Gravitational field at the center of the semicircle formed by thin wire AB of mass m and length l as shown in the figure is:-
Gml2along+xaxis
Gmπl2along+yaxis
2πGml2along+xaxis
2πGml2along+yaxis
A
Gmπl2along+yaxis
B
2πGml2along+yaxis
C
Gml2along+xaxis
D
2πGml2along+xaxis
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Solution
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The correct option is D2πGml2along+yaxis
Mass of element dm=m1xdθ
Gravitational field due to element dg=GdmR2=GmdθπR2
dgcosθ of each element will be cancelled out
δ=∫dgsinθ
=∫π0(sinθπR2)sinθ
=GmπR2∫π0sinθdθ
=2GmπR2
=2πGml2along+yaxis
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