$(a)$ In a meter bridge,the balance point is found to be at $39.5\; cm$ from the end $A$,when the resistor $Y$ is of $12.5\; \Omega$. Determine the resistance of $X$. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
$(b)$ Determine the balance point of the bridge above if $X$ and $Y$ are interchanged.
$(c)$ What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

(N/A) meter bridge with resistors $X$ and $Y$ is represented in the given figure.
$(a)$ Balance point from end $A$,$l_{1} = 39.5\; cm$.
Resistance of the resistor $Y = 12.5\; \Omega$.
Condition for the balance is given as:
$\frac{X}{Y} = \frac{l_{1}}{100 - l_{1}}$
$X = Y \times \frac{l_{1}}{100 - l_{1}} = 12.5 \times \frac{39.5}{100 - 39.5} = 12.5 \times \frac{39.5}{60.5} \approx 8.16\; \Omega$.
Therefore,the resistance of resistor $X$ is approximately $8.16\; \Omega$.
The connections between resistors in a Wheatstone or meter bridge are made of thick copper strips to minimize their resistance,which is not taken into consideration in the bridge formula.
$(b)$ If $X$ and $Y$ are interchanged,then $l_{1}$ and $100 - l_{1}$ get interchanged.
The balance point of the bridge will be at $100 - l_{1}$ from $A$.
$100 - l_{1} = 100 - 39.5 = 60.5\; cm$.
Therefore,the balance point is $60.5\; cm$ from $A$.
$(c)$ When the galvanometer and cell are interchanged at the balance point of the bridge,the galvanometer will show no deflection. Hence,no current would flow through the galvanometer.