Question 19: Units , Dimensions and Error Analysis

Question

Young's modulus of the material of wires X and Y are in the ratio 4:1 and the areas of cross sections of the wires X and Y are in the ratio $2: 1$. If the same amount of load is applied to both the wires, the ratio of elongation produced in the wires X and Y will be: (Assume length of the wires X and Y initially are the same)

Accepted Answer

To find the elongation $\Delta L$ of a wire under a load $F$, YoungÃ¢ÂÂs modulus $E$, crossÃ¢ÂÂsectional area $A$, and length $L$, we use:
$$ \Delta L = \frac{F\,L}{A\,E} $$
For wires X and Y (same $F$ and $L$), the ratio of their elongations is
$$ \frac{\Delta L_X}{\Delta L_Y} = \frac{F\,L/(A_X E_X)}{F\,L/(A_Y E_Y)} = \frac{A_Y\,E_Y}{A_X\,E_X}. $$
Given
Ã¢ÂÂ¢ $E_X : E_Y = 4 : 1$
Ã¢ÂÂ¢ $A_X : A_Y = 2 : 1$
we have
$$ \frac{\Delta L_X}{\Delta L_Y} = \frac{1/\,2}{4/\,1} = \frac{1}{2} 	imes \frac{1}{4} = \frac{1}{8}. $$
Hence, the elongation ratio $\Delta L_X : \Delta L_Y$ is
$ 1 : 8 $
Option A.

Answer

$1: 8$

Answer

$1: 1$

Answer

$8: 1$

Answer

$1: 2$

Units , Dimensions and Error Analysis

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