In an experiment to determine the Young's modulus of wire of a length exactly 1 m, the extension in the length of the wire is measured as 0.4 mm with an uncertainty of ± 0.02 mm when a load of 1 kg is applied. The diameter of the wire is measured as 0.4 mm with an uncertainty of ± 0.01 mm. The error in the measurement of Young's modulus (Δ Y) is found to be x × 10¹⁰ Nm⁻². The value of x is _________________. (take g=10 ms⁻² )

Solution

Physics sir · 4 steps

Find Y first, then add the fractional errors, and count the diameter's error twice because Y depends on d².

1. Write Y. With load F = mg = 10 N, cross-section A = (π d²)/4 and extension l, the Young's modulus is Y = FL/Al = 4FL/(π d² l).
2. Put in numbers. With d = l = 0.4 × 10⁻³ m and L = 1 m, Y = (4 × 10 × 1)/(π × 0.16 × 10⁻⁶ × 0.4 × 10⁻³) = (6.25 × 10¹¹)/π ≈ 2 × 10¹¹ N m⁻².

   (Δ Y)/Y = 2(Δ d)/d + (Δ l)/l = 2 × 0.01/0.4 + 0.02/0.4 = 0.05 + 0.05 = 0.1

3. Why only these. L is exact, and F and π carry no error, so only d and l add to the uncertainty. The power of d in the formula becomes the factor 2 on its fractional error.
4. So. Δ Y = 0.1 × 2 × 10¹¹ = 2 × 10¹⁰ N m⁻², so x = 2.

The common slip: 1.5. Leaving out the factor 2 on the diameter's error gives Δ Y/Y = 0.075 and x = 1.5, not 2.

The idea to keep. For a product of powers, fractional errors add, and each one is multiplied by its power. Find Δ Y/Y this way, then multiply by Y to get the absolute error.

Answer from the NTA final answer key. Solution by PrepWiser.