Young 's modulus of steel is `2.0 xx 10^(11)N/m^(2) `. Express it is `"dyne"/(cm^(2))`.
Text Solution
Verified by Experts
(i)For steel, `Y_(S)=("Stress")/("Strain")` `"Stress"=(4Mg)/(piD^(2))=(4W_(s))/(piD_(s)^(2))` `"Strain"=(DeltaL_(S))/(L_(S))thereforeDeltaL_(S)=(4W_(S)L_(S))/(piY_(S)D_(S)^(2))` Putting values and simplifying we get, `DeltaL_(S)=(4xx10xx9.8xx1.5)/(3.14xx2.0xx10^(11)xx(2.5xx10^(-3))^(2))`, where `W_(S)=(4+6)9.8N` or `DeltaL_(S)=1.5xx10^(-4)m` (ii) Similarly, for brass `DeltaL_(b)=(4xx6xx98xx1.0)/(3.14xx(2.5xx10^(-3))^(2)xx9.1xx10^(10))`, were `W_(S)=6xx9.8N orDeltaL_(b)=1.3xx10^(-4)m`
