A) \[6\Omega \]
B) \[9\Omega \]
C) \[14\Omega \]
D) \[15\Omega \]
Correct Answer: C
Solution :
Let \[\rho \]be the resistance per unit length of the potentiometer wire AB. If D is the balance point, then from the condition of balance of wheat stone's bridge \[\frac{P}{Q}=\frac{R}{S}\]or \[\frac{6}{R}=\frac{30\rho }{(100-30)\rho }=\frac{30}{70}=\frac{3}{7}\] or, \[R=\frac{7\times 6}{3}=14\Omega \] Hence, the correction option is [c].You need to login to perform this action.
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