Question

The mass of 2 kg is hung from a steel wire of radius 0.5mm and length 3m. Compute the extension produced. What should be the minimum radius of wire so that elastic limit is not exceeded?

Answer 1

Answer:

Extension in the wire is given as

$\Delta L = 3.82 \times 10^{-4} m$

Minimum possible radius of the wire is

$r = 4.48 \times 10^{-5} m$

Explanation:

As per the formula of elasticity we know that

$\frac{\Delta L}{L} = \frac{T}{AY}$

so we have

$\Delta L = \frac{T L}{A Y}$

so we have

$T = mg$

$T = 2 \times 10 = 20 N$

$A = \pi r^2$

$A = \pi (0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7} m^2$

$Y = 2 \times 10^{11} N/m^2$

so we have

$\Delta L = \frac{20 \times 3}{(7.85 \times 10^{-7})(2 \times 10^{11})}$

$\Delta L = 3.82 \times 10^{-4} m$

Eleastic limit of steel wire is given as

$E = 3.17 \times 10^9 N/m^2$

so we have

$3.17 \times 10^9 = \frac{mg}{\pi r^2}$

$3.17 \times 10^9 = \frac{20}{\pi r^2}$

$r = 4.48 \times 10^{-5} m$

#Learn

Topic : Elasticity

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