A mass of 11.0 kg is suspended from a rigid support by an aluminum wire of length 2.0 m, diameter 2.0 mm and Young's modulus 7.0 x 10\(^{11}\) Nm\(^{-2}\). Determine the extension produced. [g = 10 ms\(^{-2}\); \(\pi\) = 3.142]
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
E = \(\frac{Fl}{Ae}\)
E = 7.0 x 10\(^{-11}\) Nm\(^{-2}\);
F = mass x acceleration due to gravity
= 11 x 10 = 110N
l = 2.0m
A = \(\pi\) r\(^2\)
d = 2.0 mm \(\implies\) r = 1.0 mm
A = 3.142 x \((\frac{10}{1000})^2\)
e = \(\frac{Fl}{EA} = \frac{110 \times 20}{7.0 \times 10^{11} \times 3.142(\frac{10}{1000})^2}\)
= 1.0 x 10\(^{-5}\)m
E = \(\frac{Fl}{Ae}\)
E = 7.0 x 10\(^{-11}\) Nm\(^{-2}\);
F = mass x acceleration due to gravity
= 11 x 10 = 110N
l = 2.0m
A = \(\pi\) r\(^2\)
d = 2.0 mm \(\implies\) r = 1.0 mm
A = 3.142 x \((\frac{10}{1000})^2\)
e = \(\frac{Fl}{EA} = \frac{110 \times 20}{7.0 \times 10^{11} \times 3.142(\frac{10}{1000})^2}\)
= 1.0 x 10\(^{-5}\)m