Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App