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Calculate elongation if $l_{steel}=2m$ and $l_{brass}=1.5m$,$a_{steel}=2mm^2$ and $a_{brass}=5mm^2$,$Y_{steel}=2\times 10^{11}N/m^2$ and $Y_{brass}=1\times 10^{11}N/m^2$.

$\begin{array}{1 1}(A)\;.28mm\\(B)\;.33mm\\(C)\;.24mm\\(D)\;\text{None}\end{array} $

1 Answer

For brass wire $f=3\times 9.8N$
So $\Delta l_1=(\large\frac{f_1}{a_1})\frac{l_1}{Y_1}$
$\Rightarrow \large\frac{3\times 9.8}{5\times 10^{-6}}\times \frac{1.5}{1\times 10^{11}}$m
$\Rightarrow 8.82\times 10^{-5}$m
$\Rightarrow 0.0882mm$
For steel wire
$f=(3+2)\times 9.8N$
So $\Delta l_1=(\large\frac{f_2}{a_2})\frac{l_2}{Y_2}$
$\Rightarrow \large\frac{5\times 9.8\times 2}{2\times 10^{-6}\times 2\times 10^{11}}$m
$\Rightarrow 24.5\times 10^{-5}m$
$\Rightarrow$ .245mm
$\Delta l=\Delta l_1+\Delta l_2$
$\Rightarrow .088+.245$
$\Rightarrow$ .333mm
Hence (B) is the correct answer.
answered May 7, 2014 by sreemathi.v

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