A steel wire of original length 1 m and cross-sectional area 4.00 mm^2 is clamped at the two ends so that it lies horizontally and without tension. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression ?
`Y of the steel `=2.0x10^11Nm^-2. Take `g=10 ms^-2`

Answer

A, C

Explanation

From `cos theta=X/(sqrt(X^2+I^2))`
`=X/I{1+X^2/I^2}^(-1/2)`
`=X.I`….1
increase in length
`Delta L=(AC+CB)-AB`
here `AC=`(`I^2`+`X^2`)^(1/2)``
So, `DeltaL=2(I^2+X^2)^(1/20`
given that `A=4mm^2`
`=4xx10^-2cm^2`
Substituing the values we have
`Y=F/A L/(Delta L)`
`rarr 2xx10612`
`=(Txx100)/((4dxx10^-2)xx[2(50^2+X^2)^(1/2)-100])`........2
from equation i, ii and the freebody diagram ltbr. `2Tcostheta=mg`
`2T(X/50)=2.16xx10^3xx980`
`=(2xx(2xx10^12)xx(4xx10^-2)xx[2(50^2+X^21/2)-100xxX])/(100xx50)`
`(2.16)xx10^3xx980`
Solving for X, we get X=1.5 cm.