The resistances in the left and right gaps of a metre bridge are $40 \Omega$ and $60 \Omega$ respectively. When the bridge is balanced, the distance of the null point from the centre of the wire towards the left is (in $\text{ cm}$)

In a meter bridge experiment,the balance point is obtained if the gaps are closed by $2 \Omega$ and $3 \Omega$ resistors. $A$ shunt of $X \Omega$ is added to the $3 \Omega$ resistor to shift the null point by $22.5 \text{ cm}$. The value of $X$ is: (in $Omega$)

Two known resistances are connected in two gaps of a meter bridge. The null point is obtained at $20 \ cm$ from the zero end. $A$ resistance of $15 \ \Omega$ is connected in series with the smaller of the two. The null point shifts to $40 \ cm$. The smaller resistance is (in $Omega$)

In a metre bridge experiment,with a standard resistance in the right gap and a resistance coil dipped in water (in a beaker) in the left gap,the balancing length obtained is $l$. If the temperature of water is increased,the new balancing length is:

In a meter bridge,the gaps are enclosed by resistances of $2 \ \Omega$ and $3 \ \Omega$. The value of shunt to be added to the $3 \ \Omega$ resistor to shift the balancing point by $22.5 \ cm$ is (in $Omega$)

In a meter bridge experiment,$S$ is a standard resistance and $R$ is a resistance wire. It is found that the balancing length is $l = 25 \; cm$. If $R$ is replaced by a wire of half the length and half the diameter of $R$ but of the same material,then the new balancing length $l^{\prime}$ (in $cm$) will be: