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Question

# The figure shows three rotating, uniform discs that are coupled by belts. One belt runs around the rims of discs A and C. Another belt runs around a central hub on the disc A and the rim of the disc B. The belts move smoothly without slippage on the rims and hub. Disc A has radius R; its hub has radius 0.500R; disc B has radius 0.250R and disc C has radius 2.00R. Discs B and C have the same density (mass per unit volume) and thickness. What is the ratio of the magnitude of the angular momentum of the disc C to that of the disc B ?

A
1224:1
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B
1024:1
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C
1424:1
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D
1624:1
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Solution

## The correct option is B 1024:1Given: Radius of disc A, rA=R Radius of hub of disc A, rAh=0.500R Radius of disc B, rB=0.250R Radius of disc C, rC=2.00R The motion of the all three discs can be related by linear velocity of the belts. Let the angular velocities of the discs A, B and C be ωA, ωB and ωC respectively. Then, for the hub of disc A and disc B, ⇒ωArAh=ωBrB ⇒ωA×0.500R=ωB×0.250R ⇒2ωA=ωB ……(1) Now for disc A and C, ωArA=ωCrC ⇒ωAR=ωC×2.00R ⇒ωA=2ωC ……(2) From equation (1) and (2), ⇒ωB=4ωC ……(3) Let the thickness and density [of each disc B and C] be t and ρ respectively. Angular momentum of a disc is given by L=Iω=mr22ω Now, Angular momentum of disk CAngular momentum of disk B=0.5(ρπr2Ct)×r2CωC0.5(ρπr2Bt)×r2BωB ⇒LCLB=r4CωCr4BωB ⇒LCLB=(2R)4ωC(0.25R)4×4ωC [from (3)] ⇒LCLB=1024:1

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