Q.

For the arrangement shown in the figure find the net gravitational field at point ‘O’ is

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a

GMR212−π2πtowards left

b

GMR22−π2πtowards right

c

GMπR24−π12towards left

d

Zero

answer is A.

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Detailed Solution

Gravitational field due to arc  E1=2G2Msinπ/6π3R2E1=6GMπR2towards leftGravitational field due to a rod at the axial point E2=GMR1R−12R E2=GM2R2 towards right Net gravitational fieldE=E1−E2   towards left  ∵E1>E2=6GMπR2−GM2R2=GMR26π−12E=GMR212−π2π towards left
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