The length of a wire of a potentiometer is 100 cm, and the e.m.f. of its standard cell is E volt. It is employed to measure the e.m.f. of a battery whose internal resistance is `0.5 Omega`. If the balance point is obtained at I = 30 cm from the positive end, the e.m.f. of the battery is .
where i is the current in the potentiometer wire.

Options

A. ` (30 E ) /( 100.5 ) `
B. ` ( 30 E ) /( 100 - 0.5 ) `
C. ` (30 ( E - 0.5i))/( 100 ) `
D. ` ( 30 E) /(100) `

Answer

D

Explanation

From the principle of potentionmeter
` V prop l rArr ( V) / ( E) lt ( l)/(L ) `
where l = balance point
L = length of potentiometer wire
`or V = (l )/(L ) E = ( 30 xx E )/ (100 ) = ( 30 E) /(100) `