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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.
| 1. |
A tight rope walker in a circus holds a long flexible pole to help stay balanced on the rope. Honding the pole horizontally and perpendicular to the rope helds the performer. .A. by lowering the overall centre of gravityB. by increasing the rotation inertiaC. in the ability to adust the centre gravity to the be over the ropeD. in achieving the centre of gravity to be under the rope. |
| Answer» Correct Answer - C | |
| 2. |
A sphere of mass `M` and radius `R` is attached by a light of length `1` to a point `P`. The sphere rolls without slipping on a circular track as shown. It is released from the horizontal position. The angular momentum of the system about `P` when the rod becomes vertical is : .A. `M sqrt((10)/(7)) gl[l + R]`B. `M sqrt((10)/(7)) gl[ l + (2)/(5) R]`C. `M sqrt((10)/(7)) gl [l +(7)/(5) R]`D. none of the above |
| Answer» Correct Answer - D | |
| 3. |
A solid sphere rolling on a rough horizontal surface. Acceleration of contact point is zero. A solid sphere can roll on the smooth surface.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) During rolling motion, there is radial acceleration towards the centre. A solid sphere can roll on smooth surface hence reason is false. |
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| 4. |
A sphere is performing pure rolling on a rough horizontal surface with constant angular velocity. Frictional force acting on the sphere is zero. Velocity of contact point is zero.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) If angular velocity is constant then frictional force acting on sphere is zero. In case of pure rolling velocity of contact point in zero. |
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| 5. |
A sphere is performing pure rolling on a rough horizontal surface with constant angular velocity. Frictional force acting on the sphere is zero. Velocity of contact point is zero.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 - B If angular velocity is constant then frictional force acting on sphere is zero. In case of pure rolling velocity of contact point is zero. |
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| 6. |
A mass M moving with a constant velocity parlale to the X-axis. Its angular momentum with respect to the originA. is zeroB. remains constantC. goes on increasingD. goes on decreasing |
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Answer» Correct Answer - B (b) Perpendicular distance `l` will remain same, so angular momentum `L = mul` will remain constant. |
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| 7. |
A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following wil not be affected?A. Moment of inertiaB. Angular momentumC. Angular velocityD. Rotational kinetic energy |
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Answer» Correct Answer - B (b) In free space, there will be no torque acting on the sphere. Therefore the angular momentum of the sphere will not be affected. |
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| 8. |
A circular platform is mounted on a vertical frictionless axle. Its radius is `r = 2 m` and its moment of inertia`I = 200 kg m^2`. It is initially at rest. A `70 kg` man stands on the edge of the platform and begins to walk along the edge at speed `v_0 = 1 m s^-1` relative to the ground. The angular velocity of the platform is.A. `1.2 rad s^-1`B. `0.4 rad s^-1`C. `0.7 rad s^-1`D. `2 rad s^-1` |
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Answer» Correct Answer - C ( c) As the system in initially at rest, therefore, initial angular momentum, `L_i = 0`. According to the law of conservation of angular momentum, final angular momentum, `L_f = 0` :. Angular momentum of man = angular momentum of platform in opposite direction i.e., `mv_0 r = I omega` or `omega = (m v_0 r)/(I) = (70(1.0)(2))/(200) = 0.7 rad s^-1`. |
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| 9. |
A ring of radius `R` is rotating with an angular speed `omega_0` about a horizontal axis. It is placed on a rough horizontal table. The coefficient of kinetic friction is `mu_k`. The time after it starts rolling is.A. `(omega_0 mu_k R)/(2 g)`B. `(omega_0 g)/(2 mu_k R)`C. `(2 omega_0 R)/(mu_k g)`D. `(omega_0 R)/(2 mu_k g)`] |
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Answer» Correct Answer - D (d) Acceleration produced in the centre of mass due friction. `a = (f)/(M) = (mu_k Mg)/(M) = mu_k g` where `M` is the mass of the ring Angular retardation produced by the torque due friction `prop = (tau)/(I) = (f R)/(I) = (mu_k M g R)/(I)` As `v = u + at` :. `v = 0 + mu k gt (because u = 0)` (Using (i)) As `omega = omega_0 + prop t` :. `omega = omega_0 - (mu_k M gR_1)/(I)` (Using (ii)) m For rolling without slipping `v = R omega` :. `(v)/(R) = omega_0 - (mu_k Mg T_1)/(I) t` `(mu_k gt)/(R) = omega_0 - (mu_k M gR)/(I) t rArr (mu_k gt)/(R) [1 + (MR^2)/(1)] =omega` `(mu_k gt)/(R) = (omega_0)/(1+ (MR^2)/(I)) rArr t = (R omega_0)/(mu_k g((1 + MR^2)/(I)))` For ring, `I = MR^2` :. `t = (R omega_0)/(mu_k g(1 + (MR^2)/(MR^2)))=(R omega_0)/(2 mu_k g)`. |
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| 10. |
Two bodies `A and B` initially at rest are attrached towards each other due to gravitation. Given that `A` is much heavier. Than `B`. Which of the followings correctly describes the relative motion of the centre of mass of the bodies ?A. it moves towards `A`B. it moves towards `B`C. it moves perpendicular to the line joining the particlesD. it remains at rest. |
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Answer» Correct Answer - D (d) As there is no external force, so centre of mass remain at rest. |
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| 11. |
If the earth suddenly contracted to half of its present radius without losing any mass, how many hours would there be in a day? (hint: Apply the principle of conservation of angular momentum). |
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Answer» Correct Answer - 6 hours |
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| 12. |
A wire of uniform cross-section is bend in the shape shown in figure. The co-ordinate of the centre of mass of each side are shown in (figure). The co-ordinates of the centre of mass of the system are. .A. `((15 l)/(14), (6l)/(7))`B. `((15 l)/(14),l)`C. `(l, (l)/(2))`D. `(l, l)` |
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Answer» Correct Answer - A (a) `X_(cm) = (m xx 0 + ml + m xx 2l + (m)/(2)(3 l)/(2))/(3.5 m)` `rArr X_(cm) = (15 ml)/(4 xx (7)/(2) m) =(15 l)/(14)` `Y_(cm) = (m xx 0 + ml + ml + (m)/(2) 2l)/(7/(2)m) ` `rArr Y_(cm) = (3 ml)/((7 m)/(2)) = (6l)/(7)`. |
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| 13. |
A solid unifrom disk of mass `m` and radius `R` is pivoted about a horizontal axis through its centre and a small body of mass `m` is attached to the rim of the disk. If the disk is released from rest with the small body, initially at the same level as the centre, the angular velocity when the small body reaches the lowest point of the disk is.A. `sqrt((12 g)/(13 r))`B. `sqrt((11 g)/(12 r))`C. `sqrt((12 g)/(11 r))`D. `sqrt((7 g)/(11 r))` |
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Answer» Correct Answer - C ( c) The conservation of energy yields `Delta K.E. = Delta P.E.` `(1)/(2) I omega^2 = mg (2 r) + mgr` `rArr (1)/(2)m(2r)^2 omega^2 + (1)/(2){((mr^2)/(2))+ mr^2} omega^2 = 3mgr` `rArr omega = sqrt((12 g)/(11 r))`. |
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| 14. |
A mass `m` is attached to a pulley through a cord as shown in figure. The pulley is a solid disk with radius `R`. The cord does not slip on the disk. The mass is released from rest at a height `h` from the ground and at the instant the mass reaches the ground, the disk is rotating with angular velocity `omega`. Find the mass of the disk. . |
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Answer» Correct Answer - (i) `M = 2m((2gh)/(R^(2) omega^(2))-1)` By work Energy Theorem `WD_(g) = Delta KE` `mgh = (1)/(2) mv^(2) + (1)/(2) omega^(2)` `mgh = (1)/(2) m(R omega)^(2) + (1)/(2) xx((1)/(2) MR^(2)) omega^(2)` (M = mass of disc) `mgh = (1)/(2) mR^(2) omega^(2) + (1)/(4) MR^(2) omega^(2)` `mgh -(1)/(2) MR^(2) omega^(2) = (1)/(4) MR^(2) omega^(2)` `M = (4(mgh -(1)/(2) mR^(2) omega^(2)))/(R^(2) omega^(2))` `M = 2m((2gh)/(R^(2)omega^(2)) -1)`. |
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| 15. |
A person pulls along a rope wound up around a pulley with a constant force `F` for a time interval of `t` seconds. If `a` and `b` are the radii of the inner and the outer circumference `(a lt b)`, then find the ratio of work done by the person in the two cases shown in the figuer is `W_(1)//W_(2)`. . |
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Answer» Correct Answer - `a^(2)//b^(2)` case `- 1` `F xx a = I alpha rArr alpha = (Fa)/(I)` As `alpha` is constant `omega = alpha t, omega = ((Fa)/(I)) alpha` `W_(1) = Delta KE rArr WD_(1) =(1)/(2) I omega^(2) rArr WD_(1) = (1)/(2) (I) (alpha^(2) t^(2))` `WD_(1) =(1)/(2)I((Fa)/(I))^(2) t^(2)`.....(1) similarly in case II `WD_(2) = (1)/(2) I ((Fb)/(I))^(2) t^(2)` ....(2) from (1) & (2) `(WD_(1))/(WD_(2)) = (a^(2))/(b^(2))`. |
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| 16. |
Assertion : The moment of inertia of a rigid body reduces to its minimum value, when the axis of rotation passes through its centre of gravity. Reason : The weight of a rigid body always acts through its centre of gravity.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 - A (a) By theorem of parallel axis both statements are correct and reason is the correct explanation of assertion. |
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| 17. |
Assertion: The velocity of a body at the bottom of an inclind plane of given height is more when is slides down the plane, compared to, when it rolling down the same plane. Reason: In rolling down a body acquires both, kinetic energy of translation and rotation.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 - A (a) In sliding down, the entire potential energy is converted into kinetic energy. While in rolling down same part of potential energy is converted into `K.E.` of rotation, therefore linear velocity acquired is less. |
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| 18. |
A basketball rolls a ramp sloping upward without slipping. With its centre of mass moving at a certain initial speed. A block of ice of the same mass is set sliding up the ramp with the same speed along a parallel line. Which object will travel farther up the ramp ?A. basketballB. the ice blockC. They will travel equally far up the ramp.D. cannot be decided |
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Answer» Correct Answer - A (a) The basketball-Earth system has more kinetic energy than the ice-Earth system due to the rotational kinetic energy of the basketball. Therefore, when the kinetic energy of both systems has transformed to gravitational potential energy when the objects momentarily come to rest at their highest point on the ramp, the basketball will be at a higher location, corresponding to the larger gravitational potential energy. |
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| 19. |
A child is standing with folded hands at the center of a platform rotating about its central axis. The kinetic energy of the system is `K`. The child now strectches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is.A. 2 KB. `K//2`C. `K//4`D. 4 K |
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Answer» Correct Answer - B From conservation of angular momentum `(I omega = constant)`, angular velocity will remain half. As, `K = (L^(2))/(2 I)` The rotational kinetic energy will become half. Hence, the correct option `(B)`. |
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| 20. |
A solid cylinder of weight 20 kg and radius 7.5 cm is placed on a inclined plane with inclination `30^(@)` to the horizontal. A light thin tape is wound around the cylinder and is taken tangentially over the upper surface parallel to the plane and finally attached to a 5-kg body after passing over a light smooth pulley. Find the tension in the tape and linear acceleration of the cylinder up the plance, assuming no slip. |
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Answer» Correct Answer - 49 N;0 |
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| 21. |
A sphere rolls up an inclined plane whose inclination is `30^(@)`. At the bottom of the inclined plane, the center of mass of the sphere has a translational speed of `5ms^(-1)` (a) How far does the sphere travel up the plane? (b) How long does it take to return to the bottom? |
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Answer» Correct Answer - 3.57 m; 2.8 s |
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| 22. |
A flywheel of mass 5 kg and radius 10 cm rolls down a plane inclined at `30^(@)` to the horizontal. It rolls from rest through 100 cm in 10s. Neglecting friction find a) the kinetic energy of the flywheel at the end of 10s, (b) the momentum of intertia of the flywheel about its axis. |
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Answer» Correct Answer - 12.2 kg`m^(2)` |
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| 23. |
In the pulley system shown, if radii of the bigger and smaller pulley are `2 m` and `1 m` respectively and the acceleration of block `A` is`5 m//s^2` in the downward direction, then the acceleration of block `B` will be : .A. `0 m//s^2`B. `5 m//s^2`C. `10 m//s^2`D. `5//2 m//s^2` |
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Answer» Correct Answer - D (d) Given `a_A = 2 prop = 5 m//s^2` `rArr prop = (5)/(2) rad//s^2` `rArr a_B = 1. (prop) = 5//2 m//s^2`. |
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| 24. |
At `t = 0`, the positions and velocities of two particles are as shown in the figure. They are kept on a smooth surface and being mutually attracted by gravitational force. Find the position of centre of mass at `t = 2s`. .A. X = 5 mB. X = 7 mC. X = 3 mD. X = 2 m |
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Answer» Correct Answer - B (b) As no external force are acting on the system. The velocity of centre of mass will be constant. `v_(cm) = (1 xx 5 - 1 xx 3)/(1 + 1) = 1 m//s` Displacement of centre of mass `Delta x_(cm) = 1 xx 2 = 2 m` Hence position of centre of mass `Delta x_(cm) = 5 + 2 = 7 m`. |
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| 25. |
A sphere cannot roll on a smooth inclined surface. The motion of a rigid body which is pivoted or fixed in some way is rotation.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) A sphere cannot roll a smooth inclined plane, because force of friction is zero for smooth surface and in case of sphere friction provides the torque for rolling. |
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| 26. |
A ring of mass `m` and radius `r` is projected with velocity `v_(0)` while spinning with angular velocity `(v_(0))/(4 r)` at the bottom of an inclined plane of angle of inclination `theta - 60^@` as shown in fig. The ring begins to roll without slipping after colliding with the inclined plane. Before collision sence of rotation is depicyed in the fig. . If the ring after collision rolling up the incline then maximum height attained by it is :A. `(V_(0)^(2))/(2 g)`B. `(V_(0)^(2))/(128 g)`C. `(V_(0)^(2))/(64 g)`D. `(V_(0)^(2))/(32 g)`. |
| Answer» Correct Answer - B | |
| 27. |
A ring of mass `M` and radius `R` lies in `x-y` plane with its centre at origin as shown. The mass distribution of ring is non uniform such that, at any point `P` on the ring, the mass per unit length is given by `lamda = lamda_0 cos^2 theta` (where `lamda_0` is a positive constant). Then the moment of inertia of the ring about z-axis is : .A. `MR^2`B. `(1)/(2) MR^2`C. `(1)/(2) (M)/(lamda_0) R`D. `(1)/(pi) (M)/(lamda_0) R` |
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Answer» Correct Answer - A (a) Divide the ring into infinitely small lengths of mass `dm_i`. Even through mass distribution is non-uniform, each mass `dm_i` is at same distance `R` from origin. :. `MI` of ring about z-axis is =`dm_ R^2 + dm_2 R^2 + …+ dm_n R^2 = MR^2`. |
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| 28. |
A solid aluminimum sphere of radius `R` has moment of inertia `I` about an axis through its centre. The moment of inertia about a central axis of a solid aluminimum sphere of radius `2 R` is.A. 4 IB. 8 IC. 16 ID. 32 I |
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Answer» Correct Answer - D (d) The sphere of twice the radius has eight times the volume and eight times the mass, and the `r^2` term in `I = (2)/(5) mr^2` also becomes four times larger. |
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| 29. |
A wheel has angular acceleration of `3.0 rad//s^2` and an initial angular speed of `2.00 rad//s`. In a tine of `2 s` it has rotated through an angle (in radian) ofA. 6B. 10C. 12D. 4 |
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Answer» Correct Answer - B (b) Using `theta = omega_0 t + (1)/(2) prop t^2` Given, `prop = 3.0 rad//s^2, omega_0 = 2.0 rad//s^2, omega_0 = 2.0 rad//s, t = 2 s` Hence, `theta = 2 xx 2 + (1)/(2) xx 3 xx (2)^2` or `theta = 4 + 6 = 10 rad. |
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| 30. |
A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is `k`. If radius of the ball be `R`, then the fraction of total energy associated with its rotation will be.A. `(K^2)/(R^2)`B. `(K^2)/(K^2 + R^2)`C. `(R^2)/(K^2 + R^2)`D. `(K^2 + R^2)/(R^2)` |
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Answer» Correct Answer - B (b) Rotational energy `= (1)/(2) I (omega)^2 = (1)/(2) (mK^2) omega^2` Linear kinetic energy `= (1)/(2) m omega^2 R^2` =`((1)/(2) (mK^2) omega^2)/((1)/(2) (mK^2) omega^2 + (1)/(2) m omega^2 R^2) = (K^2)/(K^2 + R^2)`. |
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| 31. |
A grindstone increases in angular speed from `4.00 rad//s` to `12.00 rad//s` in `4.00 s`. Through what angle does it turn during that time interval if he angular acceleration is constant ?A. 8.00 radB. 12.0 radC. 16.0 radD. 32.0 rad |
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Answer» Correct Answer - D (d) The angular displacement will be `Delta theta = omega_(avg) Delta t = ((omega_f + omega_i)/(2)) Delta t` =`((12.00 rad//s + 4.00 rad//s)/(2))(4.00 s) = 32.0 rad`. |
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| 32. |
A hoop of radius `2 m` weight `100 kg`.It rolls along a horizontal floor so that its centre of mass has a speed of `20 cm s^-1`. How much work has to be done to stop it ?A. 2 JB. 4 JC. 6 JD. 8 J |
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Answer» Correct Answer - B (b) Here, `R = 2 m, M = 100 kg, v = 200 cm s^-1` =`20 xx 10^-2 m s^-1` Total kinetic of the loop `= K_T + K_R` =`(1)/(2) Mv^2 + (1)/(2) I omega^2 ["because For a hoop", I = MR^2]` =`(1)/(2) Mv^2 + (1)/(2) MR^2omega^2` =`(1)/(2) Mv^2 + (1)/(2) Mv^2` =`Mv^2` `"Work required to stop the hoop = Total kinetic energy of the hoop"`. =`Mv^2 = (100 kg) (20 xx 10^2 ms^-1) = 4 J`. |
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| 33. |
A disc of radius `R` rolls on a horizontal ground with linear acceleration `a` and angular acceleration `alpha` as shown in Fig. The magnitude of acceleration of point `P` as shown in the figure at an instant when its linear velocity is `v` and angular velocity is `omega` will be a A. `sqrt((a + r prop)^2 + (r omega^2)^2)`B. `(ar)/( R)`C. `sqrt(r^2 prop^2 + r^2 omega^4)`D. `r prop` |
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Answer» Correct Answer - A (a) `vec a_o = a hat i, vec a_(p//o) = prop r hat i + omega^2 r(- hat j)` `vec a_p = vec a_o + vec a_(p//o) =(a + prop r) hat i - omega^2 r hat j` `a_p = sqrt((a + prop r)^2 + (omega^2 r)^2)`. |
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| 34. |
Assertion: In an elasticcollision of two billard balls, the total `KE` is conservation during the short times of collision of the balls`(i.e., when they are in constant). Reason: Energy spend against friction does not follow the law of conservation of energy.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 - A (a) Since the balls are in contact in the deformed state their total energy is in the form of potential energy and kinetic energy, so kinetic energy is less than the total energy. The energy spent against friction is dissipated in the form of heat which is not available for doing work. |
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| 35. |
Two particles of equal mass have velocities `vec v_1 = 2 hat i= m//s^-1` and `vec v_2 = 2hat j m//s^-1`. First particle has an acceleration `vec a_1= (3 hat i+ 3 hat j) ms^-2` while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of.A. circleB. parabolaC. straight lineD. ellipse |
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Answer» Correct Answer - C ( c) `vec v_(cm) = (m 2 hat i+ m 2hat j)/(m + m) = hat i+ hat j m//s` `vec a_(cm) = (m(3 hat i+ 3 hat j)+m xx 0)/(m+m) = (3)/(2) hat i+ (3)/(2) hat j` Since acceleration and velocity are in same direction, so path should be straight line. |
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| 36. |
The velocities of two particles `A and B` of same mass are `vec V_A = a hat i and vec V_B = b hat j` where `a and b` are constants. The acceleration of particle `A is (2 a hat i+ 4 b hat j)` and acceleration of particle `B is (a hat i- 4 b hat j)` (in m//s^2)`. The centre of mass of two particle will move in :A. straight lineB. parabolaC. ellipseD. circle |
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Answer» Correct Answer - A (a) `vec V_(com) = (m_1 vec V_1 + m_2 vec V_2)/(m_1 + m_2) = (vec V_1 + vec V_2)/(2) =(a hat i + b hat j)/(2)` `vec a_(com) = (a_1 + a_2)/(2) =(3)/(2) (a hat i + b hat j)` `vec V` is parallel to `vec a_(com)`. Hence path will be a straight line. |
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| 37. |
A square frame, made of `4` equal rods of length `l`, mass `m` each is hinged at corner `A` so that it can rotate about a horizontal axis paralle to line `DB` as shown. It is slightly disturbed and it starts falling about hinge axis. Calculate following parameters when the frame is completely inverted and point `C` reached the lower point of its motion. Velocity of point `B` :A. `sqrt((3)/(sqrt(2))gl)`B. `sqrt(3 sqrt(2) gl)`C. `sqrt(3 (g)/(l))`D. `sqrt(6 (g)/(l))` |
| Answer» Correct Answer - A | |
| 38. |
A square frame, made of `4` equal rods of length `l`, mass `m` each is hinged at corner `A` so that it can rotate about a horizontal axis paralle to line `DB` as shown. It is slightly disturbed and it starts falling about hinge axis. Calculate following parameters when the frame is completely inverted and point `C` reached the lower point of its motion. Angular velocity of frame :A. `sqrt(3 sqrt(2) (g)/(l))`B. `sqrt((3g)/(l))`C. `sqrt((3)/(sqrt(2)) (g)/(l))`D. `sqrt((6 g)/(l))` |
| Answer» Correct Answer - A | |
| 39. |
A particle of mass m moves in the `xy-plane` with velocity of `vec v = v_x hat i+ v_y hat j`. When its position vector is `vec r = x vec i + y vec j`, the angular momentum of the particle about the origin is.A. `m(xv_y + yv_x) hat k`B. `-m(xv_y + yv_x) hat k`C. `m(yv_x - xv_y) hat k`D. `m(xv_y - yv_x) hat k`. |
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Answer» Correct Answer - D (d) We use `vec L = vec r xx vec p` : `vec L = |{:(hati,hatj,hatk),(x,y,0),(mv_(x),mv_(y),0):}|` =`hat i(0 - 0) - hat j (0 - 0) + hat k(mxv_y - myv_x)` `vec L = m(xv_y - yv_x) hat k`. |
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| 40. |
For a particle of a mass `100 gm`, position and velocity at any instant are given as `10 hat i+ 6 hat j cm` and `vec v = 5 hat I cm//s`. Calculate the angular momentum about the point (1,1) cm.A. `25 xx 10^-5 (- hat k) kgm^2 s^-1`B. `30 xx 10^-5 (- hat k) khm^2 s^-1`C. `3 xx 10^-3 (- hat k) kgm^2 s^-1`D. `25 xx 10^-4 (- hat k) kgm^2 s^-1` |
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Answer» Correct Answer - A (a) `vec L = m(vec r xx vec c) = (100)/(1000) [(9 hat i+ 5 hat j)/(100)] xx (5 hat i)/(100)` =`25 xx 10^-5 (-hat k) khm^2 s^-2`. |
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| 41. |
Four identical thin rods each of mass `M` and length `l`, from a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane isA. `(4)/(3) Ml^2`B. `(2)/(3) Ml^2`C. `(13)/(3) Ml^2`D. `(1)/(3) Ml^2` |
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Answer» Correct Answer - A (a) The moment of inertia of rod about an axis through its centre of mass and perpendicular to rod + (mass of rod) xx (perpendicular distance between two axes) =`(Ml^2)/(12) + M((l)/(2))^2 = (Ml^2)/(3)` Moment of inertia of the system `= (Ml^2)/(3) xx 4` =`(4)/(3) Ml^2`. |
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| 42. |
Two bodies of mass `1 kg` and `3 kg` have position vectors `hat i+ 2 hat j + hat k` and `- 3 hat i- 2 hat j+ hat k`, respectively. The centre of mass of this system has a position vector.A. `-2 hat i+ 2 hat k`B. `- 2 hat i- hat j + hat k`C. `2 hat i- hat j - 2 hat k`D. `-1 hat i+ hat j + hat k` |
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Answer» Correct Answer - A (b) The position vector of centre of mass `vec r = vec r = (m_1 vec r_1 + m_2 vec r_2)/(m_1 + m_2)` ltbrtgt =`(1(hat i + 2 hat j + hat k)+3(-3 hat i - 2 hat j + hatk))/(1 + 3)` =`(1)/(4) (-8 hat i - 4 hat j + 4 hat k)` =`-2 hat i- hat j + hat k`. |
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| 43. |
A flywheel rotating at `420 rpm` slows down at a constant rate of `2 rad s^-2` The time required to stop the flywheel is.A. 22 sB. 11 sC. 44 sD. 12 s |
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Answer» Correct Answer - A (a) Here, `v_0 = 420 rpm = 7 rps` :. `omega_0 = 2 pi v_0 = 2 xx (22)/(7) xx 7 = 44 rad s^-1` `omega = 0, prop = -2 rad s^-2` `t = (omega - omega_0)/(prop) = (-44)/(-2) = 22 s`. |
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| 44. |
A wheel of mass `5 kg` and radius `0.40 m` is rolling on a road without sliding with angular velocity `10 rad s^-1`. The moment of ineria of the wheel about the axis of rotation is `0.65 kg m^2`. The percentage of kinetic energy of rotate in the total kinetic energy of the wheel is.A. `22.4 %`B. `11.2 %`C. `88.8 %`D. `44. 8 %` |
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Answer» Correct Answer - D (d) Here, `m = 5kg, I = 0.65 kg m^2` `omega = 10 rad s^-1, R = 0.40 m` Linear velocity `v = R omega = 0.40 xx 10 = 4 m s^-1` Translation `KE` =`(1)/(2) mv^2 = (1)/(2) xx 5 xx 16 = 40 J` Rotational `KE` =`(1)/(2) I omega^2 = (1)/(2) xx 0.65 xx 100 = 32.5 J` Total `KE` = Translational `KE + Rotational KE` =`40 + 32.5 = 72.5 J` percentage of rotational `KE = (Rotational KE)/(Total KE) xx 100` =`(32.5)/(72.5) xx 100 = 44.8 %`. |
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| 45. |
Two discs of moments of inertia `I_1` and `I_2` about their respective axes, rotating with angular frequencies, `omega_1` and `omega_2` respectively, are brought into contact face to face with their axes of rotation coincident. The angular frequency of the composite disc will be `A`.A. `(I_1 omega_1 + I_2 omega_2)/(I_1 + I_2)`B. `(I_2 omega_1 + I_1 omega_2)/(I_1 + I_2)`C. `(I_1 omega_1 + I_2 omega_2)/(I_1 - I_2)`D. `(I_2 omega_1 + I_1 omega_2)/(I_1 - I_2)` |
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Answer» Correct Answer - A (a) Total initial angular momentum of the two discs is `L_i = I_1 omega_1 + I_2 omega_2` When two discs are brought into contact face to face (one on top of the other) and their axes of rotation coincide, the moment of inertia `I` of the system is equal to the sum of their individual moments of inertia. i.e., `I = I_1 + I_2` Let `omega` be the final angular speed of the system. The final angular momentum of the system is. `L_f = I omega = (I_1 + I_2) omega` As no external torque acts on the system, therefore according to law of conservation of angular momentum, we get `L_i = L_f` `I_i omega_1 + I_2 omega_2 = (I_1 + I_2) omega` `omega = (I_1 omega_1 + I_2 omega_2)/(I_1 + I_2)`. |
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| 46. |
Calculate the vertical and horizontal reactions of the pivot when the rod of the problem 15 has turned through `theta=60^(@)`. |
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Answer» Correct Answer - `(7/16)mg`, `(3sqrt(3))/(16)`mg |
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| 47. |
A tim,ber of mas m rests on two rollers, each of mass `m/2` and radius r and is pulled along a horizontal plane by a force P. Assuming there is no slipping and treating the rollers as right circular cylinders, find the acceleration of the timber. |
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Answer» Correct Answer - `(8)/(11)(P)/(m)` |
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| 48. |
A bobbin is pushed along on a rough stationary horizontal surface as shown in the figure. The board is kept horizontal and there is no slipping at any contact points. The distance movedby the board when distance moved by the axis of the bobbin is `l` is .A. `l(1 + (r)/(2 R))`B. `l(2 + (r)/(2 R))`C. `l(1 + (r)/(R))`D. `l(1 + (2r)/(R))` |
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Answer» Correct Answer - C ( c) For no slipping , `v_(cm) = omega R` `v_B = omega r + v_(cm) = v_(cm)(1 + (r)/(R))` `(l _B)/(l) = (v_B)/(v_(cm)) = (1 + (r)/(R)) rArr l_B = l (1 + (r)/(R))`. |
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| 49. |
A gun fires a bullet of mass `50 g` with a velociy of `30 m//s`. Due to this, the gun is pushed back with a velocity of `1 m//s`, then the mass of the gun is :A. 5.5 kgB. 1.5 kgC. 0.5 kgD. 3.5 kg |
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Answer» Correct Answer - B (b) If no external force acts upon a system of two (or more) bodies then the total momentum of the system remains constant. Hence, momentum before collision = momentum after constant `m_1 u_1 = m_2 v_2` Given, `u_1 = 1 m//s, m_2 = 0.05 kg, v_2 -= 30 m//s` `rArr m_1 xx 1 = 0.05 xx 30` `rArr m_1 = 1.5 kg`. |
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| 50. |
In previous problem, the tangential acceleration of a point on the rim is :A. `(T)/(M)`B. `(MR^2)/(T)`C. `(2 T)/(M)`D. `(MR^2)/(2 T)` |
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Answer» Correct Answer - C ( c) Tangential acceleration `a = r prop = R. ((2 T)/(MR)) = (2 T)/(M)`. |
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