In the shown arrangement of the experiment of the meter bridge if AC corresponding to null deflection of galvanometer is x, what would be its value if the radius of the wire AB is doubled?

Options

A. 4x
B. 2x
C. x
D. x/2

Answer

3

Explanation

Potential across `R_(1)`=potential across x.
From wheatston e Bridge principle .
`(R_(1))/(R_(2))=(R_(AC))/(R_(C B))=(AC)/(CB)`
If radius is doubled, then using `(R=(rhol)/(A))`
`R_(AC)= 4R_(AC)`
`R_(CB)=4R_(CB)`
From (i) and (ii), `(R_(1))/(R_(2))=(4R_(AC))/(4R_(CB))=(R_(AC))/(R_(CB))=(AC)/(CB)`
Thus new balancing length AC=x is same as b efore.