If the balance point is obtained at the $35^{th} cm$ in a meter bridge the resistances in the left and right gaps are in the ratio of
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(a) Using Wheatstone principle $\frac{P}{Q} = \frac{R}{S} = \frac{R}{{100 - l}}$
$ = \frac{{35}}{{100 - 35}} = \frac{{35}}{{65}} = \frac{7}{{13}}$
$ = \frac{{35}}{{100 - 35}} = \frac{{35}}{{65}} = \frac{7}{{13}}$
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