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