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(A) In a potentiometer wire,the potential difference is directly proportional to the length,i.e.,$V = K \cdot l$,where $K$ is the potential gradient.L

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$2R$

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(A) In a potentiometer wire,the potential difference is directly proportional to the length,i.e.,$V = K \cdot l$,where $K$ is the potential gradient.Let $I$ be the current flowing through the circuit containing resistors $R$ and $X$.Case $1$: When the galvanometer is connected at point $A$,the potential drop across resistor $R$ is balanced by the potentiometer wire of length $10 \ cm$.$IR = K 	imes 10$  ............ $(1)$Case $2$: When the galvanometer is connected at point $C$,the potential drop across the series combination of resistors $R$ and $X$ is balanced by the potentiometer wire of length $30 \ cm$.$I(R + X) = K 	imes 30$  ......... $(2)$Dividing equation $(2)$ by equation $(1)$:$\frac{I(R + X)}{IR} = \frac{K 	imes 30}{K 	imes 10}$$\frac{R + X}{R} = 3$$1 + \frac{X}{R} = 3$$\frac{X}{R} = 2$$X = 2R$

Similar Questions

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