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If the ratio of lengths , radii and Young's modulli of two wires in the figure shown are 2,4, and 1 respectively. Then the corresponding ratio of increase in their length will be

\[\begin {array} {1 1} (a)\;32 \\ (b)\;\frac{3}{16} \\ (c)\;1 \\ (d)\;\frac{3}{64}\end {array}\]

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Let ratio of length be a
Let ratio of radii be b
Let ratio of Young's modulus be c
We know $y=\large\frac{F/A}{\Delta L/L}$
$\therefore \Delta L= \large \frac{FL}{AY}$
$\large\frac{\Delta Ls}{\Delta _B}=\bigg(\large\frac{F_S}{F_B}\bigg)\bigg(\large\frac{L_S}{L_B}\bigg)\bigg(\large\frac{A_B}{A_S}\bigg)\bigg(\large\frac{Y_B}{Y_S}\bigg)$
$\qquad= \bigg(\large\frac{3M}{2M}\bigg)$$(a) \bigg(\large\frac{1}{b^2}\bigg)\bigg(\large\frac{1}{c}\bigg)$
$\qquad=\large\frac{3a}{2b^2c}$
$\qquad=\large\frac{3 \times 2}{2 \times 4^2 \times 1}$
$\qquad=\large\frac{3}{16}$
Hence b is the correct answer.

 

answered Oct 7, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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