Question

An ideal cell having a steady emf of $2volts$ is connected across the potentiometer wire of length $10m$. The potentiometer wire is of magnesium and having resistance of $11.5\Omega /m$. An another cell gives a null point at $6.9m$. If a resistance of $5\Omega$ is put in series with potentiometer wire, the new position of the null point is given as $\frac{x}{10}m$. Find $x$.

Answer 1

When there is no any deflection in the galvanometer $\Rightarrow$ voltage of secondary circuit = voltage drop across the wire AJ
voltage drop on the balancing length of the wire $\Rightarrow$i$\rho$X
where i= currnt flowing in primary circuit
$\rho$=resistance per unit length of the wire
X=balancing length of wire AB when deflection in galvanometer is zero
When deflection is zero in 1st case $\Rightarrow E=\left(\frac{2}{115}\right)\times 11.5\times 6.9$ Eq.(1)
When deflection is zero in 2nd case $\Rightarrow E=\left(\frac{2}{115+5}\right)\times 11.5\times y$ ($y$ is the length of wire AJ) Eq.(2)
Now $\frac{Eq.\left(1\right)}{Eq.(2}\Rightarrow \frac{6.9}{115}=\frac{y}{120}$
$y=\left(\frac{6.9}{115}\times 120\times 10\right)\times \frac{1}{10}=\frac{72}{10}$
$x=72$