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A golden ring of radius 0.5cm and cross section area $2mm^2$ is shifted to a wooden of disk of radius 8cm.If young modulus is $3\times 10^{10}N/m^2$,what is force along its length?

$\begin{array}{1 1}(A)\;24KN\\(B)\;36KN\\(C)\;18KN\\(D)\;20KN\end{array} $

1 Answer

$\large\frac{\Delta l}{l}=\frac{2\pi r_2-2\pi r_1}{2\pi r_1}$
$\Rightarrow \large\frac{.8-.5}{.5}$
$\Rightarrow \large\frac{3}{5}$
As stress=Y$\times$ strain
$F=3\times 10^{10}\times \large\frac{3}{5}$$\times 2\times 10^{-6}$
$\;\;\;=\large\frac{18}{5}$$\times 10^4$
$\;\;\;=36KN$
Hence (B) is the correct answer.
answered May 8, 2014 by sreemathi.v
 

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