In a meter bridge experiment,when a nichrome wire is in the right gap,the balancing length is $60 \ cm$. When the nichrome wire is uniformly stretched to increase its length by $20 \%$,and again connected in the right gap,the new balancing length is nearly: (in $cm$)

$A$ student obtained the following observations in an experiment of a meter bridge to find the unknown resistance of the circuit. The most accurate value of the unknown resistance is ............ $\Omega$.

$S.No.$	$R$	$l$	$100-l$	$S = \left( \frac{100-l}{l} \right)R$
$1$	$20\,\Omega$	$43$	$57$	$26.51\,\Omega$
$2$	$30\,\Omega$	$51$	$49$	$28.82\,\Omega$
$3$	$40\,\Omega$	$59$	$41$	$27.80\,\Omega$
$4$	$60\,\Omega$	$70$	$30$	$25.71\,\Omega$

Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{1}$ is $40\,cm$. Now an unknown resistance $x$ is connected in series with $P$ and the new balancing length is found to be $80\,cm$ measured from the same end. Then the value of $x$ will be $.......\Omega$.

In a metre bridge experiment, the resistance in the left gap is $15 \text{ } \Omega$ and in the right gap is $45 \text{ } \Omega$. The bridge is balanced. The distance of the null point from the centre of the wire is (in $\text{ cm}$)

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 $cm$)?

The meter bridge shown is in a balanced position with $\frac{P}{Q} = \frac{l_{1}}{l_{2}}$. If we now interchange the positions of the galvanometer and the cell,will the bridge work? If yes,what will be the balance condition?