Question

In Searl's experiment, which is used to find Young's Modulus of elasticity, the diameter of experimental wire is D = 0.05cm (measured by a scale of least count 0.001cm) and length is L = 110cm (measured by a scale of least count 0.1cm). A weight of 50N causes an extension of X = 0.125 cm (measured by a micrometer of least count 0.001cm). find the maximum possible error in the values of Young's modulus. Screw gauge and meter scale are free error.

Answer 1

The maximum possible error in the values of Young's modulus is ΔY/ Y  = 0.0489

Explanation:

Young's modulus of elasticity is given by

Y = stress / strain

Y = F/A ÷ l / L

Y = FL ÷ lA = FL / l(πd^2 ÷ 4)

Substituting the values, we get

Y = 50 × 1.1 × 4 / (1.25 × 10^−3) × π × (5.0 × 10^−4)^2

Y = 2.24 × 10^11 N/m^2

Now, ΔY/ Y = ΔL / L + Δl / l + 2Δd / d

ΔY/ Y  = (0.1 / 110) + (0.001 / 0.125) + 2 (0.001 / 0.05)

ΔY/ Y  = 0.0489

Hence the maximum possible error in the values of Young's modulus is ΔY/ Y  = 0.0489