IIT JEE past question Thermodynamics 2

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.

7 Answers

21
tapanmast Vora ·

Can v use this ??

Stored Elastic Potential energy in Wire = M*C*∂T

1
ANKIT MAHATO ·

elastic 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 = ρ * l * A * 420 * ΔT

→ ΔT = (1/2) (F)2 / (7860 A2 Y 420 )

→ ΔT = .. substitute and calculate

4.567295859 * 10-3 celsius

21
tapanmast Vora ·

Ans : 4.567*10-3K [7]

3
iitimcomin ·

unique way of solvin

1
ANKIT MAHATO ·

what is so unique .. i have done the same !! [50] yaar iitcomin :)

3
iitimcomin ·

no ive related it to a spring u see.......

3
iitimcomin ·

Y = FL / AdL ..........

F/dL = YA/L ..................

F/dL = K ................

although it ends up being same...............

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