The upper end of a wire of radius $4 m m$ and length $100 c m$ is clamped and its other end is twisted through an angle of $30$ . The angle of shear is:

12

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1.2

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0.12

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0.012

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Solution

The correct option is C $0.12$
According to relation
$r θ = l ϕ$
So, $ϕ = \frac{r θ}{l}$
$\Rightarrow ϕ = \frac{0.4 \times 30}{100} = 0.12$

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

The correct option is C 0.12According to relationrθ=lϕSo, ϕ=rθl⇒ϕ=0.4×30100=0.12

The upper end of a wire of radius $4 m m$ and length $100 c m$ is clamped and its other end is twisted through an angle of $30$ . The angle of shear is:

The correct option is C $0.12$
According to relation
$r θ = l ϕ$
So, $ϕ = \frac{r θ}{l}$
$\Rightarrow ϕ = \frac{0.4 \times 30}{100} = 0.12$