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 $cm$)

$A$ balanced bridge is shown in the circuit diagram. The metre bridge wire has a resistance of $1 \Omega \text{ cm}^{-1}$. The current drawn from the battery is (Internal resistance of the battery is negligible). (in $\text{ A}$)

In a meter bridge,a $30 \Omega$ resistance is connected in the left gap and a pair of resistances $P$ and $Q$ in the right gap. Measured from the left,the balance point is $37.5 \text{ cm}$ when $P$ and $Q$ are in series,and $71.4 \text{ cm}$ when they are in parallel. The values of $P$ and $Q$ (in $\Omega$) are:

The resistances in the two arms of a meter bridge are $5\,\Omega$ and $R\,\Omega$,respectively. When the resistance $R$ is shunted with an equal resistance,the new balance point is at $1.6\,l_1$. The resistance $R$ is ................. $\Omega$.

$A$ resistance wire connected in the left gap of a meter bridge balances a $10\, \Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3: 2$. If the length of the resistance wire is $1.5\, m$,then the length of $1\, \Omega$ of the resistance wire is $....... \times 10^{-2}\, m$.

In the gaps of a meter bridge,two unknown resistances $A$ and $B$ are connected. $A$ null point is obtained when the wire is divided in the ratio $3:4$. When each resistance is increased by $40 \Omega$,the null point divides the wire in the ratio $7:8$. The values of resistances $A$ and $B$ are respectively: