Question

a composite wire is prepared by joining a tungsten wire and steel wire end to end.Both the wires are of the same length and same area of cross section.If this composite wire is suspended to a rigid support and a force is applied to its free end, it gets extended by 3.25mm.Calculate the increase in length of tyngsten wire and steel wire seperately.​

Answer 1

Answer:

Extension in tungsten wire= 1.182mm, extension in steel wire= 2.068mm

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Answer 2

$\Delta l_s=2.393\ mm$ is the increment in the length of steel wire.

$\Delta l_t=0.857\ mm$ s the increment in the length of tungsten wire.

Explanation:

Since the wires are connected end to end i.e. in series so the force acting on them will be equal.

According to given, the length and cross-sectional area of both the tungsten wire and the steel wire are equal.

$\Delta l_s+\Delta l_t=3.25\ mm$ .................................(1)

We have,

Young's modulus of steel, $E_s=215000\ MPa$

Young's modulus of tungsten, $E_t=600000\ MPa$

So accordingly the stress in both the wires will be equal:

$\sigma_s=\sigma_t$

$E_s\times \epsilon_s=E_t\times \epsilon_t$

$215000\times \frac{\Delta l_s}{l} =600000\times \frac{\Delta l_t}{l}$

$\Delta l_s=\frac{120}{43} (3.25-\Delta l_s)$

$\Delta l_s=2.393\ mm$ is the increment in the length of steel wire.

$\Delta l_t=0.857\ mm$ s the increment in the length of tungsten wire.

#Learnmore TOPICS: stress, strain

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