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Four particles each of mass m are placed...

Four particles each of mass `m` are placed at the corners of a square of side length `l`. The radius of gyration of the system about an axis perpendicular to the plane of square and passing through its centre is

A

`(5)/(2)M(4r^(2)+5R^(2))`

B

`(2)/(5)M(4r^(2)+5R^(2))`

C

`(2)/(5)M(4r^(2)+5r^(2))`

D

`(5)/(2)M(4r^(2)+5r^(2))`

Text Solution

Verified by Experts

The correct Answer is:
B
M. I. of sphere A about its diameter `I_(O)=(2)/(5)Mr^(2)`
Now M.I. of sphere A about an axis perpendicular to the plane of square and passing through its centre will be
`I_(O)=I_(O).+M((R)/(sqrt(2)))^(2)=(2)/(5)Mr^(2)+(MR^(2))/(2)` [by the theorem of parallel axis]
Moment of inertia of system (i.e. four sphere) = `4I_(O)=4[(2)/(5)Mr^(2)+(MR^(2))/(2)]`
`=(2)/(5)M[4r^(2)+5R^(2)]`
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Knowledge Check

  • Four particles each of mass m are placed at the corners of a square of side length l. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is

    A
    `(l)/(sqrt(2))`
    B
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    C
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    D
    `(sqrt(2))l`

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    A
    `sqrt(2)`
    B
    `2`
    C
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    D
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    A
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    B
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