First slide

Elastic Modulie

Question

An aluminium and steel wires of same length and cross-section are attached end to end. The compound wire is hung from a rigid support and a load is suspended from the
free end. Y of steel is ( 20/7) times of aluminium. The ratio of increase of length of steel wire to aluminium wire is:

Easy

A

20 : 3
B

10 : 7
C

7 : 20
D

1 : 7

Solution

$∆ l = \frac{FL}{AY} \Rightarrow ∆ l \infty \frac{1}{y} \Rightarrow \frac{{(∆ l)}_{steel}}{{(∆ l)}_{Al}} = \frac{Y_{Al}}{Y_{steel}} = \frac{7}{20}$

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