A mild steel wire of length 1.0 m and cross-sectional area 0.50 xx 10^(-2) cm^(-2) is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.

A mild steel wire of length 1.0 m and cross-sectional area 0.50 xx 10^

1.	A mild steel wire of length 1.0 m and cross-sectional area 0.50 xx 10^(-2) cm^(-2) is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.
Answer» Solution :`X=[(Mg)/(YA)]^(1//3)` Here `2l=1m, l=0.5 mA=0.50xx10^(-6)m^(2), M=0.1 kg` `Y=2xx10^(11)N//m^(2) and g=10m//s^(2) "" :. "Dpression " x=1.074xx10^(-2)m`