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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)` thenA. `R_(1)gtR_(2)` when balanced point is in first halfB. `R_(1) lt R_(2)` when balanced point is in first halfC. `R_(1)gtR_(2)` when balanced point is in second halfD. `R_(1)gtR_(2)` always |
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Answer» Correct Answer - A 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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