1.

A Leclanche cell of emf 1.46 V balances against 292 cm of a potentio meter wire. If the current through the wire is 400 mA, the resistance per unit length of the potentiometer wire is

Answer»

`2 Omega//m`
`2 Omega//cm`
`12.5 Omega//m`
`12.5 Omega//cm`

Solution :`(E_(1))/(l_(1))=I LAMDA" "therefore lamda =(E_(1))/(Il_(1))=(1.46)/((0.4)(0.292))=(1)/(0.08)=12.5 Omega//m.`


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