Two wires $A$ and $B$ of equal lengths are connected in left and right gap respectively of a metre bridge, null point is obtained at $40 \text{ cm}$ from left end. Diameters of the wires $A$ and $B$ are in the ratio $3:1$ respectively, the ratio of specific resistance of $A$ to that of $B$ is (in $: 1$)

$A$ 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 meter bridge shown,the resistance $X$ has a negative temperature coefficient of resistance. Neglecting the variation in other resistors,when current is passed for some time in the circuit,the balance point should shift towards:

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 $\Omega$)

The resistance per centimeter of a meter bridge wire is $r$. $A$ resistance of $X \ \Omega$ is placed in the left gap and a resistance of $25 \ \Omega$ is placed in the right gap. The balancing length from the left end is $40 \ cm$. Now,the wire is replaced by another wire of $2r$ resistance per centimeter. The new balancing length for the same settings will be at: (in $cm$)

Assertion $(A):$ In a meter bridge experiment,the unknown resistance is placed inside an enclosure maintained at a higher temperature. The null point can be obtained at the same position as before by decreasing the value of the standard resistance.
Reason $(R):$ The resistance of a metal increases with an increase in temperature.