How much resistance should be connected to `15Omega` resistor shown in the circuit in following figure so that the points M and N are at the same potential :

Explanation

In balance condition of Wheatstone.s bridge
`(P)/(Q)=(R)/(S)`
`(3)/(X)=(30)/(60)`
`X=(3xx60)/(30)=6 Omega`
Since, the resultant of two resistances is less than one of the resistances therefore, the two resistances must be connected in parallel.
Let resistance R is connected in parallel with resistance of `15 Omega`. Then
`(1)/(X)=(1)/(15)+(1)/(R)`
`(1)/(6)=(1)/(15)+(1)/(R)`
`(1)/(R)=(1)/(6)-(1)/(15)`
`(1)/(R)=(5-2)/(30)=(3)/(30)=(1)/(10)`
`R=10 Omega`
Hence, `10 Omega` resistance should be connected in parallel to `15Omega` resistor.