Null method is superior over deflection method because

In the given arrangement of experiment on metre bridge, if $AD$ corresponding to null deflection of the galvanometer is $X$, what would be its value if the radius of the wire $AB$ is doubled?

What is end error in a metre bridge? How is it overcome? The resistances in the two arms of the metre bridge are $\mathrm{R}=5 \mathrm{\Omega }$ and $\mathrm{S}$ respectively. 

When the resistance $\mathrm{S}$ is shunted with an equal resistance, the new balance length found to be $1.5{\mathrm{l}}_1$, where ${\mathrm{l}}_1$ is the initial balancing length. Calculate the value of $\mathrm{S}$.

During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at $40.0 \mathrm{cm }$using a standard resistance of $90\mathrm{ }\mathrm{\Omega }$ ,as shown in the figure. The least count of the scale used in the metre bridge is $1\mathrm{mm}$. The unknown resistance is

Consider a $72 \mathrm{cm}$ long wire $AB$ as shown in the figure. The galvanometer jockey is placed at $P$ on $AB$ at a distance $x \mathrm{cm}$ from $A$. The galvanometer shows zero deflection.

The value of $x$, to the nearest integer, is

In a meter bridge, the wire of length $1 m$ has a non-uniform cross-section such that, the variation $\frac{dR}{dl}$ of its resistance $R$ with length $l$ is $\frac{dR}{dl}\propto \frac{1}{\sqrt{l}}.$ Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point $P.$ What is the length $AP?$

In order to measure the internal resistance $r_1$ of a cell of emf $\epsilon ,$ a meter bridge of wire resistance $R_0=50 \mathrm{\Omega }$, a resistance $\frac{R_0}{2}$, another cell of emf $\frac{\epsilon }{2}$ (internal resistance $r$) and a galvanometer $G$ are used in a circuit, as shown in the figure. If the null point is found at $l=72 \mathrm{cm},$ then the value of $r_1=\_\_\_\_\mathrm{\Omega }.$