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Question

Four particles each of mass $$M$$, are located at the vertices of a square with side $$L$$. The gravitational potential due to this at the centre of the square is


A
32GML
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B
64GML2
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C
zero
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D
32GML
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Solution

The correct option is A $$-\sqrt{32}\cfrac{GM}{L}$$

$${ \mu  }_{ p }={ \mu  }_{ 1p }+{ \mu  }_{ 2p }+{ \mu  }_{ 3p }+{ \mu  }_{ 4p }$$
     $$=\dfrac { -GM }{ L/\sqrt { 2 }  } -\dfrac { GM }{ L/\sqrt { 2 }  } -\dfrac { GM }{ L/\sqrt { 2 }  } -\dfrac { GM }{ L/\sqrt { 2 }  } $$
     $$=\dfrac { -1 }{ L/\sqrt { 2 }  } $$ $$(GM+GM+GM+GM)$$ $$=\dfrac { -4GM }{ L/\sqrt { 2 }  } =\dfrac { -4\sqrt { 2 } GM }{ L } $$

1239847_1106937_ans_d04952fd2ca94c4d877aebc6730144b9.png

Physics

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