21Can v use this ??
Stored Elastic Potential energy in Wire = M*C*∂T
A 5 m long cylindrical steel wire with radius 2x10-3m is suspended vertically from a rigid support and carries a bob of mass 100Kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring radiation losses.
For the steel wire: Young's Modulus=2.1x1011 Pa. Density =7860 Kg/m3, Specific heat =420 J/Kg-K.
-
UP 0 DOWN 0 1 7
7 Answers
21
1elastic potential energy stored = 1/2 (YA/l) (Δl)2 = 1/2 (YA/l) (Fl/AY)2 = 1/2 (l/AY) (F)2
F = 1000 N
l =5 m
A = Î * (2x10-3)2
Y=2.1x10^11 N/m2
elastic potential energy stored = heat generated = mass of wire * Specific heat * ΔT
1/2 (l/AY) (F)2 = Ï * l * A * 420 * ΔT
→ ΔT = (1/2) (F)2 / (7860 A2 Y 420 )
→ ΔT = .. substitute and calculate
4.567295859 * 10-3 celsius
1what is so unique .. i have done the same !! [50] yaar iitcomin :)
3Y = FL / AdL ..........
F/dL = YA/L ..................
F/dL = K ................
although it ends up being same...............