Q#5 (Past Exam Paper – November 2018 Paper 13 Q19)
In an experiment to measure the Young modulus of a metal, a wire of the metal of diameter 0.25 mm is clamped, as shown.
The wire passes from a clamp, around a frictionless pulley, and then to a second frictionless pulley where loads F are applied to it. A marker is attached to the wire so that the total length of wire between the clamp and the marker is initially 3.70 m. A scale is fixed near to this marker.
The graph shows how the reading on the scale varies with F.
What is the Young modulus of the metal?
A 5.5 × 1010 Pa
B 9.4 × 1010 Pa
C 1.6 × 1011 Pa
D 2.2 × 1011 Pa
Solution:
Answer: D.
Young modulus E = stress / strain
Stress = Force / Area = F / A
Strain = extension / original length = e / L
E = (F/A) / (e/L) = FL / Ae
The original length L (= 3.70 m) is initially at the marker position of 4.5 mm on the scale (refer to the graph on the y-axis when F = 0 N).
When the force F = 10 N, extension e = 8.0 – 4.5 = 3.5 mm = 3.5×10-3 m
Cross-sectional A = πd2 / 4
E = FL / Ae
E = 4FL / πd2e
E = 4 × 10 × 3.70 / (π × (0.25×10-3)2 × (3.5×10-3))
Young modulus E = 2.2×1011 Pa
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