A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickne

Question

A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickne

A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness `t//4`. The relation between the moments of inertia `I_(A)` and `I_(B)` is (about an axis passing through centre and perpendicular to the disc)
A. `I_(A)=I_(B)`
B. `I_(A)gt I_(B)`
C. `I_(B)gt I_(A)`
D. data is insufficient

Monjur Alahi

8 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - C
`(I_(A))/(I_(B))=(M_(A)R_(A)^(2))/(M_(B)R_(B)^(2))=(rho A_(A)xx t_(A)R_(A)^(2))/(rho A_(B)xx t_(B)R_(B)^(2))`
`= (pi R_(A)^(2)xx t_(A)xx R_(A)^(2))/(pi R_(B)^(2)xx t_(B)xx R_(B)^(2))`
`= (t_(A))/(t_(B))xx((R_(A))/(R_(B)))^(4)`
`= ((x)/(4))xx ((R )/(4R))^(4)`
`= 4xx (1)/(4^(4)`
`=(I_(A))/(I_(B))=(1)/(4^(3))`
`= (1)/(64)`
`I_(B)gt I_(A)`.