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The upper end of wire 1 m long and 2mm radius is clamped, the lower end twisted through an angle of 45⁰ the angle (in degree) of shear is
Option: 1 0.09

Option: 2 0.9

Option: 3 9

Option: 4 90

Answers (1)

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As we have learned

Shearing strain -

$\phi = \frac{x}{L}$

- wherein

$\phi = Shearing \: Strain$

$x= deformed \: position$

$L= length$

Deformation= $\Delta x = \theta R$

$\Delta x = \theta R = 45 \times 2 mm = 90 mm$

Angle of shear is= $\phi = \frac{\Delta x}{l}$ $\phi = \frac{\Delta x}{l} = \frac{90 \times 10 ^{-3}}{1}$

Angle of shear is= $\phi = \frac{\Delta x}{l} =\frac{R\theta }{l}= \frac{45^0 \times 2\times 10 ^{-3}}{1}$

Angle of shear is= (0.09)⁰

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shivangi.bhatnagar

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