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Sahay Sir > Question Answers > Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) > Given F = (xy^2)i + (x^2y)j N. The work done by F when a particle is taken along the semicircular path OAB where the coordinates of B are (4,0) is

Given F = (xy^2)i + (x^2y)j N. The work done by F when a particle is taken along the semicircular path OAB where the coordinates of B are (4,0) is

0) is Given F = (xy^2)i + (x^2y)j N. The work done by F when a particle is taken along the semicircular path OAB where the coordinates of B are (4