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Question

Given F=(xy2)^i+(x2y)^jN. The work done by F when a particle is taken along the semicircular path OAB where the coordinates of B are (4,0) is


A

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B

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C

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D

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Solution

The correct option is D


Given F=(xy2)^i+(x2y)^j

W = Fxdx+Fydy

= xy2dx+x2ydy

= 12d(x2y2)=[x2y22](4,0)(0,0)=0


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