A standard resistance of 11 is connected in the left gap of a meter bridge. A uniform wire of length 40 cm and resistivity 1.1 10⁻⁶ m is connected in the right gap. The balancing length is found to be 44 cm from the left end. The radius of the wire is (Take = 22 7 ):

Step-by-step solution

From the meter bridge balancing condition, R₁ l = R₂ 100 - l 11 44 = R₂ 100 - 44 1 4 = R₂ 56 R₂ = 14 The resistance of the wire is given by R₂ = L r^2 . Substituting the given values: 14 = 1.1 10⁻⁶ 0.4 22 7 r^2 14 = 0.44 10⁻⁶ 7 22 r^2 14 = 3.08 10⁻⁶ 22 r^2 308 r^2 = 3.08 10⁻⁶ r^2 = 10⁻⁸ m ^2 r = 10⁻⁴ m = 0.1 mm Answer: 0.1 mm