A body of mass 11kg is suspended from a ceiling fan by an aluminum wire of length 2m and diameter 2mm. Calculate the elastic energy stored in the wire [Young’s modulus of aluminum is 7.0 x \(10^10Nm^{-2}\), g = 10m\(s^{-2}\), \(\pi\) = 3.142]`
- 1.1 x 10-1J
-
5.5 x 10-2J
- 1.1 x 10-4J
- 5.5 x 10-8J
The correct answer is: B
Explanation
e = \(\frac{4FI}{Y \pi d^2} = \frac{4 \times 11 \times 10 \times 2}{7 \times 10^{10} \times 3.14(2 \times 10^{-3})^2}\)
= 0.001m
E = \(\frac{1}{2} Fe = \frac{1}{2} \times 110 \times 0.001\)
= 5.5 x \(10^{-2}\)J