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A stone of 0.5kg mass is attached to one end of a 0.8m long aluminium wire, 0.7mm in diameter and suspended vertically. The stone is now rotated in a horizontal plane at a rate, such that the wiire makes an angle of 85 o with the vertical. Find increase in length of wire. Young’s modulus of Al = 7*10 10 Nm -2 .

A stone of 0.5kg mass is attached to one end of a 0.8m long aluminium wire, 0.7mm in diameter and suspended vertically. The stone is now rotated in a horizontal plane at a rate, such that the wiire makes an angle of 85o with the vertical. Find increase in length of wire. Young’s modulus of Al = 7*1010 Nm-2 .

Grade:12

1 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
9 years ago
Dear student,
The net force acting ib thewire would be given by:
T=mgcos85^{\circ}=0.5\times 9.8\times 0.087=0.427 N
Hence the stress on the rod would be given by:
Stress=\frac{0.427}{\pi\times \left (\frac{0.7}{2\times 1000} \right )^{2}}=1109091 N/m^{2}
And the Strain would be given by:
Strain=\frac{\Delta L}{L}=\frac{\Delta L}{0.8}
Hence we get:
Strain=\frac{Stress}{Y}=\frac{1109091}{7\times 10^{10}}=1.58*10^{-5}
Thus,
\Delta L=L\times 1.58*10^{-5}=1.26*10^{-5} m=0.0126 mm
Regards
Sumit

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