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Mechanical Properties Of Fluids

Question
CBSEENPH11018112

One end of a wire 1m long is fixed at the centre of rotating horizontal table and other end is attached to 2kg load. Find the increase in the length of wire when table rotates with angular velocity 2.5 rad/s. The radius of wire is 2 x 10–4 m and Young's modulus of the material of wire 1.8 x1010 N/m2.

Solution

Given,
Length of one end of wire, L = 1m
Angular velocity,<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>=12 rad/s
Mass of the load, M=2kg
Therefore tension in the string is,
straight T equals MLω squared
space space equals 2 cross times 1 cross times left parenthesis 2.5 right parenthesis squared
space space equals 12.5 straight N
Now comma
Young apostrophe straight s space modulus space is comma space straight Y space equals fraction numerator TL over denominator straight A space straight ell end fraction equals fraction numerator TL over denominator πr squared straight ell end fraction
rightwards double arrow space space space space space space space space space straight ell equals fraction numerator TL over denominator πτ squared straight Y end fraction

Here comma space space space straight T equals 12.5 straight N comma space space space space space space space space space space space straight L equals 1 straight m

space space space space space space space space space space space space straight r equals 2 straight x 10 to the power of negative 4 end exponent straight m comma space space space space space space space straight Y equals 1.8 cross times 10 to the power of 10 space straight N divided by straight m squared space

therefore space space space space space space space space space straight ell space equals fraction numerator 12.5 cross times 1 over denominator straight pi open parentheses 2 cross times 10 to the power of negative 4 end exponent close parentheses squared cross times 1.8 cross times 10 to the power of 10 end fraction

space space space space space space space space space space space space space space space equals space 0.553 cross times 10 to the power of negative 2 end exponent straight m

space space space space space space space space space space space space space space space equals 5.53 mm
space space space space 
Increase in the length of the wire = (5.53 - 1) = 4.53 m