First slide

Elastic Modulie

Question

The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights addded to the steel and brass wires must be in the ratio of

Moderate

A

1 : 1
B

1 : 2
C

2 : 1
D

4 : 1

Solution

Young modulus Y = $\frac{W . I}{A ∆ I}$
I, A and $∆ l$ are the same for both the wires.
Y $\propto W$
$\frac{Y_{s}}{Y_{b}} = \frac{W_{s}}{W_{b}} = \frac{2}{1}$
$W_{s} : W_{b} = 2 : 1$

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