In the circuit shown, a meter bridge is in its balanced state. The meter bridge wire has a resistance of $0.1 \, \Omega/cm$. The value of the unknown resistance $X$ and the current drawn from the battery of negligible internal resistance are:

When a wire is connected in the left gap of a metre bridge,the balancing point is at $40 \ cm$ from the left end of the bridge wire. If the wire in the left gap is stretched so that its length is doubled and again connected in the same gap,then the balancing point from the left end of the bridge wire is

$A$ wire of length $10 \ cm$ and radius $\sqrt{7} \times 10^{-4} \ m$ is connected across the right gap of a meter bridge. When a resistance of $4.5 \ \Omega$ is connected in the left gap using a resistance box,the balance length is found to be at $60 \ cm$ from the left end. If the resistivity of the wire is $R \times 10^{-7} \ \Omega \ m$,then the value of $R$ is:

In a meter bridge experiment,a null point is obtained at $20 \ cm$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$,then where will be the new position of the null point from the same end,if one decides to balance a resistance of $4X$ against $Y$ (in $cm$)?

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

The figure shows a rough sketch of a meter bridge. The galvanometer $(G)$ shows zero deflection at a length $l \, cm$. Now,$R_1$ and $R_2$ are interchanged,and the balancing length increases by $25 \, cm$. Find the ratio $R_1/R_2$.