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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 disc of moment of inertia `4kgm^2` is spinning freely at `3rads^(-1)`. A second disc of moment of inertia `2kgm^2` slides down the spindle and they rotate together. (a) What is the angular velocity of the combination ? (b) What is the change in kinetic energy of the system? |
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Answer» (a) Since there are no external torques acting, we may apply the conservation of angular momentum, `I_(f)omega_(f)=I_(i)omega_(i)rArr(6 k g m^(2))omega_(f)=(4kgm^(2))(3 rads^(-1))` Thus `omega_(f)=2rads^(-1)` (b) The kinetic energies before and after the collision are `K_(i)=1/2I_(i)omega_(i)^(2), K_(f)=1/2I_(f)omega_(f)^(2)=12 J` The change is `DeltaK=K_(f)-K_(i)=-6 J` In order for the two discs to spin together at the same rate, there had to be friction between them . the loss in kinetic energy is converted into thermal energy. |
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| 2. |
A uniform rod of mass `m` is supported on two rollers each of mass `m//2` and radius `r` and rolls down the inclined rough plane as shown in the figure. Assuming no slipping at any contanct and treating the rollers as uniform solid cylinders: If acceleration of rod down in the plane is `a_(R)` and acceleration of centre of mass of roller is `a_(o)` then:A. `a_(R)gta_(0)`B. `a_(R)lta_(0)`C. `a_(R)=a_(0)`D. information is insuficieng to decide |
| Answer» Correct Answer - A | |
| 3. |
A rod leans against a stationary cylindrical body as shown in figure, and its right end sides to the right on the floor with a constant speed `v`. Choose the correct option (s) A. the angular speed `omega` is `(_Rv^(2)(2x^(2)-R^(2)))/(x^(2)(x^(2)-R^(2))^(3//2))`B. the angular acceleration `alpha` is `(Rv)/(xsqrt(x^(2)-R^(2)))`C. the angular speed `omega` is `(Rv)/(xsqrt(x^(2)-R^(2)))`D. the angular acceleration `alpha` is `(-Rv^(2)(2x^(2)-R^(2)))/(x^(2)(x^(2)-R^(2))^(3//2))` |
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Answer» Correct Answer - C::D From the geometry, `x=R/(sin theta)` Also, `omega=-(d theta)/(dt)`. Therefore, `v=(dx)/(dt)=d/(dt)(R/(sin theta))` `=(-R (d theta//dt)cos theta)/(sin^(2) theta)=(omegaR cos theta)/(sin^(2) theta)` ltbgt `omega=(v sin^(2) theta)/(R sin theta)=(Rv)/(xsqrt(x^(2)-R^(2)))` `alpha=(d omega)/(dt)=d/(dt)((Rv)/(x(sqrt(x^(2)-R^(2))))=-(Rv^(2)(2x^(2)-R^(2)))/(x^(2)(x^(2)-R^(2))^(3//2))` |
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| 4. |
The uniform `120 N` board shown in figure is supported by two ropes. A `400 N` weight is suspended one-fourth of the way from the left end. Choose the correct options A. `T_(1)=185N`B. `T_(2)=371N`C. `T_(2)=185N`D. `tan theta=0.257` |
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Answer» Correct Answer - A::B::D Taking moments about the left edge and resolving `T_(1)` into `x` and `y` components, `sum tau=0` yields `LT_(1) cos 30^(@)-(0.25 L)(400)` `-(0.5 L)(120)=0` Dividing throught by `L` and solving, we get `T_(1)=185 N` Substituting into our earlier equations, we get `T_(1)=185 N` `T_(2) = sin theta=92.5 N` and `T_(2)=cos theta=360 N` Dividing the equations yields `tan theta=0.257, `or `theta=14.4^(@)` Than `0.249 T_(2)=92.5` and `T_(2)=371 N` One can always chek moment problem results by taking moments about another point, such as the right end of the bar for this problem |
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| 5. |
Three particles, each of mass `m` are placed an the points `(x_(1),y_(1),z_(1)),(x_(2),y_(2),z_(2))` and `(x_(3),y_(3),z_(3))` on the inner surface of a paraboloid of revolution obtained by rotating the parabola `x^(2)=4ay` about the `y`-axis. Neglected the mass of the paraboloid. (`y`-axis. is along the vertical (a) the moment of inertia of the system about the axis of the paraboloid is `I=4ma(y_(1)+y_(2)+y_(3))` A. the moment of inertia of the system about the axis of the paraboloid is `I=4ma(y_1+y_(2)+y_(3))`B. if potential energy at `O` is taken to be zero, the potential energy of the system is `mg(y_(1)+y_(2)+y_(3))`C. if the particle at`(x_(1),y_(1),z_(1))` slides down the smooth surface, its speed at `O` is `sqrt(2gy_(1))`D. if the parabola spins about `OY` with an angular speed, `omega`, the kinetic energy of the system will be `2ma(y_(1)+y_(2)+y_(3))omega^(2)`. |
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Answer» Correct Answer - A::B::C::D For any point on the sufrace of paraboloid,`(x^(2)+z^(2))=4ay` `I=sum_(i=1)^(3)m(x_(i)^(2)+z_(i)^(2))` (distance of `m_(1)` from `y`-axis is) `sqrt(x_(i)^(2)+z_(i)^(2))=4ma(y_(1)+y_(2)+y_(3))` `mg(x_(1)+x_(2)+x_(3))=mg(y_(1)+y_(2)+y_(3))` `mgy_(1)=1/2mupsilon_(1)^(2)rArrupsilon_(1)sqrt(2gy_(1))` distance `y`-mass `m_(p)` from `y`-axis `r_(i)=sqrt(x_(i)^(2)+z_(i)^(2))=sqrt(4ay_(i))` `KE=1/2 m omega^(2)(r_(1)^(2)+r_(2)^(2)+r_(3)^(2))` `=1/2momega^(2)4a(y_(1)+y_(2)+y_(3))` |
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| 6. |
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. `2I`B. `I`C. `I//2`D. `I//4` |
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Answer» Correct Answer - A `I=(MR^(2))/2` and `I^(1)=MR^(2)=2I` |
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| 7. |
Ratio of densities of materials of two circular discs of same mass and thickness `5:6` the ratio of their `M.I.` about natural axes isA. `5:6`B. `6:5`C. `25:26`D. `1:1` |
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Answer» Correct Answer - B `I=1/2pitR^(4)rho` |
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| 8. |
A ring of mass `m` and radius `r` is melted and then moulded into a sphere. The moment of inertia of the sphere will beA. more than that of the ringB. less than that of the ringC. equal to that of the ringD. none of the above |
| Answer» Correct Answer - B | |
| 9. |
Three rings, each of mass `m` and radius `r`, are so placed that they touch each other. Find the moment of inertia about the axis as shown in Fig. A. `3MR^(2)`B. `3/2 MR^(2)`C. `5MR^(2)`D. `7/2 MR^(2)` |
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Answer» Correct Answer - D `I=(MR^(2))/2+2[3/2MR^(2)]=(MR^(2))/2+3MR^(2)=(7MR^(2))/2` |
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| 10. |
As shown in figure from a uniform rectangular sheet a triangular sheet is removed from one edge. The shift of centre of mass is A. `4.2 cm`B. `-4.2 cm`C. `6.67 cm`D. `-6.67 cm` |
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Answer» Correct Answer - D shift `=("-mass of removed part x d")/("Mass of remaining part")` here `d=20 cm` |
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| 11. |
A uniform sphere of mass `m` radius `r` and moment of inertia `I` about its centre moves along the `x`-axis as shown in Fig. Its centre of mass moves with velocity `= v_(0)`, and it rotates about its centre of mass with angular velocity `=omega_(0)`. Let `vecL=(Iomega_(0)+mv_(0)r)(-k)`. The angular momentum of the body about the origin `O` is A. `L`, only if `v_(0)=omega_(0)r`B. greater than `L`, if `v_(0)gt omega_(0)r`C. less than, `L` if `v_(0)gt omega_(0)r`D. `L`, for all value of `v_(0)` and `omega_(0)` |
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Answer» Correct Answer - D `L=RxxMv_(0)+Iomega_(0),=Mv_(0)r+Iomega`, which is constant |
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| 12. |
A uniform square plate and a disc having same mass per unit area are kept in contact as shown in Fig. The side of square and diameter of circle are both equal to `L`. Locate the position of centre of mass of the system w.r.t. the centre of the square. A. at point of contactB. inside the discC. inside the squareD. outside the system |
| Answer» Correct Answer - B | |
| 13. |
Three particles each of mass `2kg` are at the corners of an equilateral triangle of side `sqrt 3 m`. If one of the particles is removed, the shift in the centre of mass isA. `0.2m`B. `0.5 m`C. `0.4 m`D. `0.3m` |
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Answer» Correct Answer - B shift `=(md)/(M-m)` |
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| 14. |
The moment of inertia of a solid sphere about an axis passing through its centre is `0.8kgm^2`. The moment of inertia of another solid sphere whose mass is same as mass of first sphere, but the density is `8` times density of first sphere, about an axis passing through its centre isA. `0.1 kgm^(2)`B. `0.2 kgm^(2)`C. `0.4kgm^(2)`D. `0.5 kgm^(2)` |
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Answer» Correct Answer - B Mass is same and `D prop1/(R^(3)), (I_(1))/(I_(2))=((R_(1))/(R_(2)))^(2)=((D_(2))/(D_(1)))^(2/3)` |
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| 15. |
The ratio of moments of inertia of solid sphere about axes passing through its centre and tangent respectively isA. `2:5`B. `2:7`C. `5:2`D. `7:2` |
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Answer» Correct Answer - B `(I_("centre"))/(I_("tangent"))=(2/5MR^(2))/(7/5MR^(2))=2/7` |
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| 16. |
The ratio of moments of inertia of two solid spheres of same mass but densities in the ratio `1:8` isA. `1:4`B. `4:1`C. `2:1`D. `8:1` |
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Answer» Correct Answer - B `I_(rms)=2/5 MR^(2)` |
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| 17. |
A circular plate of diameter d is kept in contact with a square plate of edge d as shown in figure. The densit of the material and the thickness are same . Everywhere. The centre of mass of the composite system will beA. at their point of contanctB. inside the circular plateC. inside the square plateD. outside the combination |
| Answer» Correct Answer - C | |
| 18. |
Figure shows a square plate of uniform thickness and side length `sqrt 2 m`. One fourth of the plate is removed as indicated. The distance of centre of mass of the remaining portion from the centre of the original square plate is A. `1//3m`B. `1//2m`C. `1//6 m`D. `1//8m` |
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Answer» Correct Answer - C m be the mass of each part `3mxxc c_(2)=mxx c c_(1)` or `X=(-ad)/(A-a)` a -area of removed plate `A`-area of original plate d-distance centres |
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| 19. |
From a uniform wire, two circular loops are made (`i`) `P` of radius `r` and (`ii`) `Q` of radius `nr`. If the moment of inertia of `Q` about an axis passing through its center and perpendicular to tis plane is `8` times that of `P` about a similar axis, the value of `n` is (diameter of the wire is very much smaller than `r` or `nr`)A. `8`B. `6`C. `4`D. `2` |
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Answer» Correct Answer - D Moment of inertia `P=Mr^(2)` Moment of inertia `Q=(nM)(nr^(2))` given `I_(Q)=8I_(P)rArrn=2` |
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| 20. |
The moment of inertia of a hollow sphere of mass `M` having internal and external radii `R` and `2R` about an axis passing through its centre and perpendicular to its plane isA. `3/2 MR^(2)`B. `13/32MR^(2)`C. `31/35 MR^(2)`D. `62/35 MR^(2)` |
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Answer» Correct Answer - D If `I_(1)` and `I_(2)` are moment of inertia of hollow spheres of radii `R` and `2R` respectively, then `I=I_(2)-I_(1)` and mass `alphaR^(3)` |
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| 21. |
The ratio of radii of two solid spheres of same material is `1:2`. The ratio of moments of inertia of smaller and larger spheres about axes passing through their centres isA. `1:4`B. `1:8`C. `1:16`D. `1:32` |
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Answer» Correct Answer - D `I=2/5MR^(2),MalphaR^(3), IalphaR^(5)` |
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| 22. |
Thickness of a wooden circular plate is same as the thickness of a metal circular plate but density of metal plate is `8` times density of wooden plate. If moment of inertia of wooden plate is twice the moment of inertia of metal plate about their natural axes, then the ratio of radii of wooden plate to metal plate isA. `1:2`B. `1:4`C. `4:1`D. `2:1` |
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Answer» Correct Answer - D `I_(1)=2I_(2), M_(1)R_(1)^(2)=2M_(2)R_(2)^(2)` `((R_(1))/(R_(2)))^(2)=2(M_(2))/(M_(1)) ` , but `m prop DxxR^(2)` `((R_(1))/(R_(2)))^(2)=2 (D_(2))/(D_(1))((R_(2))/(R_(1)))^(2)=((R_(1))/(R_(2)))^(4)=2(D_(2))/(D_(1))` |
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| 23. |
A metallic thin wire has uniform thickness. From this wire, two circular loops of radii `r`, `2r` are made. If moment of inertia of `2^(nd)` loop about its natural axis is `n` times moment of inertia of `1st` loop about its natural axis. The value of `n` isA. `2`B. `4`C. `2sqrt(2)`D. `8` |
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Answer» Correct Answer - D `IpropMR^(2)` and `MpropLpropR :. IpropR^(3)` and `(I_(2))/(I_(1))=((R_(2))/(R_(1)))^(3)` |
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| 24. |
The moment of inertia of two spheres of equal masses about their diameters are the same. One is hollow, then ratio of their diametersA. `1:5`B. `1:sqrt(5)`C. `pi:1`D. `sqrt(5):sqrt(3)` |
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Answer» Correct Answer - D `M.I.` of solid sphere about diameter `=2/5 mr^(2)` `M.I.` of hollow sphere about diamter `=2/3mr^(2)` |
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| 25. |
Four point size bodies each of mass `m` are fixed at four corners of light square frame of side length `1m`. The radius of gyration of these four bodies about an axis perpendicular to the plane of frame passing through its centre isA. `sqrt(2)`B. `2`C. `1/(sqrt(2))`D. `1/2` |
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Answer» Correct Answer - C `I=2ml^(2), k=sqrt(I/(4m))` |
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| 26. |
Three identical spheres each of mass `m` and radius `R` are placed touching each other so that their centres `A,B` and `C` lie on a straight line. The position of their centre of mass from centre of `A` isA. `(2R)/3`B. `2R`C. `(5R)/3`D. `(4R)/3` |
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Answer» Correct Answer - B `x_(cm)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))` |
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| 27. |
Four identical solid spheres each of mass `M` and radius `R` are fixed at four corners of a light square frame of side length `4R` such that centres of spheres coincides with corners of square, moment of inertia of `4` spheres about an axis passing through any side of square isA. `(21MR^(2))/5`B. `(42MR^(2))/5`C. `(84 MR^(2))/5`D. `(168 MR^(2))/5` |
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Answer» Correct Answer - D `I=2(I_(c)+I_(1))` and `I_(1)=I_(c)+Md^(2)` `I_(c)=2/5 MR^(2)` and `d=4R` |
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| 28. |
Four identical solid spheres each of mass `M` and radius `R` are fixed at four corners of a light square frame of side length `4R` such that centres of spheres coincides with corners of square. The moment of inertia of `4` spheres about an axis perpendicular to the plane to frame and passing through its centre isA. `(21MR^(2))/5`B. `(42MR^(2))/5`C. `(84 MR^(2))/5`D. `(168 MR^(2))/5` |
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Answer» Correct Answer - D `I=4I_(1)` where `I_(1)` is `M.I.` of each sphere `I_(1)=I_(c)+Md^(2)` and `I_(c)=2/5 MR^(2), d=L/(sqrt(2)), L=4R` |
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| 29. |
Two spheres each of mass `M` and radius `R//2` are connected at their centres with a mass less rod of length `2 R`. What will be the moment of inertia of the system about an axis passing through the centre of one of the sphere and perpendicular to the rod ?A. `21/5 MR^(2)`B. `2/5 MR^(2)`C. `5/2 MR^(2)`D. `5/21 MR^(2)` |
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Answer» Correct Answer - A `I=2[2/5M(R/2)^(2)]+M(2R)^(2)` |
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| 30. |
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. `2omega`B. `omega`C. `(omega)/2`D. `0` |
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Answer» Correct Answer - A Aa no external torque acts on the system, angular moemtum should be conserved. Hence, `I omega=constant`....(i) 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) `rArrI omega=I/2 xxomega_(2)` (as mass reduce to half, hence, moment of inertia also reduced to half) `Iomega_(2)=2 omega` |
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| 31. |
`vecF = a hat i + 3hat j+ 6 hat k` and `vec r = 2hat i-6hat j -12 hat k`. The value of `a` for which the angular momentum is conserved isA. `-1`B. `0`C. `1`D. `2` |
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Answer» Correct Answer - A `vec(tau)=vec(r)xxvec(F)` and `tau=(dL)/(dt)=0` |
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| 32. |
A thin uniform rod of length `l` and mass `m` is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is `omega`. Its centre of mass rises to a maximum height of - |
| Answer» `mgh=1/2 Iomega^(2)=1/2(ml^(2))/3 omega^(2)rArrh=(l^(2)omega^(2))/(6g)` | |
| 33. |
The angular velocity of the seconds hand in a watch isA. `0.053 rad//s`B. `0.210 rad//s`C. `0.105 rad//s`D. `0.42 rad//s` |
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Answer» Correct Answer - C `omega=(2pi)/60` |
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| 34. |
A ring of mass M and radius R is rotating with angular speed `omega` about a fixed vertical axis passing through its centre O with two point masses each of mass `(M)/(8)` at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is `(8)/(9) omega` and one fo the masses is at a distance of `(3)/(5) R` from O. At this instant the distance of the other mass from O is |
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Answer» Correct Answer - 4 Using conservation of angular momentum `mR^(2)omega=(mR^(2)xx(8 omega)/9)+(m/8xx(9R^(2))/25xx(8 omega)/9)+(m/8xxx^(2)xx(8 omega)/9)rArrx=(4R)/5` |
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| 35. |
If the earth were to suddenly contract to `1//n^(th)` of its present radius without any change in its mass, the duration of the new day will be nearlyA. `24//n` hoursB. `24 n` hoursC. `24//n^(2)` hoursD. `24n^(2)` hours |
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Answer» Correct Answer - C `I_(1)omega_(1)=I_(2)omega_(2), (I_(1))/(T_(1))=(I_(2))/(T_(2)),R_(2)=(R_(1))/n` |
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| 36. |
If most of the population on earth is migrated to poles of the earth then the duration of a dayA. increasesB. decreasesC. remains sameD. first increases then decreases |
| Answer» Correct Answer - B | |
| 37. |
If radus of earth shrinks by `0.1%` without change in its mass, the percentage change in the duration of one dayA. decreases by `0.1 %`B. increases by `0.1%`C. decreases by `0.2%`D. increases by `0.2 %` |
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Answer» Correct Answer - C `Iomega=2/5MR^(2)xx((2pi)/T)=constant` `TpropR^(2)` and `(DeltaT)/T=2(DeltaR)/R` |
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| 38. |
Three point masses `m_(1), m_(2)` and `m_(3)` are located at the vertices of an equilateral triangle of side `alpha`. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through `m_(1)?`A. `(m_(1)+m_(2)+m_(3))a^(2)`B. `((m_(2)+m_(3))a^(2))/6`C. `((m_(2)+m_(3))^(2)a^(2))/2`D. `((m_(2)+m_(3))a^(2))/4` |
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Answer» Correct Answer - D `I=summr^(2), r_(1)=0,r_(2)=r_(3)=a/2` |
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| 39. |
A ballet dancer spins about a vertical axis at 60 rpm with his arms closed. Now he stretches his arms such that M.I. Increases by `50%`. The new speed of revolution isA. 80 rpmB. 40 rpmC. 90 rpmD. 30 rpm |
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Answer» Correct Answer - B `I_(1)omega_(1)=I_(2)omega_(2), I_(1)n_(1)=I_(2)n_(2)` |
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| 40. |
A cylinder of mass `10 kg` is rolling perfectly on a plane of inclination `30^(@)`. Find the force of friction between the cylinder and the surface of inclined plane.A. `49 N`B. `24.5N`C. `49//3N`D. `12.25N` |
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Answer» Correct Answer - C `f=(mg sin theta)/(1+(R^(2))/(K^(2)))` where `(R^(2))/(K^(2))=2` |
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| 41. |
When the following bodies having same radius starts rolling down on same inclined plane, identify the increasing order of their accelerations (I) hollow cylinder ,(II) Solid cylinder (III) solid sphere, (IV) hollow sphereA. `I,IV,III,II`B. `IV,I,II,III`C. `I,IV,II,III`D. `I,IV,III,II` |
| Answer» Correct Answer - C | |
| 42. |
A ring of mass `m` and radius `R` is rolling down on a rough inclined plane of angle `theta` with horizontal. Plot the angular momentum of the ring about the point of contact of ring and the plane as a function of timeA. B. C. D. |
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Answer» Correct Answer - D `L=(mg sin theta)Rt`, since , `tau=(DeltaL)/(Deltat)`, so, `L=tau(Deltat)` The curve between `L` and time `t` will be a straight line. |
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| 43. |
The handle of a door is at a distance `40cm` from axis of rotation. If a force `5N` is applied on the handle in a direction `30^@` with plane of door, then the torque isA. `0.8 Nm`B. `1 Nm`C. `1.6 Nm`D. `2Nm` |
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Answer» Correct Answer - B `tau =r F sin theta` |
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| 44. |
A door can just be opened with `10N` force on the handle of the door. The handle is at a distance of `50cm` from the hinges. Then, the torque applied on the door (in `Nm`) isA. `5`B. `10`C. `15`D. `20` |
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Answer» Correct Answer - A `tau=rF` |
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| 45. |
One end of a uniform rod of mass `M` and length `L` is supproted by a frictionless hinge which can with stand a tension of `1.75 Mg`. The rod is free to rotate in a vertical plane. The maximum angle should the rod be rotated from the vertical position so that when left, the hinge does not brek is `(pi)/n` |
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Answer» Correct Answer - 3 `1.75 Mg =Mg+(2M)/L((Lomega)/2)^(2)....(i)` `MgL/2(1-cos theta)=1/2(ML^(2))/3 omega^(2)....(ii)` Solving (i) and (ii) we get `theta=60^(@)` `a_(x)=0` |
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| 46. |
As shown in figure, the hinges `A` and `B` hold a uniform `400N` door in place. The upper hinge supports the entire weight of the door. Find the resultant force exerted on the door at the hinges, the width of the door is `h/2`, where `h` is the distance between the hinges. A. `312N`B. `280N`C. `412N`D. `480N` |
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Answer» Correct Answer - C Only a horizontal force acts at hinge `B`, because hinge `A` is assumed to support the door, s weight. Let us take torques about `A` as axis. `sumF_(x)=0`, or `F_(2)-H=0` `sum F_(y)=0` or `V-400 N=0` We find from these that `H=100 N` and `V=400 N` To find the resultant force `vec(R)` on hinge at `A`, we have `R=sqrt(((400)^(2)+(100)^(2))=412 N)` |
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| 47. |
A string is wrapped several time on a cylinder of mass `M` and radius `R`, the cylinder is pivoted about its axis of symmetery . A block mass of `m` tied to the string rests on support so that the string is black. The block is lifted upto a height `h` and the support is removed. (shown in figure) If `M=m`, what friction of `KE` is lost due to the jerk developed in the stringA. `1//2`B. `2//3`C. `1//3`D. `1//4` |
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Answer» Correct Answer - C For `M=m,K_(1)=(2K_(0))/3` so the fraction lost is `((K_(0)-K_(1))/(K_(0)))=1/3` |
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| 48. |
A stationary wheel starts rotating about its own axis at uniform rate amgular acceleration `8rad//s^(2)`.The time taken by its to complete `77` rotation isA. `14`B. `21`C. `28`D. `35` |
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Answer» Correct Answer - C `theta=2piN, omega^(2)=2alphatheta` |
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| 49. |
A uniform cube of mass `m` and side a is placed on a frictionless horizontal surface. A vertical force `F` is applied to the edge as shown in Fig. Match the following (most appropriate choice) : (a)`mg//4 lt F lt mg//2` (i) Cube will move up. (b) `F gt mg//2` (ii) Cube will not exhibit motion. (c ) `F gt mg` (iii) Cube will begin to rotate and slip at `A`. (d) `F = mg//4` (iv) Normal reaction effectively at `a//3` from A, no motion. A. `Ato2,Bto3,Cto1,Dto4`B. `Ato1,Bto2,Cto3,Dto4`C. `Ato4,Bto3,Cto2,Dto1`D. `Ato3,Bto4,Cto1,Dto2` |
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Answer» Correct Answer - A Cube has no motion if `Fxxa=(mga)/2` `:. F=(mg)/2` Cube rotates when `Fxxa gt mga/2` `:. F gt (mg)/2` When `(mg)/2 lt F lt (mg)/2` there is no motion when `F.(mg)/2` cube moves When `F gt mg` cube moves up when `F=(mg)/2` no motion |
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A rigid massless beam is balanced by a particle of mass `4m` in left hand side and a pulley particle system in right hand side. The value of `x/y` is A. `7/6`B. `5/6`C. `1:1`D. `11//12` |
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Answer» Correct Answer - D `a=((2m-m)g)/(3m)=g/3` for `T` in lower string, `T=2mg-2mxxg/3=4 mg//3` `& T^(1)=2T=(8mg)/3` `T^(1)+mg=(8mg)/3+mg=(11mg)/3` Let has take movent about hinge `4mgxxx=(11 mg)/3xxyrArrx/y=11/12` |
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