The length of a potentiometer wire is l. A ce | vedclass

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(B) Let $V$ be the potential difference applied across the potentiometer wire of length $l$. The potential gradient is $K = \frac{V}{l}$.For the cell

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$\frac{l}{2}$

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(B) Let $V$ be the potential difference applied across the potentiometer wire of length $l$. The potential gradient is $K = \frac{V}{l}$.For the cell of emf $E$,the balancing length is $\frac{l}{3}$. Thus,$E = K \cdot \frac{l}{3} = \frac{V}{l} \cdot \frac{l}{3} = \frac{V}{3}$.When the length of the wire is increased by $\frac{l}{2}$,the new length becomes $l' = l + \frac{l}{2} = \frac{3l}{2}$.The potential difference $V$ across the wire remains the same. The new potential gradient is $K' = \frac{V}{l'} = \frac{V}{\frac{3l}{2}} = \frac{2V}{3l}$.Let the new balancing length be $l_{new}$. Then $E = K' \cdot l_{new}$.Substituting the values,$\frac{V}{3} = \left(\frac{2V}{3l}\right) \cdot l_{new}$.Solving for $l_{new}$,we get $l_{new} = \frac{V}{3} \cdot \frac{3l}{2V} = \frac{l}{2}$.

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The $e.m.f.$ of a standard cell balances across $150 \ cm$ length of a potentiometer wire. When a resistance of $2 \ \Omega$ is connected as a shunt across the cell,the balance point is obtained at $100 \ cm$. The internal resistance of the cell is .............. $\Omega$.

$A$ potentiometer has a uniform wire of length $5 \,m$. $A$ battery of emf $10 \,V$ and negligible internal resistance is connected between its ends. $A$ secondary cell connected to the circuit gives a balancing length at $200 \,cm$. The emf of the secondary cell is: (in $\,V$)

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