In a Searle's experiment, the diameter of the wire as measured by a screw gauge is 0 . 050   cm . The pitch of screw gauge is 2   mm and number of divisions on circular scale is 200 . The length, measured by a scale of least count 0 . 1   cm , is 100   cm . When a weight of 50   N is suspended from the wire, the elongation is measured to be 0 . 125   cm by a micrometer of least count

Step-by-step solution

First, recall the formula of Young's modulus in terms of force applied, cross- sectional area of the wire and strain produced, y = m g A l Δ l = m g π r 2 l Δ l , So, the percentage error in Young's modulus, Δ y y × 100 = 2 Δ r r + Δ l l + Δ Δ l Δ l × 100 = 2 × 0 . 001 0 . 05 + 0 . 1 100 + 0 . 001 0 . 125 × 100 = 2 50 + 1 1000 + 1 125 × 100 = 0 . 04 + 0 . 001 + 0 . 008 × 100 = 4 . 90 .