Q: A cylindrical wire of radius 1 mm , length 1 m , Young’s modulus Y = 2 × 10^{11 } N/m^{2} , Poisson’s ratio μ = π/10 is stretched by a force of 100 N . Its radius will become

(a) 0.99998 mm

(b) 0.99999 mm

(c) 0.99997 mm

(d) 0.99995 mm

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Sol: $\displaystyle Stress = \frac{F}{A} = \frac{100}{\pi r^2}$

$\displaystyle Stress = \frac{100}{\pi (10^{-3})^2}= \frac{10^8}{\pi}$

$\displaystyle Strain , \frac{\Delta l}{l}= \frac{Stress}{Y} $

$\displaystyle \frac{\Delta l}{l}= \frac{10^8 / \pi}{2 \times 10^{11}} = \frac{5}{\pi} \times 10^{-4}$

$\displaystyle \mu = – \frac{\Delta r /r}{\Delta l/l} $

$\displaystyle \frac{\pi}{10} = – \frac{\Delta r /r}{\frac{5}{\pi} \times 10^{-4}} $

$\displaystyle \Delta r = -0.00005 mm $

$\displaystyle r_f = 1 – 0.00005 mm $

= 0.99995 mm