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A body of mass 500 gm is suspended from the end of a steel wire 1m and having cross sectional area of 1mm^2 . the other end of the wire is fixed to a rigid support . The body is moved side-ways, till the wire makes an angle of 60 with the vertical and then released. Find the increase in length of the wire when the body is vertically below the point of suspension.[ Y = 2 x 10^11 N/m^2 ]

A body of mass 500 gm is suspended from the end of a steel wire 1m and having cross sectional area of 1mm^2 . the other end of the wire is fixed to a rigid support . The body is moved side-ways, till the wire makes an angle of 60 with the vertical and then released. Find the increase in length of the wire when the body is vertically below the point of suspension.[ Y = 2 x 10^11 N/m^2 ]


 

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1 Answers

Aishwarya Muralidharan
34 Points
10 years ago

snce

Y=(mg/pi*r2)/(l/L)=(mg*L)/(l*pi*r2)

therefore,

l=(mg*L)/(pi*r2*Y)

=9.8*500*1/100*1*10-6*2*1011

=4.9*10-5/2

=2.45*10-5m.

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