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(A) The circuit represents a balanced Wheatstone bridge condition because the galvanometer shows no deflection.According to the Wheatstone bridge prin

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$150$

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(A) The circuit represents a balanced Wheatstone bridge condition because the galvanometer shows no deflection.According to the Wheatstone bridge principle,the ratio of resistances in the two arms must be equal:$\frac{R_{AC}}{R_{CB}} = \frac{3 \, \Omega}{4 \, \Omega} = \frac{3}{4}$Since the resistance of a uniform wire is directly proportional to its length $(R \propto L)$,we have:$\frac{AC}{CB} = \frac{3}{4}$Given the total length $AB = AC + CB = 350 \, cm$,we can express $CB$ as $350 - AC$.Substituting this into the ratio:$\frac{AC}{350 - AC} = \frac{3}{4}$$4 	imes AC = 3 	imes (350 - AC)$$4 	imes AC = 1050 - 3 	imes AC$$7 	imes AC = 1050$$AC = \frac{1050}{7} = 150 \, cm$.

The galvanometer has no deflection and the length of wire $AB$ is $350 \, cm$. Then $AC$ equals to ............... $cm$.

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