Total mass of cycle & cyclist is m. Then the magnitude and direction of force acting on the cycle from inclined plane is :A. `mg cos alpha` perpendicularly into the inclined planeB. `mg cos alpha` perpendiculary outward of the inclined planeC. mg perpendicularly outward of the inclined planeD. mg vertical upwards |
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Answer» Correct Answer - d |
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| 77. |
The inclined surface is rough `mu = (1)/(2)`. For different values of m and M system slides down or up the plane or remains stationary. Match the appropriate entries of column-1 with those of column-2 `{:(,"Column-1",,"Column-2"),("(A)","Minimum value of " (m)/(M) "so theta m slides down","(P)",(5)/(3)),("(B)","Minimum valie of " (M)/(m)"so that m slides up","(Q)",1),("(C)","Value of" (m)/(M) "so that friction force on m is zero","(R)",(3)/(5)),("(D)",underset("and acceleration of M")"Ratio of vertical component of acceleration of m","(S)",5):}` |
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Answer» Correct Answer - A-S; B-Q; C-P; D-R |
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| 78. |
A small block slides without friction down an iclined plane starting form rest. Let `S_(n)` be the distance traveled from time `t = n - 1` to `t = n`. Then `(S_(n))/(S_(n + 1))` is:A. `(2n-1)/(2n)`B. `(2n+1)/(2n-1)`C. `(2n-1)/(2n+1)`D. `(2n)/(2n+1)` |
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Answer» Correct Answer - c |
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| 79. |
See the diagrams carefully in Column-1 and match each with the obeying relation (S) in column-2. The string is massless, inextensible and pulley is frictionless in each case. a=g/3, m= mass of block T = tension in a given string, `a_("pulley") =` acceleration of movable pulley in each case, acceleration due to gravity is g. |
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Answer» Correct Answer - A-Q, R, S; B-P, Q, R; C-P, Q, R, S; D-P, Q |
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| 80. |
In the figure shown `m_(1)=1 kg,m_(2)=2kg`, pulley is ideal. At `t=0`, both masses touches the ground and string is taut. A force `F=2t` is applied to pulley (t is in second) then `(g = 10 m//s^(2))` : A. `m_(2)` is lifted off the ground at `t=20` secB. acceleration of pulley when `m_(2)` is about to lift off is `5 m//s^(2)`C. `m_(1)` is lifted off the ground at `t=0` secD. both blocks are lifted off simultaneously |
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Answer» Correct Answer - a,b,c |
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| 81. |
Two blocks of masses `m_(i) = 2kg` and `m_(2) = 4kg` hang, over a massless pulley as shown in the figure. A force `F_(0)=100 N` acting at the axis of the pulley accelerates the system upwards Then : A. acceleration of 2kg mass is `15 m//s^(2)`B. acceleration of 4kg mass is `2.5 m//s^(2)`C. acceleration of both masses is sameD. acceleration of both the masses is upward |
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Answer» Correct Answer - a,b,d |
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| 82. |
The force acting on the block is given by `F = 5 - 2t`. The frictional force acting on the block after time `t = 2` seconds will be: `( mu = 0.2)` A. 2 NB. 3 NC. 1 ND. Zero |
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Answer» Correct Answer - a |
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| 83. |
A particle stays at rest as seen in a frame. We can concude that :A. Resultant force on the particle is zeroB. The frame may be inertial but the resultant force on the particle is zeroC. The frame is inertialD. The frame may be non-inertial but there is a non-zero resultant force |
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Answer» Correct Answer - b,d |
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| 84. |
A particle of mass m rotates with a uniform angular speed `omega`. It is viewed from a frame rotating about the Z-axis with a uniform angular speed `omega_0`. The centrifugal force on the particler isA. `m omega^(2)a`B. `m omega_(0)^(2)a`C. `m((omega+omega_(0))/(2))^(2) a`D. `m omega omega_(0)a` |
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Answer» Correct Answer - b |
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| 85. |
Suppose `F,F_(N) & f` are the magnitudes of the contact force, normal force and the frictional force exerted by one surface on the other, kept in contact, if none of these is zero :A. `F gt f`B. `F_(N) gt f`C. `F gt F_(N)`D. `(F_(N)-f)lt (F_(N)+f)` |
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Answer» Correct Answer - a,c,d |
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| 86. |
A particle is observed from two frames `S_(1) and S_(2)`. The frame `S_(2)` moves with respect to `S_(1)` with an acceleration a. Let `F_(1) and F_(2)` be the pseudo forces on the particle when seen from `S_(1) and S_(2)` respectively. Which of the following are not possible ?A. `F_(1) ne 0,F_(2) =0`B. `F_(1) ne 0, F_(2) ne 0`C. `F_(1) =0,F_(2) ne 0`D. `F_(1) = 0,F_(2) =0` |
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Answer» Correct Answer - d |
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| 87. |
A particle is moving along a circular path. The angular velocity, linear velocity, angular acceleration and centripetal acceleration of the particle at any instant are `vec(omega),vec(v),vec(a),vec(a)_(c)` respectively. Which of the following relations are correct ?A. `vec(omega) bot vec(v)`B. `vec(omega) bot vec(a)`C. `vec(omega) bot vec(a)_(c)`D. `vec(v) bot vec(a)_(c)` |
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Answer» Correct Answer - a,c,d |
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| 88. |
In the figure the block A,B and C of mass `m` each, have acceleration `a_(1),a_(2) & a_(3)` respectively. `F_(1) & F_(2)` are external forces of magnitude 2 mg and mg respectively : A. `a_(1) gt a_(3) gt a_(2)`B. `a_(1) = a_(2),a_(2) = a_(3)`C. `a_(1)=a_(2)=a_(3)`D. `a_(1) gt a_(2),a_(2)=a_(3)` |
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Answer» Correct Answer - a |
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| 89. |
Block A and C starts from rest and move to the right with acceleration `a_(A)=12tms^(-2)` and `a_(C )=3ms^(-2)`. Here t is in second. The time when block B again comes to rest is A. `1s`B. `(3)/(2)s`C. `2s`D. `(1)/(2)s` |
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Answer» Correct Answer - d |
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| 90. |
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration `a_(c)` is varying with time t as `a_(c) = k^(2)rt^(2)`, where k is a constant. The power delivered to the particle by the forces acting on it is :A. `2pi mk^(2)r^(2)t`B. `mk^(2)r^(2)t`C. `(mk^(4)r^(2)t^(5))//3`D. Zero |
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Answer» Correct Answer - b |
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| 91. |
A block of mass 0.1 kg is kept on an inclined plane whose angle of inclination can be varied from `theta = 30^(@)` to `theta = 90^(@)`. The coefficient of friction between the block & the inclined plane is `mu =1`. A force of constant magnitude `(1)/(2)mg` newton always acts on the block directed up the inclined plane and parallel to it. Then : A. B. C. D. |
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Answer» Correct Answer - b |
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| 92. |
A block kept on an inclined surface, just begins to slide if the inclination is `30^(@)`. The block is replaced by another block B and it is just begins to slide if the inclination is `40^(@)`, then :A. Mass of `A gt ` mass of BB. Mass of `A lt` mass of BC. Mass of `A=` mass of BD. All the three are possible |
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Answer» Correct Answer - d |
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| 93. |
As shown below AB represents an infinite wall tangential to a horizontal semi-circular track. O is a point source of light on the ground at the centre of the circle. A block moves along the circular track with a speed `v` starting from the point where the wall touches the circle. If the velocity and acceleration of shadow along the length of the wall is respectively `V` and a, then : A. `V=vcos((vt)/(R))`B. `V=v sec^(2)((vt)/(R))`C. `a=((v^(2))/(R))sec^(2)((vt)/(R))tan((vt)/(R))`D. `a=((2v^(2))/(R))sec^(2)((vt)/(R))tan((vt)/(R))` |
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Answer» Correct Answer - b,d |
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| 94. |
Figure shows a heavy block kept on a frictionless surface and being pulled on the left rope is withdrawn but the force on the right end continues to act. Let `F_1` and `F_2` be the magnitude of the forces by the right rope and the left rope on the block respectively. A. `F_(1)=F_(2)=F+mg` for `t lt 0`B. `F_(1) = F,F_(2)=F` for `t gt0`C. `F_(1)=F_(2)=F` for `t lt 0`D. `F_(1) lt F,F_(2)=F` for `t gt 0` |
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Answer» Correct Answer - c |
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| 95. |
A spring block system is placed on a rough horizontal floor. The block is pulled towards right to gives spring some elongation and released, Then : A. the block may stop before the spring attains its natural lengthB. the block must stop with spring having some compressionC. the block may stop with spring having sme compressionD. it is not possible that the block stops at mean position |
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Answer» Correct Answer - a,c |
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| 96. |
Column-1 shows certain siruations and column-2 shows information about forces. |
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Answer» Correct Answer - A-P, Q; B-P, Q, S; C-P, Q, R; D-P, Q,R |
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| 97. |
Which of the following is/are incorrect :A. If net normal force on a surface is zero, friction will be zeroB. Value is static friction is given by `mu_(s)N`C. Static friction opposes relative motion between two surfaces is contactD. kinetic friction reduces velocity of an object. |
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Answer» Correct Answer - a,b,c,d |
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