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A 50KN force is applied on a table,the length of four legs of table is 2m each and of radius 5cm.If young modulus of material of table is $5\times 10^{10}N/m^2$.Find the decrease in length in a leg of table.

$\begin{array}{1 1}(A)\;1mm\\(B)\;2mm\\(C)\;3mm\\(D)\;4mm\end{array} $

1 Answer

Force on 1 leg=$\large\frac{50\times 10^3}{4}$
Area(a)=$\pi r^2$
$\Rightarrow \pi(5\times 10^{-2})^2$
$\Rightarrow 25\pi\times 10^{-4}$
$l=2m$
As $Y=(\large\frac{f}{a})\frac{l}{\Delta l}$
$\Rightarrow \Delta l=(\large\frac{f}{a})\frac{l}{Y}$
$\Rightarrow \large\frac{50\times 10^3}{4\times 25\pi\times 10^{-4}}\times \frac{2}{5\times 10^{10}}$
$\Rightarrow 100\times 10^{-5}m$
$\Rightarrow$ 1mm
Hence (A) is the correct answer.
answered May 7, 2014 by sreemathi.v
 

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