The probable error of the adjusted bearing at the middle is
\(\frac { 1 } { 2 }\) r\(\sqrt{n}\)
\(\frac { 1 } { 3 }\) r\(\sqrt{n}\)
\(\frac { 1 } { 4 }\) r\(\sqrt{n}\)
\(\frac { 1 } { 5}\) r\(\sqrt{n}\)
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Answer : Option A
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