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Question

One of the two forces is double and the other resultant is equal to the greater force. The angle between then is 


A
cos1(1/2)
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B
cos1(1/2)
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C
cos1(1/4)
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D
cos1(1/4)
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Solution

The correct option is D $$cos^{-1} (-1/4)$$
Let the smaller force is $$F$$
so bigger force is  $$2F$$
resultant $$(R)=2F$$
$${R^2} = {F^2} + {\left( {2F} \right)^2} + 2 \times F \times 2F\cos \left( x \right)$$  (its a formula)
$$\begin{array}{l} 4{ F^{ 2 } }={ F^{ 2 } }+4{ F^{ 2 } }+4{ F^{ 2 } }\cos  \left( x \right)  \\ 4{ F^{ 2 } }\cos  \left( x \right) ={ F^{ 2 } } \\ \cos  \left( x \right) =\frac { { -1 } }{ 4 }  \\ x={ \cos ^{ -1 }  }\left( { \frac { { -1 } }{ 4 }  } \right)  \end{array}$$
$$\therefore$$ Option $$D$$ is correct.

Physics

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