A) \[4x\]
B) \[2x\]
C) \[x\]
D) \[x/2\]
Correct Answer: C
Solution :
According to Wheatstone bridge principle \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{R}_{AC}}}{{{R}_{CB}}}=\frac{AC}{CB}\] ?. (i) Now, radius of the wire AB is doubled As \[R=\frac{\rho l}{A}=\frac{\rho l}{\pi {{r}^{2}}}\] If radius of wire is doubled Then new resistance, \[R=\frac{R}{4}\] Similarly \[R{{}_{AC}}=\frac{{{R}_{AC}}}{4}\] ... (ii) and \[R{{}_{CB}}=\frac{{{R}_{CB}}}{4}\] ?. (iii) \[\therefore \] From Eqs. (i), (ii) and (iii), we get \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{4R{{}_{AC}}}{4R{{}_{CB}}}=\frac{R{{}_{AC}}}{R{{}_{AC}}}=\frac{{{R}_{AC}}}{{{R}_{CB}}}=\frac{AC}{CB}\] Thus, new balancing length AC = x is same as before.
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