1.

Consider the meter bridge circuit without neglecting and corrections (alpha, beta) If the unknown Resistance calculated without using the end corraction, is R_(1) and using the end corrections is R_(2) then

Answer»

`R_(1)gtR_(2)` when balanced point is in first HALF
`R_(1) LT R_(2)` when balanced point is in first half
`R_(1)gtR_(2)` when balanced point is in second half
`R_(1)gtR_(2)` always

Solution :If balance point is firsthalf say `l=40`
If balance point is second half say `l=70`
MAXIMUM permissible error in `p`: the specific resistivity of wire, from meter bridge is
`p=(piD^(2)S)/(4L)(l)/(100-l)`
Assume that know resistance in `RB(S)`, and total length of wire is presicely known, then lets find maximum permissible error in `p` due to error in `p` due to error in measurement of `l` (Balance length and `D`(diameter of wire).
In `p=In((piS)/(4L)+2InD+Inl-In(100-l))`
`(DP)/(p)=2(dD)/(D)+(dl)/(l)-(d(100-l))/((100-l))`
`=2(dD)/(D)+(dl)/(l)-(dl)/(100-l)`
`((dp)/(p))_(max)=2(DeltaD)/(D)+(Deltal)/(l)-(Deltal)/(100-l)`
`((dp)/(p))_(max)` due to error in `l` only is
`=(Deltal)/(l)+(Deltal)/(100-l)=(Deltal(100))/(l(100-l))`
`((dp)/(p))_(max)` will be least when `l(100-l)`
if max imum, i.e. `l=50cm`


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