InterviewSolution
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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. |
Value of radius of gyration of a body depends on axis of rotation. Radius of gyration is root mean square distance of particle of the body from the axis of rotation.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - a Radius of gyration of a body about a given axis is equal to K = `sqrt(r_1^(2) + r_2^(2) + …..r_n^(2)) / n`. It thus depends upon shape and size of the body, position and configuration of the axis of rotation and also on distribution of mass of body w.r.t the axis of rotation. |
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| 2. |
The radius of gyration of an uniform rod of length `l` about an axis passing through one of its ends and perpendicular to its length is.A. `l/sqrt 2`B. `l/3`C. `l/sqrt 3`D. `l/2` |
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Answer» Correct Answer - c Moment of inertia of rod of mass M and length l about its axis passing through one of its ends and perpendicular to its is `I =1/3Ml^(2)`, As I`=Mk^(2)` where k is the radius of the gyration `therefore Mk^(2)=1/3Ml^(2)` or `k=l/sqrt(3)` |
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| 3. |
Which of the following relations is correct?A. Mechanical advantage = Effort/LoadB. Load arm `xx` Effort = Effort arm `xx` LoadC. Load arm `xx` Load = Effort arm `xx` EffortD. None of these |
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Answer» Correct Answer - c Mechanical advantage `= ("Load")/("Effort")` Load arm x load = Effort arm x Effort |
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| 4. |
Moment of couple is calledA. angular momentumB. forceC. torqueD. impulse |
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Answer» Correct Answer - c Moment of couple is called torque. |
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| 5. |
A disc is rotating with angular velocity `omega`. A force F acts at a point whose position vector with respect to the axis of rotation is r. The power associated with torque due to the force is given byA. `(vecr xx vecF).vecomega`B. `(vecr xx vecF) xx vecomega`C. `vecr . (vecF xx vecomega)`D. `vecr xx (vecF. vecomega)` |
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Answer» Correct Answer - a Torque, `vectau=vecr xx vecF` `therefore` Power associated with the torque is `P=vectau.vecomega=(vecr xx vecF).vecomega` |
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| 6. |
When a disc rotates with uniform angular velocity, which of the following is not true ?A. The sense of rotation remains same.B. The orientation of the axis of rotation remains same.C. The speed of rotation is non-zero and remains same.D. The angular acceleration is non-zero and remains same. |
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Answer» Correct Answer - d When a disc rotates with uniform angular velocity, angular acceleration of the disc is zero. Hence, option(d) is not true. |
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| 7. |
A particle of mass m is moving in YZ-plane with a uniform velocity `v` with its trajectory running parallel to `+ve` Y-axis and intersecting Z-axis at `z=a` in figure. The change in its angular momentum about the origin as it bounces elastically form a wall at y=constant is |
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Answer» Correct Answer - b The initial velocity is `vecv_(i) = vhatey` After reflection from the wall, the final velocity is `vecv_(f) = -vhatey` The trajectory is given as `vecr= yhate_(y) + ahate_(z)` Hence, the change in angular momentum is `DeltavecL=vecr xx m(vecv_(f) - vecv_(i))` `=(yhate_(y) + avece_(z)) xx (-2mhatey)` `=2"mva"vece_(x) [therefore hate_(y) xx hate_(y) = 0` and `hate_(z) xx hate_(y) = -hate_(x)]` |
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| 8. |
Match the Column I and Column II.,B A. A-p, B-q, C-r, D-sB. A-q, B-r, C-s. D-pC. A-r, B-q, C-p, D-pD. A-r, B-s, C-p, D-q |
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Answer» Correct Answer - d For translational equilibrium, `sigmavecF=0`, A- r For rotational equilibrium, `sigmavectau=0`, `B-s` Torque is required to produce angular accelerations, D-q |
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| 9. |
(1) Centre of gravity (C.G.) of a body is the point at which the weight of the body acts, (2) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius, (3) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the body can be cosidered to be concentrated at its C.G.., (4) The radius of gyration of any body rotating about ab axis is the length of the perpendicular dropped from thr C.G. the body to the axis. which one of the following paries of statements is correct ?A. (1) and (4)B. (1) and (2)C. (2) and (3)D. (3) and (4) |
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Answer» Correct Answer - a Statement (1) and (4) are true. |
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| 10. |
Assertion: The centre of gravity of a body coincides with its centre of mass only if the gravitational field does not vary form one part of the body to the other. Reason: Centre of gravity is independent of the gravitational field.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - c Centre of gravity of an object depends on the gravitational field on the body. |
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| 11. |
A body is rolling down an inclined plane. If kinetic energy of rotation is `40 %` of kinetic energy in translatory start then the body is a.A. ringB. cylinderC. hollow ballD. solid ball |
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Answer» Correct Answer - d Rotational kinetic energy is `K_(R) = 1/2Iomega^(2) = 1/2Mk^(2) (v/r)^(2)` `(therefore I=Mk^(2)` and `v=Romega)` `=1/2Mv^(2)(k^(2)/R^(2))` Translation kinetic energy is `K_(T) = 1/2Mv^(2)` As per question, `K_(R) = 40% K_(T)` `therefore 1/2Mv^(2)(k^(2)/R^(2)) = 40% " of " 1/2 Mv^(2)` or `k^(2)/R^(2) = 40/100 = 2/5` For solid sphere, `k^(2)/R^(2) = 2/5` Hence, the body is solid ball. |
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| 12. |
The position of a particle is given by `vecr=(hati+2hatj-hatk)` and momentum `vecp=(3hati+4hatj-2hatk)`. The angular momentum is perpendicular to theA. x-axisB. y-axisC. z-axisD. yz-plane |
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Answer» Correct Answer - a Here, `vecr=hati+2hatj-hatk, vecp=3hati+4hatj-2hatk` `vecL= vecr xx vecp= |{:[hati, hatj, hatk],[1,2,-1],[3,4,-2]:}|=hati(-4+4)+hatj(-3+2)+hatk(4-6)=0hati-1hatj-2hatk` `vecL` has component along `-y` axis and `-z` axis but it has no component along the x-axis. The angular momentum `vecL` is in yz-plane. i.e., perpendicular to x-axis. |
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| 13. |
Consider a particle of mass m having linear momentum `vecp` at position `vecr` relative to the origin O. Let `vecL` be the angular momentum of the particle with respect to the origin. Which of the following equations correctly relate(s) `vecr`, `vecp` and `vecL`?A. `(dvecL)/(dt) + vecr xx (dvecp)/(dt)=0`B. `(dvecL)/(dt) + (dvecr)/(dt) xx vecp=0`C. `(dvecL)/(dt) - (dvecr)/(dt) xx vecp=0`D. `(dvecL)/(dt)- vecr xx (dvecp)/(dt)=0` |
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Answer» Correct Answer - d As `vecL= vecr xx vecp` Differentiate both sides with respect to time, we get `(dvecL)/(dt) - d/(dt) (vecr xx vecp) = (dvecr)/(dt) xx vecp + vecr xx (dvecp)/(dt)` `=vecr xx (dvecp)/(dt)` `therefore (dvecr)/(dt) xx vecp=0` `(dvecL)/(dt) - vecr xx (dvecp)/(dt)=0` |
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| 14. |
Moment of inertia of body depends uponA. mass of the bodyB. axis of rotation of the bodyC. shape and size of the bodyD. all of these |
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Answer» Correct Answer - d Moment of inertia of a body depends on the mass of the body, its shape and size, distribution of mass about the axis of rotation, and the position and orientation of the axis of rotation. |
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| 15. |
Assertion: No real body is truly rigid. Reason: A rigid body is a body with a perfectly definite and unchanging shape. The distances between different pairs of particles of such a body do not change.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - a It is evident from the second statement that no real body is truly rigid, since real bodies deform under the influence of forces. But in many situations the deformations are negligible. |
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| 16. |
The moment of inertia of rigid body depends only on the mass of the body, its shape and size. Moment of inertia `I = MR^2` where `M` is the mass of the body and `R` is the radius vector.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - d The moment of inertia of a rigid body depends on the mass of the body, its shape and size, distribution of mass about the axis rotation, and the position and orientation of the axis of rotation. |
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| 17. |
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 true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - a By theorem of parallel axis both statements are correct and reason is the correct explanation of assertion. |
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| 18. |
The position of the centre of mass of a cube of uniform mass density will be atA. the centre of one faceB. the centre of the interaction of diagonals of one face.C. the geometric centre of the cubeD. the edge of a cube |
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Answer» Correct Answer - c As the mass density of the cube is uniform, therefore its center of mass lies at its geometric center. |
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| 19. |
The reduce mass of two particles having masses m and 2 m isA. 2 mB. 3 mC. 2 m/3D. m/2 |
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Answer» Correct Answer - c Reduced mass `=(m xx 2m)/(m+2m)=(2m)/(3)` |
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| 20. |
Angular momentum of the particle rotating with a central force is constant due toA. constant torqueB. constant forceC. constant linear momentumD. zero torque |
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Answer» Correct Answer - d Torque due to central force is zero. `therefore tau=(dL)/(dt)=0 rArr L="constant"` |
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| 21. |
Total angular momentum of a rotating body remains constant, if the net torque acting on the body isA. zeroB. maximumC. minimumD. unity |
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Answer» Correct Answer - a According to law of conservation of angular momentum, if the net torque acting on the body is zero, then the total angular momentum of the body is constant. |
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| 22. |
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along.A. the radiusB. the tangent the orbitC. the line at angle of `45^(@)` to the plane of rotationD. the axis of rotation |
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Answer» Correct Answer - d When a mass is rotating in a plane about a fixed point, its angular momentum is directed along the axis of rotation. |
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| 23. |
A solid sphere of mass `m` and radius `R` is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic emergies of rotation `(E_("sphere")//E_("cylinder"))` will be.A. 0.085416666666667B. 0.045138888888889C. 0.044444444444444D. 0.12569444444444 |
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Answer» Correct Answer - b `(E_("sphere"))/(E_("cylinder"))= (1/2 I_(s) omega_(s)^(2))/(1/2 I_(c)omega_(c)^(2)) = (I_(s)omega_(s)^(2))/(I_(c)omega_(c)^(2))` Here, `I_(s) = 2/5 mR^(2), I_(c) = 1/2mR^(2)` `omega_(c) = 2omega_(s)` `(E_("sphere"))/(E"cylinder") = (2/5 mR^(2) xx omega_(s)^(2))/(1/2 mR^(2) xx (2omega_(s))^(2)) = 4/5 xx 1/4 = 1/5` |
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| 24. |
A person with outstretched arms, is spinning on a rotating stool. He suddenly brings his arms down to his sides. Which of the following is true about his kinetic energy K and angualr momentum L?A. Both K and L increaseB. Both K and L remain unchangedC. K remains constant, L increasesD. K increases but L remains constant |
| Answer» Correct Answer - d | |
| 25. |
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 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, `L_(1) = I_(1)omega_(1)` Final moment of inertia, `I_(2) = 6+2 xx 5 xx(0.2)^(2) = 6.4 kg m^(2)` Final angular momentum , `L_(2) = I_(2)omega_(2)` According to law of conservation of angualr momentum, `L_(1) = L_(2)` `omega_(2) = (I_(1)omega_(1))/(I_(2)) = 16 kg m^(2)(1.2 rev s^(-1))/(6.4 kg ms^(2)) = 3 rev s^(-1)` |
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| 26. |
A child is standing with his two arms outstretched at the centre of a turntable that is rotating about its central axis with an angular speed `omega_0`. Now, the child folds his hands back so that moment of inertia becomes `3` times the initial value. The new angular speed is.A. `3omega_o`B. `omega_o`/3C. `6omega_o`D. `omega_o`/6 |
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Answer» Correct Answer - b Here, Initial angular speed, `omega_(i) = omega_(0)` Initial moment of inertia =`I_(i)` Final moment of inertia `I_(f)=3I_(i)` According to the law of conservation of angualr momentum, we get `L_(i)=L_(f)`or `I_(i)omega_(i)=I_(f)omega_(f)` `omega_(f) =(I_(i)omega_(i))/(I_(f)) = (I_(i))/(3I_(i))omega_(0)=omega_(0)/3` |
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| 27. |
A cylinder of radius `R` and mass `M` rolls without slipping down a plane inclined at an angle `theta`. Coeff. of friction between the cylinder and the plane is `mu`. For what maximum inclination `theta`, the cylinder rolls without slipping ?A. tan`theta` gt `3 mu_s`B. tan`theta` `le` `3mu_s`C. tan`theta` lt `3 mu_s`D. None of these |
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Answer» Correct Answer - b The condition for rolling of cylinder without slipping is `tanthetale 3mus_(s)` |
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