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**Hint:**Meter Bridge is the simplest practical application of Wheatstone bridge. Initial balance point of Wheatstone bridge substitute in the balance equation. By doubling the value of R and X in the balance equation, a new balance point is calculated.

**Complete step by step answer:**

Meter bridge working is based on the principle of Wheatstone bridge. After taking out a suitable resistance R from the resistance box, the jockey is moved along the wire AB till there is no deflection in the galvanometer. This is the balance condition of the Wheatstone bridge. If P and Q are the resistance of the parts AJ and JB of the wire, then for the balanced condition of the bridge, we have

\[\dfrac{P}{Q} = \dfrac{R}{X}\]

Now, let us write the information given in the question.

Balance point\[AJ = 40cm\], now R and X are doubled and interchanged then the new balance point we have to find.

Let us use the balancing condition for Wheatstone bridge \[\dfrac{P}{Q} = \dfrac{R}{X}\]

Initial balance point is 40cm.

\[\dfrac{R}{X} = \dfrac{{40}}{{100 - 40}} = \dfrac{{40}}{{60}}.....\left( 1 \right)\]

Now, according to the question, when the vales of both the X and \[R\]are doubled and interchanged, the balance point (L) is calculated below.

$\Rightarrow$ \[\dfrac{{2X}}{{2R}} = \dfrac{L}{{100 - L}}....\left( 2 \right)\]

Let us equate equation (1) and (2).

$\Rightarrow$ \[\dfrac{{40}}{{60}} = \dfrac{L}{{100 - L}}\]

Taking cross multiplication we get,

$\Rightarrow$ \[60L = 40\left( {100 - L} \right)\]

On multiply the bracket term and we get,

\[60L = 4000 - 40L\]

Let us subtract we get,

\[ \Rightarrow L = 40cm\]

Hence, the new balancing point is the same as earlier.

If the galvanometer and battery are also interchanged there will be no effect on the position of balance point.

**Note:**The deflection of galvanometer not getting affected by interchanging position of battery and galvanometer. In a slide wire, if a balance point is obtained at L cm from the zero ends, then the unknown resistance X is calculated as by below expression.

$X = \dfrac{{\left( {100 - L} \right)}}{L}R$

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