First slide

Elastic Modulie

Question

The Young's modulus of steel is twice that of brass. Two wires of same length and of 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 added 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

$Y = \frac{Fl}{A ∆ l} \Rightarrow ∆ l = \frac{Fl}{AY}$
${(∆ l)}_{steel} = {(∆ l)}_{Brass} \Rightarrow \frac{W_{s}^{}}{Y_{S}} = \frac{W_{B}}{Y_{B}} \frac{W_{S}}{W_{B}} = \frac{Y_{S}}{Y_{B}} = 2 : 1$

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