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A mass m is to be placed on the rod of length L and pivoted at its centre at a distance x from the left end of the rod so that the system shown in figure remains in equilibrium . Find x (Ignore the force between Q (beaneth q ) and 2q and the force between Q (beneath 2q) and q .

$(a)\;\large\frac{L}{2}\;(1-\large\frac{k\;q\;Q}{m\;g\;h^2})\qquad(b)\;\large\frac{L}{2}\;(1+\large\frac{k\;q\;Q}{m\;g\;h^2})\qquad(c)\;L\;(1-\large\frac{k\;q\;Q}{m\;g\;h^2})\qquad(d)\;\large\frac{L}{2}\;(1+\large\frac{k\;q\;Q}{2 m\;g\;h^2})$

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Answer : (b)\; $\large\frac{L}{2}\;(1+\large\frac{k\;q\;Q}{m\;g\;h^2})$
Explanation :
F .B . D of rod
Since the system is in equilibrium the torque about the centre O should be zero . Therefore
$\large\frac{F_{1}L}{2}-mg(\large\frac{L}{2}-x)-\large\frac{F_{2}L}{2}=0$
$F_{1}=\large\frac{k\;q\;Q}{h^2}$
$F_{2}=\large\frac{2\;k\;q\;Q}{h^2}$
$\large\frac{k\;q\;Q}{h^2}\;\large\frac{L}{2}-mg(\large\frac{L}{2}-x)-\large\frac{k\;2\;q\;Q\;L}{2 h^2 }=0$
$mgx-\large\frac{mgL}{2}=\large\frac{k\;q\;Q}{h^2}\;\large\frac{L}{2}$
$x-\large\frac{L}{2}=\large\frac{k\;q\;Q}{m\;g\;h^2}\times\large\frac{L}{2}$
$x=\large\frac{L}{2}\;(1+\large\frac{k\;q\;Q}{m\;g\;h^2})\;.$
answered Feb 18, 2014 by yamini.v
edited Feb 20, 2014 by yamini.v
 

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