In a metre bridge experiment, the resistance in the left gap is $20 \text{ } \Omega$ and in the right gap is $60 \text{ } \Omega$. 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,the ratio of the left gap resistance to the right gap resistance is $2:3$. The balance length from the left end is (in $\,cm$)

In a meter bridge experiment, the balance point is obtained at length $\ell_1 \, cm$ from the left end when resistances in the left gap and right gap are $5 \, \Omega$ and $R \, \Omega$ respectively. When the resistance $R$ is shunted with an equal resistance, the new balance point is at $1.6 \ell_1$. The resistance $R$ in ohm is

When the right gap of a meter bridge consists of two equal resistors in series,the balancing point is at $50 \ cm$. When one of the resistors in the right gap is removed and is connected in parallel to the resistor in the left gap,the balancing point is at: (in $cm$)

$A$ meter bridge with two resistances $R_{1}$ and $R_{2}$ as shown in the figure was balanced (null point) at $40 \text{ cm}$ from the point $P$. The null point changed to $50 \text{ cm}$ from the point $P$, when a $16 \ \Omega$ resistance is connected in parallel to $R_{2}$. The values of resistances $R_{1}$ and $R_{2}$ are . . . . . . .

$A$ 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 $....... \times 10^{-2}\, m$.