A cell in the secondary circuit gives a null | vedclass

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(C) Let $V$ be the $EMF$ of the cell in the secondary circuit. The potential gradient $k$ of the potentiometer wire is given by $k = \frac{V_p}{L}$, w

Vedclass · Accepted Answer

$2.75$

Vedclass · Answer

(C) Let $V$ be the $EMF$ of the cell in the secondary circuit. The potential gradient $k$ of the potentiometer wire is given by $k = \frac{V_p}{L}$, where $V_p$ is the potential drop across the wire and $L$ is the total length of the wire.Initially, $L_1 = 10 \,m$ and the null point is at $l_1 = 2.5 \,m$. The potential drop across $l_1$ is $V = k_1 l_1 = \left(\frac{V_p}{L_1}ight) l_1$.When the length of the wire is increased to $L_2 = 10 + 1 = 11 \,m$, the potential drop across the wire remains the same $(V_p)$ because the primary circuit is unchanged.The new potential gradient is $k_2 = \frac{V_p}{L_2} = \frac{V_p}{11}$.For the same cell in the secondary circuit, the new null point $l_2$ satisfies $V = k_2 l_2$.Equating the two expressions for $V$: $\frac{V_p}{10} 	imes 2.5 = \frac{V_p}{11} 	imes l_2$.Solving for $l_2$: $l_2 = \frac{2.5 	imes 11}{10} = 2.75 \,m$.

$A$ cell in the secondary circuit gives a null deflection for $2.5 \,m$ length of a potentiometer wire having a total length of $10 \,m$. If the length of the potentiometer wire is increased by $1 \,m$ without changing the cell in the primary circuit, the new position of the null point is: (in $m$)

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