Q 1 :    

A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is x5. The value of x is _______ .          [2024]



(2)       KERotKETrans=12Iω212Iω2+12mv2=(12)(23mR2)ω2(12)(23mR2)ω2+12m(Rω)2

           KERotKETrans=2323+1=25

          Hence, x=2

 



Q 2 :    

A circular disc reaches from top to bottom of an inclined plane of length l. When it slips down the plane, if takes t s. When it rolls down the plane then it takes (α2)1/2 t s, where α is _______ .              [2024]



(3)         If disc slips on inclined plane, then it's acceleration, a1=gsinθ

             l=12a1t12t=2lgsinθ                       (i)

            If disc rolls on inclined plane its acceleration, 

            a2=gsinθ1+ImR2=gsinθ1+mR22mR2=23gsinθ

           Now l=12a2t22t2=2l23gsinθ

            t2=322lgsinθ=32t             (ii)

           On comparing, 32t=α2tα=3

 



Q 3 :    

A solid sphere and a hollow cylinder roll up without slipping on the same inclined plane with the same initial speed v. The sphere and the cylinder reach up to maximum heights h1 and h2, respectively, above the initial level. The ratio h1:h2 is n10. The value of n is _____.      [2024]



(7)

Energy conservation, Gain in P.E. = Loss in K.E.

12mv2+12Iω2=mgh

For pure rolling; v=ωR

12mv2+12Iv2R2=mgh

12v2(m+mK2R2)=mgh

mgh=12mv2(1+K2R2)h1+K2R2

h1h2=1+251+1=75×2=710

Hence, n=7



Q 4 :    

A solid sphere of mass 'm' and radius 'r' is allowed to roll without slipping from the highest point of an inclined plane of length 'L' and makes an angle 30° with the horizontal. The speed of the particle at the bottom of the plane is v1. If the angle of inclination is increased to 45° while keeping L constant. Then the new speed of the sphere at the bottom of the plane is v2. The ratio of v12:v22 is          [2025]

  • 1:2

     

  • 1 : 3

     

  • 1 : 2

     

  • 1:3

     

(1)

Loss in P.E. = gain in K.E.

mgL sinθ=12(75)mv2  v2 sinθ

 v12v22=sin 30°sin 45°=12



Q 5 :    

A uniform solid cylinder of mass 'm' and radius 'r' rolls along an inclined rough plane of inclination 45°. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder axis will be:          [2025]

  • 12g

     

  • 132g

     

  • 2g3

     

  • 2g

     

(3)

a=g sin θ1+Imr2

a=g21+12=2g23=2g3



Q 6 :    

A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be t1 and t2, respectively, then          [2025]

  • t1<t2

     

  • t1=t2

     

  • t1=2t2

     

  • t1>t2

     

(1)

The time taken to reach the bottom, t=2lacm

Acceleration, acm=g sin θ1+IcmMR2

a1=acm1=5g sin θ7          ... Solid

a2=acm2=3g sin θ5          ... Hollow

As a1>a2 hence, t1<t2



Q 7 :    

A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is        [2025]

  • 3.5 m/s2

     

  • 0.35 m/s2

     

  • 2.5 m/s2

     

  • 0.25 m/s2

     

(1)

Torque about bottom point, F×2r=Iα

49×2r=75mr2α

14=4rα          ... (i)

As sphere rolls without slipping

a=rα          ... (ii)

 a=144=72=3.5 m/s2



Q 8 :    

A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is x5 where x = __________.          [2025]



(4)

Applying Mechanical Energy conservation:

ki+Ui=kf+Uf

 0+mgh=12mv2(1+k2R2)+0

 V=2gh1+k2R2

So, Ratio of velocities

VRingVSolid sphere=1+251+1=710

x = 3.5  Rounding off x = 4.