A wire of radius 0.2 mm is extended by 0.5% of its length when supported by a load of 1.5 kg. Determine the Young’s modulus for the material of the wire.
[Take g = 10 ms\(^{-2}\)]
-
\(2.4×10^{10}(Nm^{-2})\)
- \(1.5×10^{10}(Nm^{-2})\)
- \(2.4×10^9(Nm^{-2})\)
- \(1.3×10^{10}(Nm^{-2})\)
The correct answer is: A
Explanation
r=0.2mm = \(2 × {10^{-4}}\) e= 0.05% of L,=0.005L, m=1.5kgg=\(10 ms^{-2}\) , y=?,
(1000 mm = 1m)
Y = \(\frac{FL}{Ae}\)
w = mg = 1.5 × 10 = 15
A = \(\pi r^2\) = \(\frac{22}{7}×(2×10^{-4})^2\) = \(1.26×10^{-7}(m^2)\)
⇒ Y = \(\frac{15L}{1.26×10^{-7}×0.005L}\)
⇒ Y = \(\frac{15}{1.26×10^{-7}×0.005}\)
⇒ Y = \(\frac{15}{6.29×10^{-10}}\)
Y = \(2.4×10^{10}(Nm^{-2})\)