InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A mild steel wire of length 1.0 m and cross-sectional are `0.5 xx 10^(-20)cm^(2)` is streached, 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. `g = 10ms^(-2), Y=2 xx 10^(11)Nm^(-2`. |
|
Answer» Let x be the depression at the mid point ie, CD = x In fig., `AC = CB = l = 0.5m , m=100 g = 0.100 kg` `AD = BD = (l^(2) + x^(2))^(1//2)` Increase in length, `Delta l = AD + DB - AB = 2 AD - AB` `=2 (l^(2)+x^(2))^(1//2) -2l` `=2l(1+(x^2)/(l^2))^(1//2) - 2l = 2l[1+(x^2)/(2l^2)] - 2l = (x^2)/(l)` `:. Strai n = (Delta l)/(2l) = (x^2)/(2l^(2))` If T is the tension in the wire, then `2 T cos theta = mg or T = (mg)/(2 cos theta)` Here, `cos theta = (x)/((l^(2)+x^(2))^(1//2)) = (x)/(l(1+(x^2)/(l^2))^(1//2)) = (x)/(l(1+1/2(x^2)/(l^2)))` As, `x lt lt l, so l=1/2 (x^2)(l^2) and 1/2 (x^2)(l^2)~~1 :. cos theta = x/l` Hence, `T = (mg)/(2(x//l)) = (mgl)/(2x)` Stress `= T/A = (mgl)/(2Ax)` `Y= (stress)/(strai n) = (mgl)/(2Ax) xx (2l^2)/(x^2) = (mgl^3)/(Ax^3)` `:. x = l[(mg)/(YA)]^(1//3) = 0.5[(0.1 xx 10)/(20 xx 10^(11) xx 0.5 xx 10^(-6))]^(1//3) = 1.074 xx 10^(-2)m = 1.074 cm`. |
|