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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. |
Moment of inertia of a body depends uponA. axis of rotationB. torqueC. angular momentum `D. angular velocity |
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Answer» Correct Answer - A Moment of inertia of a body depends on position and orientation of the axis of rotation with respect to the body. |
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| 2. |
A particle of mass 1 kg is kept at (1m,1m,1m). The moment of inertia of this particle about Z-axis would beA. `1 "kg-m"^(2)`B. `2 "kg-m"^(2)`C. `3 "kg-m"^(2)`D. None of these |
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Answer» Correct Answer - B Perpendicular distance from Z-axis would be `sqrt((1)^(2)+(1)^(2))=sqrt(2)m` `:. I=Mr^(2)=(1)(sqrt(2))^(2)=2"kg-m"^(2)`. |
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| 3. |
Two identical rods each of mass M and length L are kept according to figure. Find the moment of inertia of rods about an axis passing through O and perpendicular to the plane of rods. |
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Answer» `because` Moment of interia about an axis passing through an end `= (ML^(2))/(3)` `:. I_("given system") = (ML^(2))/(3)+(ML^(2))/(3)=(2ML^(2))/(3)` |
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| 4. |
Two uniform, thin identical rods each of mass M and length `l` are joined together to form a cross. What will be the moment of inertia of the cross about an axis passing through the point at which the two rods are joined and perpendicular to the plane of the cross ?A. `(Ml^(2))/(12)`B. `(Ml^(2))/(6)`C. `(Ml^(2))/(4)`D. `(ML^(2))/(3)` |
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Answer» Correct Answer - B Moment of inertia, `I=4 [(M)/(2)((l//2)^(2))/(3)]=(Ml^(2))/(6)` |
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| 5. |
Assertion : The angular velocity of a rigid body in motion is defined for the whole body Reason : All points on a rigid body performing pure rotational motion are having same angular velocity.A. If both Assertion and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is fasleD. If Assertion is false but Reason is true |
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Answer» Correct Answer - B Angular velocity of particle of rigid body is same in rotational motion. |
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| 6. |
Circular disc of mass 2 kg and radius 1 m is rotating about an axis perpendicular to its plane and passing through its centre of mass with a rotational kinetic energy of 8 J. The angular momentum is (Js) isA. 8B. 4C. 2D. 1 |
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Answer» Correct Answer - B Rotational kinetic energy `= (1)/(2)I omega^(2)=8 J` `rArr (1)/(2)xx(1)/(2)mr^(2)omega^(2)=8or (1)/(4)xx2xx(1)^(2)omega^(2)=8` or `omega^(2)=16 or omega=4 "rad s"^(-1)` Angular momentum, `L=Iomega=(1)/(2)mr^(2)omega` `=(1)/(2)xx2xx(1)^(2)xx4=4"J-s"`. |
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| 7. |
For the uniform T shaped structure, with mass 3M, moment of inertia about an axis normal to the plane and passing through O would be A. `(2)/(3)Ml^(2)`B. `Ml^(2)`C. `(Ml^(2))/(3)`D. None of these |
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Answer» Correct Answer - B Moment of inertia, `I = 3[(Ml^(2))/(3)]=Ml^(2)` |
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| 8. |
A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity `omega` . Another disc of same dimensions but of mass `(1)/(4)` M is placed gently on the first disc co-axially. The angular velocity of the system isA. `(2)/(3) omega`B. `(4)/(5) omega`C. `(3)/(4) omega`D. `(1)/(3) omega` |
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Answer» Correct Answer - B We have, `I_(1)omega_(1)=I_(2)omega_(2)or omega_(2)=(I_(1))/(I_(2))omega_(1)` `=((1//2MR^(2)))/((1//2MR^(2))+(1//2)(1//4MR^(2)))omega=(4)/(5)omega`. |
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| 9. |
A particle P is moving in a circle of radius a with a uniform speed u, C is the centre of the circle and AB is a diameter. The angular velocities of P about A and C are in the ratioA. `1:1`B. `1:2`C. `2:1`D. `4:1` |
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Answer» Correct Answer - B From the property of a circle, if an arc subtend an angle `theta` at any point A on circumference, then it will subtend an angle `2 theta` at centre C. So in the same time interval a particle rotating in a circle turn double the angle with respect to centre point C compared to point A. `:. omega_(c)=2 omega_(A)`. |
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| 10. |
A sphere can roll on a surface inclined at an angle `theta`, if the friction coefficient `mu gt (2//7)g sin theta`. Now suppose the friction coefficient is `(1//7)g sin theta` and the sphere is released from rest on the incline, thenA. it will stay at restB. it will make pure translational motionC. it will translate and rotate about the centreD. the angular momentum of he sphere about its centre will remains constant |
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Answer» Correct Answer - C `mu lt mu_(min)"but" mu ne0`. Therefore, sphere will roll with forward slipping. |
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| 11. |
A disc is free to rotate about a smooth horizontal axis passing through its centre of mass. A particle is fixed at the top of the disc. A slight push is given to the disc and it starts rotating. During the process. A. only mechanical energy is conservedB. only angular momentum (about the axis of rotation) is conservedC. both mechanical energy and angular momentum are conservedD. Neither the mechanical energy nor the angular momentum are conserved. |
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Answer» Correct Answer - A Angular momentum will not remain conserved due to a torque of weight of particle about axis of rotation. |
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| 12. |
A sphere cannot roll onA. a smooth inclined surfaceB. a smooth horizontal surfaceC. a rough inclined surfaceD. a rough horizontal surface |
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Answer» Correct Answer - B On inclined surface gravitationl force provides torque. On rough surface friction provides torque for rolling but on horizontal smooth surface we need external force to create torque. |
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| 13. |
A thin bar of mass `m` and length `l` is free to rotate about a fixed horizontal axis through a point at its end.The bar is brought to a horizontal position. `(theta = 90^(@))` and then released. The angular velocity when it reaches the lowest point isA. directly proportional to its length and inversely proportional to its massB. independent of mass and inversely proportional to the square root of its lengthC. dependent only upon the acceleration due to gravity and the mass of the barD. directly proportional to its length and inversely proportional to its length and inversely proportional to the acceleration due to gravity. |
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Answer» Correct Answer - B Decrease in gravitational potential energy = Increase in rotational kinetic energy about an axis passing through end point `:. "mg"(l)/(2)=(1)/(2)((ml^(2))/(3))omega^(2)` `rArr` Angular velocity, `omega=sqrt((3g)/(l)) or omegaprop sqrt((g)/(l))` |
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| 14. |
A mass of 10 kg connected at the end of a rod of negligible mass is rotating in a circle of radius 30 cm with an angular velocity of 10 rad/s. If this mass is brought to rest in 10 s by a brake, what is the magnitude of the torque applied ?A. 0.9 N-mB. 1.2 N-mC. 2.3 N-mD. 0.5 N-m |
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Answer» Correct Answer - A `omega_(1)=10 "rad s"^(-1), omega_(2)=0,t=10s` `:. alpha=(omega_(2)-omega_(1))/(t)=(0-10)/(10)=-1 "rad s"^(-2)` Negative sign means retardation Now, `I=MR^(2)=10xx(0.3)^(2)=0.9 "kg-m"^(2)` `:.` Torque, `tau=Ialpha=0.9xx(1)=0.9 "N-m"`. |
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| 15. |
A portion of a ring of radius R has been removed as shown in figure. Mass of the remainting portion is `m`. Centre of the ring is at origin O. Let `I_(A) and I_(O)` be the moment of inertia passing through points A and O are perpendicular to the plane of the ring. Then, A. `I_(O)=mR^(2)`B. `I_(O)=I_(A)`C. `I_(O) gt I_(A)`D. `I_(A) gt I_(O)` |
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Answer» Correct Answer - D Whole mass has equal distance from the centre O. Hence, `I_(0) mR^(2)`. Further centre of mass of the remaining portion will be to the left of point O. More the distance of axis from centre of mass,more is the moment of inertia. Hence, `I_(A) gt I_(0)` |
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| 16. |
A disc of mass m and radius R is rolling on horizontal ground with linear velocity v. What is the angular momentum of the disc about an axis passing through bottommost point and perpendicular to the plane of motion ?A. `(3)/(2) mvR`B. `mvR`C. `(1)/(2)mvR`D. `(4)/(3)mvR` |
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Answer» Correct Answer - A `L=mvR+I_(C) omega=mvR+(1)/(2)mvR=(3)/(2)mvR` |
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| 17. |
A particle travels in a circle of radius 20 cm at a uniformly increasing speed. If the speed changes from `8 ms^(-1)` to `9 ms^(-1)` in 2s, what would be the angular acceleration in `"rad s"^(-2)` ?A. `1.5 "rad s"^(-2)`B. `2.5 "rad s"^(-2)`C. `3.5 "rad s"^(-2)`D. `4.5 "rad s"^(-2)` |
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Answer» Correct Answer - B Given, radius of the circle `= 20 cm = 20 xx 10^(-2) m` When speed is `8 ms^(-1) and 9 ms^(-1)`, then angular speeds are respectively `omega_(1)=(8)/(20xx10^(-2))=40 "rads"^(-1)` and `omega_(2)=(9)/(20xx10^(-2))=45 "rads"^(-1)` `:.` Angular acceleration `= (omega_(2)-omega_(1))/(t)=(45-40)/(2)=2.5 "rad s"^(-2)`. |
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| 18. |
A rigid body rotates with an angular momentum L. If its rotational kinetic energy is made 4 times, its angular momentum will becomeA. 4LB. 16LC. `sqrt(2)L`D. 2L |
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Answer» Correct Answer - D Angular momentum, `L= sqrt(2KI)` (Just like `p= sqrt(2km)`) K is made 4 times. Hence, L will become 2 times as `L prop sqrt(K)`. |
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| 19. |
A solid sphere of radius r `r` is rolling on a horizontal surface. The ratio between the rotational kinetic energy and total energy.A. `(5)/(7)`B. `(2)/(7)`C. `(1)/(2)`D. `(1)/(7)` |
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Answer» Correct Answer - B The rotation kinetic energy, `K_(r)=(1)/(2)Iomega^(2)` where, `I=` moment of inertia of solid sphere. The total kinetic energy and translational kinetic energy `K_(T)=(1)/(2)mv^(2)+(1)/(2) Iomega^(2)` `=(1)/(2)m omega^(2)r^(2)+(1)/(2) I omega^(2)" "( :. "For pure rolling" v=omegar)` Now, `(K_(r))/(K_(T))=((1)/(2)I omega^(2))/((1)/(2) omega^(2)(mr^(2)+I))=(I)/(mr^(2)+I)` As `I=(2)/(5)mr^(2)` (for solid sphere) `rArr(K_(r))/(K_(T))=((2)/(5)mr^(2))/(mr^(2)+(2)/(5)mr^(2))=((2)/(5)mr^(2))/((7)/(5)mr^(2))=(2)/(5)xx(5)/(7)=(2)/(7)`. |
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| 20. |
Consider three solid spheres, sphere (i) has radius r and mass m, sphere (ii) has radius r and mass 3 m, sphere (iii) has radius 3r and mass m, All can be placed at the same point on the same inclined plane, where they will roll without slipping to the bottom, If allowed to roll down the incline, then at the bottom of the inclineA. sphere (i) will have the largest speedB. sphere (ii) will havethe largest speedC. sphere (iii) will have the largest kinetic energyD. all the sphere will have equal speeds. |
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Answer» Correct Answer - D Acceleration, `a=(g sin theta)/(1+(I)/(mv^(2)))` For a sphere, `(I)/(mv^(2))=(2)/(5)` Hence, `a=(5)/(7)g sin theta=` constant, Hence, speed of all sphere is same at the bottom. Sphere (ii) has the largest mass. Hence, it will have the maximum kinetic energy. |
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| 21. |
Assertion : Two identical solid spheres are rotated frm rest to same angular velocity `omega` about two different axes as shown in figure. More work will have to be done to rotate the sphere in case-2 Reason : Moment of inertia is case-2 is more. A. If both Assertion and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is fasleD. If Assertion is false but Reason is true |
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Answer» Correct Answer - A Moment of inertia of sphere in case 1 is `(2)/(5) MR^(2)` Moment of inertia of sphere is case 2 is `(7)/(5) MR^(2)` So, more work will have to be done to rotate the sphere in case 2. |
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| 22. |
A Merry -go-round, made of a ring-like plarfrom of radius `R and mass M`, is revolving with angular speed `omega`. A person of mass `M` is standing on it. At one instant, the person jumps off the round, radially awaay from the centre of the round (as see from the round). The speed of the round after wards isA. `2 omega`B. `omega`C. `(omega)/(2)`D. 0 |
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Answer» Correct Answer - A As no external torque acts on the system angular momentum should be conserved Hence, `I omega =` constant where, `I` is moment of inertia of the system and `omega` is angular velocity of the system From Eq. (i) `I_(1)omega_(1)=I_(2)omega_(2)` (where `omega_(1) and omega_(2)` are angular velocities before and after jumping) `rArr I omega = (I)/(2)xxomega_(2)` (as mass reduced to half, hence, moment of inertia also reduced to half) `rArr omega_(2)=2 omega`. |
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| 23. |
Calculate the kinetic energy of rolling ring of mass 0.2 kg about an axis passing through its centre of mass and perpendicular to it, if centre of mass is moving with a velocity of 3 m/s. |
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Answer» `KE=(1)/(2)mv_(CM)^(2)(1+(K^(2))/(R^(2)))` Moment of inertia of the ring about an axis passing through its centre of mass `= MR^(2)` `:. K=1` (radius of gyration) `m=0.2 kg,v_(CM) = 3 m//s` `KE=(1)/(2)xx0.2xx9(1+1)=0.2xx9=1.8 "kgm"^(2)s^(-2)`. |
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| 24. |
A horizontal force `F` acts on the sphere at its centre as shown. Coefficient of friction between ground and sphere is `mu`. What is maximum value of `F` for which there is no slipping? |
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Answer» `F-f=Ma` `f.R=(2)/(5)MR^(2)(a)/(R)` `rArrf=(2)/(5)MarArr f=(2)/(7)F` `(2)/(7) F le mu mg rArrF le(7)/(2)mu mg`. |
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| 25. |
A wheel of mass 8 kg has moment of inertia equals to `0.5 "kg-m"^(2)`. Determine its radius of gyration. |
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Answer» Given, mass , `M= 8 kg` Moment of Inertia, `I=0.5 "kg-m^"(2)` `because I=MK^(2)rArrK^(2)=(I)/(M)rArrK=sqrt((I)/(M))` `rArr` Radius of gyration, `K=sqrt((0.5)/(8))rArr K=0.25 m or K=25` cm. |
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| 26. |
On which two days are the days and nights equal all over the world and why ? What name do you give to these days ? |
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Answer» On 21st March and 23rd September the days and flights are equal all over and world due to the sun’s rays vertical on the equator. The name is Equinox of this position. |
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| 27. |
Why are days and nights equal throughout the world on 21st March and 23rd September ? |
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Answer» 21 st March and 23rd September are seasonal positions of the Earth and are known as equinoxes meaning equal nights. The Sun is overhead at the equator. Both the Hemispheres are equally inclined towards the Sun. The circle of light passes through the poles. With the result, one half of each hemisphere is in the light and the other half in darkness. So the days and nights are equal all over the world. |
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| 28. |
Assertion : A body is moving along a circle with a constant speed. Its angular momentum about the centre of the circle remains constant Reason : In this situation, a constant non-zero torque acts on the body.A. If both Assertion and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is fasleD. If Assertion is false but Reason is true |
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Answer» Correct Answer - C L about centre = mvR = constant `because tau = (dL)/(dt)` and derivative of any constant is zero `tau_("centre")=0` |
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| 29. |
A flywheel having a radius of gyration of 2m and mass 10 kg rotates at an angular speed of `5 rad s^(-1)` about an axis perpendicular to it through its centre. The kinetic energy of rotation isA. 500 JB. 2000 JC. 1000 JD. 250 J |
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Answer» Correct Answer - A `I=mK^(2)=10(2)^(2)=40 "kg-m"^(2)` `K_(R)=(1)/(2)I omega^(2)=(1)/(2)(40)(5)^(2)=500J` |
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| 30. |
A solid flywheel of 20 kg mass and 120 mm radius revolves at `600 "revmin"^(-1)`. With what force must a brake lining be pressed against it for the flywheel to stop in 3s, if the coefficient of friction is 0.1 ? |
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Answer» Given, revolutions per minute, `n_(0)=600 "rev min"^(-1)` `=(600)/(60)"rev s"^(-1)` Revolutions per second, `n_(0)=10 "rev s"^(-1)` So, initial angular velocity, `omega_(0)=2pin_(0)=(2pi)(10)=20pi "rad s"^(-1)` Let `alpha` be the constant angular retardation, then applying `omega = omega_(0)=alpha t` or `0=(20 pi)-3(alpha)` or `alpha=(20)/(3)pi "rad s"^(-2)` Further, `alpha=(tau)/(I)` Here, torque `tau=muNR" "("R=radius")` or `tau=muFR" "("F=applied force")` `=N=F and I=(1)/(2)mR^(2)` From the above equations, `(20)/(3)pi=(muFR)/((1)/(2)mR^(2))=(2muF)/(mR)` or Force, `F=(10pimR)/(3mu)` Substituting the values, we have `F=(10xx22xx20xx0.12)/(3xx7xx0.1)` or `F=2.51.43 N`. |
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| 31. |
What is meant by rotation of the earth ? Discuss the effect of the rotation of the earth. |
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Answer» (a) Rotation of the earth takes place around the axis of the earth within 24 hours making one complete round. (b) The rotation causes day and night alternately; the part of the earth facing the sun experiences day and on the opposite side the dark part experiences night. One night and one day together make a complete solar day. |
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| 32. |
Give reasons for each of the following :The speed of the rotation of the earth is greater at the Equator than at the Arctic Circle. |
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Answer» As every part of the earth takes a frill round within 24 hours, the equator being the largest circle of latitude 0°, the speed is greatest at the equator due to crossing the maximum distance per hour. i.e. it is 1660 km per hour, while it decreases to 0° at poles. |
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| 33. |
Assertion : Moment of inertia about an axis passing throught centre of mass is always minimum Reason : Theorem of parallel axis can be applied for 2-D as well as 3-D bodies.A. If both Assertion and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is fasleD. If Assertion is false but Reason is true |
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Answer» Correct Answer - C When compared with many parallel axes, `I_(CM)` is least. But it is wrong to say that parallel axis theorem can be applied to 2D body only. |
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| 34. |
What does the word equinox mean ? |
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Answer» Equal nights. |
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| 35. |
Give reasons for the following statements : (a) The Sun does not rise at the same time everywhere in the world. (b) The speed of rotation at Leningrad (60°N), Genoa (45°N) and Singapore (0°N) along the Earth’s axis is not the same. (c) We do not feel the great speeds of Earth’s rotation in day-to-day life. |
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Answer» (a) The Sun does not rise at the same time everywhere in the world because earth is not a flat disc. (b) The speed of rotation at Leningrad (60°N), Genoa (45°N) and Singapore (0°N) along the Earth’s axis is not the same because of the spherical shape of the earth. (c) We do not feel the great speeds of Earth’s rotation in day- to-day life because the land, water and the air around the Earth are moving as a whole. |
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| 36. |
What do you mean by “Rotation of Earth” ?What are its effects ? |
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Answer» The two motion of the earth are : 1. Rotation or the daily motion and, 2. Revolution or the annual motion. Rotation : The earth rotates (turns) round its axis from west to east once in twenty four hours. This motion is called the Daily Motion or Rotation. This rotation of the earth is the real cause of the apparent rising and setting of the sun which is stationary. Its effects : 1. Day and night’s are caused. This is by far the most important effect of rotation. 2. The sun, the moon and the star appear to revolve round the earth from east to west. 3. Winds and currents changes their direction. 4. Different places have different local times. 5. Tides occur regularly twice a day. Revolution : The earth revolves round the Sun once approximately 365 – 1/4 days. This motion of the earth round the Sun is called Revolution. The earth revolves round the sun in a fixed path at a speed of 29.8 km/sec. or 107,2000 km. per hour. This path is called the orbit of the earth. The earth’s orbit round the Sun cover a distance of about 965 million km. Its effects : 1. This motion of the earth is one of the cause of the seasons. 2. Days and nights are of unequal length at the same place. 3. Change in the altitude of midday Sun during the year. |
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| 37. |
The moment of inertia of ring about an axis passing through its diameter is `I`. Then moment of inertia of that ring about an axis passing through its centre and perpendicular to its plane isA. `(I)/(2)`B. `2I`C. `(I)/(4)`D. `4I` |
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Answer» Correct Answer - B For a ring, `I_(z) =MR^(2)` From perpendicular axis theorem, `I_(x)+I_(y)=I_(z)` Given, `I_(x)=I_(y)=I` From Eq. (i), we get `2I-I_(z)` . |
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| 38. |
Figure represents the moment of inertia of the solid sphere about an axis parallel to the diameter of the solid sphere and at a distance x from t. Which one of the following represents the variations of `I` with `x` ?A. B. C. D. |
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Answer» Correct Answer - D Moment of inertia, `I = I_(CM)+mx^(2)` i.e., `I-x` graph is a parabol not passing through origin. |
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| 39. |
Two discs have same mass and thickness. Their materials are of densities `pi_(1)` and `pi_(2)`. The ratio of their moment of inertia about central axis will beA. `1:rho_(1)rho_(2)`B. `rho_(1)rho_(2):1`C. `rho_(1):rho_(2)`D. `rho_(2):rho_(1)` |
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Answer» Correct Answer - D Moment of inertia of disc, `I=(1)/(2)MR^(2)=(1)/(2)M((M)/(pi rhot))` `(because rho=(M)/(V)=(M)/(piR^(2)t):. R^(2)=(M)/(pi rhot))` `I = (1)/(2) (M^(2))/(pi rho t)` `:. I prop (1)/(rho)` (As M and t are constant) `(I_(1))/(I_(2))=(rho_(2))/(rho_(1))`. |
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| 40. |
The moment of inertia `(I)` and the angular momentum `(L)` are related by the expressionA. `I=L omega`B. `L = I omega`C. `L=I^(2) omega`D. `omega=LI` |
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Answer» Correct Answer - B Angular momentum `L = I omega` |
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| 41. |
The moment of ineria `(I)` of a sphere of radius R and mass M is given byA. `I=MR^(2)`B. `I=(1//2)MR^(2)`C. `I=(4//3)MR^(2)`D. `I=(2//5)MR^(2)` |
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Answer» Correct Answer - D Moment of inertia about its diameter is `I = (2)/(5) MR^(2)` |
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| 42. |
The conservation of angular momentum demands thatA. the external force on the system must be zeroB. the external torque on the system must be zeroC. Both the external force as well as the external torque must be zeroD. Neither of them must be zero |
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Answer» Correct Answer - B From law of conservation of angular momentum, if no external torque acts on a system, then total vector sum of angular momentum of different particles of the system remains conserved. `tau_("ext") =0` then `(dL)/(dt)=0` |
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| 43. |
The moment of inertia of a circular loop of radius R, at a distance of `R//2` around a rotating axis parallel to horizontal diameter of loop isA. `MR^(2)`B. `(1)/(2)MR^(2)`C. `2MR^(2)`D. `(3)/(4)MR^(2)` |
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Answer» Correct Answer - D According to theorem of parallel axis, `I=I_(CM)+M((R)/(2))^(2)` `rArr I=(1)/(2)MR^(2)+(MR^(2))/(4)` Moment of inertia, `I = (3)/(4) MR^(2)`. |
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| 44. |
Four point masses each of value m, are placed at the corners of a square ABCD of side `l`, The moment of inertia of the system about an axis passing throught A and parallel to BD is A. `sqrt(3) ml^(2)`B. `3 ml^(2)`C. `ml^(2)`D. `2 ml^(2)` |
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Answer» Correct Answer - B AS it is clear from the figure, `AC=BD=sqrt(l^(2)+l^(2))=sqrt(2)l` Moment of inertia of four point masses about BD `I_(BD)=m((lsqrt(2))/(2))^(2)+mxx0+m((lsqrt(2))/(2))^(2)+mxx0` `I_(BD) = ml^(2)` Applying the theorem of parallel axis, `I_(xy)=I_(BD)+M(AO)^(2)=ml^(2)+4m((l)/(sqrt(2)))^(2)=3ml^(2)`. |
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| 45. |
The density of a non-uniform rod of length 1m is given by `rho (x)=a (1+bx^(2))` where, a and b are constant and `0 le x le 1`. The centre of mass of the rod will be atA. `(3(2+b))/(4(3b+b))`B. `(4(2+b))/(3(3+b))`C. `(3(3+b))/(4(2+b))`D. `(4(3+b))/(3(2b+b))` |
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Answer» Correct Answer - A Density is given as `rho (x) = a (1+bx^(2))` where a and b are constants and `0 le x le 1` Let `b rarr 0`, in this case `rho (x) = a=` constant Hence, centre of mass will be at `x=0.5` m (middle of the rod) Putting, `b=0` in all the options, only (a) given 0.5. Note : We should not check options by putting `a=0`, because `rho = 0` for a = 0. |
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| 46. |
When a disc rotates with uniform angular velocity, which of the following is not true ?A. The sense of rotation remains sameB. The orientation of the axis of rotation remain sameC. The speed of rotation is non-zero and remains sameD. The angular acceleration is non-zero and remains same |
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Answer» Correct Answer - D We know that angular acceleration `alpha=(d omega)/(dt)`, where `omega` is angular velocity of the disc given `omega=` constant `rArr alpha=(d omega)/(dt)=(0)/(dt)=0` Hence, angular acceleration is zero. |
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| 47. |
A sphere of mass m attached to a spring on incline plane as shown in figure is held in unstretched position of spring. Suddenly sphere is left free what is the maximum extension of spring if friction allows only rolling of sphere about horizontal diametre ? |
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Answer» At the point of maximum extension sphere is at rest. No work is done by frictional force in rolling `rArr` Loss in gravitational PE =Gain in PE of spring `rArr (1)/(2)kx^(2)=mhxsin theta rArr x=(2mg sin theta)/(k)` |
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| 48. |
Analogue of mass in rotational motion is.A. moment of inertiaB. angular momentumC. gyrationD. None of these |
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Answer» Correct Answer - A The role of moment of inertia in the study of rotational motion is analogous to that of mass in study of linear motion. |
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| 49. |
Consider a uniform rod of mass m and length 2l with two particles of mass `m` each at its ends. Let AB be a line perpendicular to the length of rod and passig through its centre. Find the moment of inertial of the system about AB. |
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Answer» Moment of inertia of the system about AB `I_(AB)=I_("rod")+I_("both particles")` `=(m(2l)^(2))/(12)+2(ml^(2))=(7)/(3)ml^(2)`. |
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A massless rod S having length `2l` has equal point masses attached to its two ends as shown in figure. The rod is rotating about an axis passing through its centre and making angle `alpha` with the axis. The magnitude of change of momentum of rod i.e., `|(dL)/(dt)|` equals A. `2 m l^(3) omega^(2) sin theta. cos theta`B. `ml^(2) omega^(2) sin 2 theta`C. `ml^(2) sin 2 theta`D. `m^(1//2) l^(1//2) omega sin theta. Cos theta` |
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Answer» Correct Answer - B The radius of the circular followed by the masses is `r = l sin alpha` As, angular momentum, `L = r xx p = r xx mv` `rArr |L| = l sin theta (m omega l sin theta)` On differentiating, we get `(d|L|)/(dt)=momega l^(2)2 sin theta. cos theta (d theta)/(dt)` `rArr |(dL)/(dt)|=2ml^(2) omega^(2)sin theta. cos theta =ml^(2)omega^(2) sin 2 theta`. |
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