Q 1 :

Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for the moment of inertia about their diameter axis AB as shown in the figure, is 8/x. The value of x is        [2024].

  • 17

     

  • 34

     

  • 67

     

  • 51

     

(3)

Hollow Sphere, Isphere=23MR2=Mk12

Solid Cylinder, Icylinder=112M(4R2)+14MR2+M(2R)2

Icylinder=6712MR2=Mk22

k1k2=23·1267=867x=67



Q 2 :

Three balls of masses 2 kg, 4 kg, and 6 kg respectively are arranged at the center of the edges of an equilateral triangle of side 2 m. The moment of inertia of the system about an axis through the centroid and perpendicular to the plane of the triangle will be ____ kg m2.    [2024]



(4)

rcos30°=12r=13

I=m1r2+m2r2+m3r2

I=(2+4+6)r2=12×13=4I=4 kg-m2



Q 3 :

Four particles each of mass 1 kg are placed at four corners of a square of side 2 m. Moment of inertia of system about an axis perpendicular to its plane and passing through one of its vertex is _____ kgm2.     [2024]



(16)

I=4ma2=4×1×(2)2=16

 



Q 4 :

Two identical spheres each of mass 2 kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centers is 150 cm. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is x/20 kgm2, where the value of x is _____.          [2024]



(53)

I=(25mR2+md2)×2

I=2(25×2×(12)2+2×(34)2)=5320 kg-m2

X=53



Q 5 :

A uniform circular disc of radius 'R' and mass 'M' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.          [2025]

  • 732MR2

     

  • 932MR2

     

  • 1732MR2

     

  • 1332MR2

     

(4)

I=IcompleteIremoved

Icomplete=MR22

Iremoved=(M4)(R2)212+(M4)(R2)2=3MR232

I=MR223MR232=1332MR2



Q 6 :

Moment of inertia of a rod of mass 'M' and length 'L' about an axis passing through its center and normal to its length is 'α'. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is:          [2025]

  • α

     

  • α/4

     

  • α/8

     

  • α/2

     

(2)

Moment of inertia of the rod in Ist case,

α=Ml212

Moment of inertia of the rod in IInd case,

α'=2[M2(l2)212]=Ml248=α4



Q 7 :

The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is:          [2025]

  • 12Mr2

     

     

  • 38Mr2

     

  • 32Mr2

     

  • 2 Mr2

     

(2)

I1=MR22

I'=I1+MR2=32MR2

For r = 2R

I'=38Mr2



Q 8 :

A rod of linear mass density 'λ' and length 'L' is bent to form a ring of radius 'R'. Moment of inertia of ring about any of its diameter is:         [2025]

  • λL316π2

     

  • λL312

     

  • λL34π2

     

  • λL38π2

     

(4)

L=2πR  R=L2π

I=MR22=λ×L2×(L2π)2=λL38π2

 



Q 9 :

The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is n times higher than the moment of inertia of the given ring. Here, n = __________. 

Consider all the bodies have equal masses.                                             [2025]



(4)

I1=MR124, I2=MR222, I3=2MR125

According to problem

I1I2=2.5  MR124MR222=52  R12R22=5          ... (i)

Now we are provided with information that

I3I2=n  2MR125MR222=n  4R125R22=n          ... (ii)

From Eq. (i) and (ii)  n = 4



Q 10 :

Two iron solid discs of negligible thickness have radii R1 and R2 and moment of inertia I1 and I2, respectively. For R2=2R1, the ratio of I1 and I2 would be 1/x, where x = __________.          [2025]



(16)

Given, R2=2R1

M1=σ×πR12=Mo

M2=σ×πR22=σ×π[2R1]2=σ×4πR12=4Mo

I1I2=M1R122M2R222=M1R12M2R22=14×14=116



Q 11 :

A, B and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure.

The moment of inertia of the given system about PQ is x15I, where I is the moment of inertia of the disc about its diameter. The value of x is _________.       [2025]



(199)

All bodies have same mass and same radius.

Disk, IA=mR24=I

Solid sphere, IB=75mR2

Spherical shell, IC=53mR2

IPQ=mR24+(25mR2+mR2)+(23mR2+mR2)

IPQ=15mR2+24mR2+60mR2+40mR2+60mR260

IPQ=19960mR2=19915(mR24)=19915I



Q 12 :

M and R be the mass and radius of a disc. A small disc of radius R/3 is removed from the bigger dise as shown in figure. The moment of inertia of remaining part of bigger disc about an axis AB passing through the centre O and perpendicular to the plane of disc is 4xMR2. The value of x is __________.          [2025]



(9)

Without cavity I1=MR22

Mass of removed disc = MπR2×(R3)2π=(M9)

M.I. of removed disc I2=M9(R3)22+M9×(2R3)2=MR218

I(remaining part)=I1I2=MR22MR218=4MR29  x=9



Q 13 :

ICM is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of the disc. IAB is its moment of inertia about an axis AB perpendicular to the plane and parallel to axis CM at a distance 23R from the center, where R is the radius of the disc. The ratio of IAB and ICM is x:9. The value of  'x' is ________.             [2023]



(17)

ICM=MR22

IAB=ICM+Mx2

       =MR22+M(23R)2

=MR22+M4R29=MR2(12+49)

=MR2(9+818)=17MR218

IAB:ICM=179



Q 14 :

If a solid sphere of mass 5 kg and a disc of mass 4 kg have the same radius, then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be x7. The value of x is __________ .             [2023]



(5)

I1=25m1R2+m1R2=m1R2(75)

I1=7R2

I2=m2R24+m2R2=54m2R2

I2=5R2

I2I1=57x=5



Q 15 :

Two discs of same mass and different radii are made of different materials such that their thicknesses are 1 cm and 0.5 cm respectively. The densities of the materials are in the ratio 3 : 5. The moment of inertia of these discs respectively about their diameters will be in the ratio of x6. The value of x is ________.         [2023]



(5)

m=ρπR2t

So R2=mρπt

       I=mR24=m24ρπt

So I1I2=ρ2t2ρ1t1=53×0.51=56

So x=5



Q 16 :

Moment of inertia of a disc of mass M and radius 'R' about any of its diameter is MR24. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, x2MR2. The value of x is __________ .              [2023]



(3)

I=Icm+Md2

      =MR22+MR2=32MR2

x=3



Q 17 :

Solid sphere A is rotating about an axis PQ. If the radius of the sphere is 5 cm, then its radius of gyration about PQ will be x cm. The value of x is ________.    [2023]



(110)

Icm=25MR2

IPQ=Icm+md2

IPQ=25mR2+m(10 cm)2

For radius of gyration, IPQ=mk2

k2=25R2+(10 cm)2

          =25(5)2+100=10+100=110

k=110cm

x=110



Q 18 :

A uniform solid cylinder with radius R and length L has moment of inertia I1, about the axis of the cylinder. A concentric solid cylinder of radius R'=R2 and length L'=L2 is carved out of the original cylinder. If I2 is the moment of inertia of the carved out portion of the cylinder, then I1I2= _______ .        [2023]



(32)

I1=m1R22  I2=m2(R/2)22

I1I2=4m1m2=4·ρπR2ρ·πR24×2I1I2=32



Q 19 :

Two identical solid spheres each of mass 2 kg and radii 10 cm are fixed at the ends of a light rod. The separation between the centres of the spheres is 40 cm. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is __________ ×10-3 kg·m2.             [2023]



(176)

I=2(Icm+md2)=2(25mr2+md2)

                               =45×2×(0.1)2+2(2)(0.20)2

                                =85×10-2+16×10-2

                                =(1.6+16)×10-2=17.6×10-2

I=17.6×10-3 kg m2



Q 20 :

A ring and a solid sphere rotating about an axis passing through their centers have the same radii of gyration. The axis of rotation is perpendicular to plane of the ring. The ratio of radius of the ring to that of the sphere is 2x. The value of x is ________ .                 [2023]



(5)

For ring, I=mR12=mK12

 Radius of gyration K1=R1

For solid sphere, I'=25m'R22=m'K22

  Its radius of gyration  K2=25R2

 K1=K2

 R1=25R2

 R1R2=25

 x=5



Q 21 :

The moment of inertia of a semicircular ring about an axis passing through the center and perpendicular to the plane of the ring is 1xMR2, where R is the radius and M is the mass of the semicircular ring. The value of x will be _______ .         [2023]



(1)

The moment of inertia of a semicircular ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is =MR2

So x=1



Q 22 :

A solid sphere and a solid cylinder of the same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyration respectively (ksph:kcyl) is 2:x, then the value of x is _________ .             [2023]



(5)

For solid sphere, 25mR2=mksph2

ksph=25R

For solid cylinder, mR22=mkcyl2

kcyl=R2

ksphkcyl=2512=25=2x

 x=5



Q 23 :

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length L (R<L) about an axis passing through the mid points of opposite sides, is ______. (Take the mass of the entire loop as M).             [2026]

  • 38MR2+16ML2

     

  • 38MR2+712ML2

     

  • 34MR2+712ML2

     

  • 34MR2+16ML2

     

(1)

Inet=2(I1+I2)

=2(M'R24+M'212)+2(M'R22+M'(2)2)

=M'R22+M'R26+M'R2+M'22

=3M'R22+2M'23

Given masses M'=M4

So, I=3(M/4)R22+2(M/4)23

I=38MR2+M26



Q 24 :

Two identical thin rods of mass M kg and length L m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is x12ML2 kg m2. The value of x is __________.  [2026]



(17)

I=ML23+(ML212+ML2)

=4ML2+ML2+12ML212

I=1712ML2

  x=17



Q 25 :

A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is __________ kg·m2.            [2026]

  • 0.18

     

  • 0.63

     

  • 0.36

     

  • 0.72

     

(2)

I=75[m1R12+m2R22]

=75[5(10)2+10(20)2]×10-4

I=63×10-2 kg m2

I=0.63 kg m2



Q 26 :

A circular disc has radius R1 and thickness T1. Another circular disc made of the same material has radius R2 and thickness T2. If the moment of inertia of both discs are same and R1R2=2, then T1T2=1α. The value of α is _______.           [2026]



(16)

m1=πR12T1ρ    m2=πR22T2ρ

I1=m1R122    I2=m2R222

I1=I2

πR12T1ρR122=πR22T2ρR222T1T2=116



Q 27 :

A uniform solid cylinder of length L and radius R has moment of inertia about its axis equal to I1. A small co-centric cylinder of length L2 and radius R3 carved from this cylinder has moment of inertia about its axis equals to I2. The ratio I1/I2 is _________.             [2026]



(162)

Original mass (M)

The removed mass (m)

m=ρ×π(R3)2×L2

=ρ.πR2L18=M18

I'=12·M18·R29=1324MR2

II'=12MR21324MR2=162



Q 28 :

Suppose there is a uniform circular disc of mass M kg and radius r m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by x256Mr2 The value of x is ______.   [2026]



(109)

M=σπR2

σπR2=16m

m=σπR216

Isystem=MR22-2(mR22×16+9mR216)

=MR22-2×19mR232

=MR22-1916mR2

=MR22-19256MR2  because m=M16

=(128-19)MR2256

=109MR2256



Q 29 :

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is n cm. The value of n is _____. [2026]



(265)

Let radius of gyration is k

mk2=23mR2+md2

k2=23×102+152=265

(n)2=265n=265