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

$\frac{65}{3} J$

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$\frac{75}{2} J$

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$\frac{73}{4} J$

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Zero

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Solution

The correct option is D
Zero

Given $\to F = (x y^{2})^i + (x^{2} y)^j$

W = $\int F_{x} d x + \int F_{y} d y$

= $\int x y^{2} d x + \int x^{2} y d y$

= $\frac{1}{2} \int d (x^{2} y^{2}) = {[\frac{x^{2} y^{2}}{2}]}_{(0, 0)}^{(4, 0)} = 0$

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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

The correct option is D Zero Given →F=(xy2)^i+(x2y)^j W = ∫Fxdx+∫Fydy = ∫xy2dx+∫x2ydy = 12∫d(x2y2)=[x2y22](4,0)(0,0)=0

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

The correct option is D
Zero

Given $\to F = (x y^{2})^i + (x^{2} y)^j$

W = $\int F_{x} d x + \int F_{y} d y$

= $\int x y^{2} d x + \int x^{2} y d y$

= $\frac{1}{2} \int d (x^{2} y^{2}) = {[\frac{x^{2} y^{2}}{2}]}_{(0, 0)}^{(4, 0)} = 0$