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A mild steel wire of length 1.0 m and cr...

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.

Text Solution

Verified by Experts

`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`
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