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Standard XII

Physics

Shear Strain

The upper end...

Question

The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of $30^{o}$ . Then angle of shear is

12°

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0.12°

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1.2°

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0.012°

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Solution

The correct option is B
0.12°

Angle of shear $ϕ = \frac{r θ}{L} = \frac{4 \times 10^{- 1}}{100} \times 30^{o} = {0.12}^{o}$

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The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 30o. Then angle of shear is

The correct option is B 0.12° Angle of shear ϕ=rθL=4×10−1100×30o=0.12o

The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of $30^{o}$ . Then angle of shear is

The correct option is B
0.12°

Angle of shear $ϕ = \frac{r θ}{L} = \frac{4 \times 10^{- 1}}{100} \times 30^{o} = {0.12}^{o}$