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A steel wire is 1 m long and 1 mm^2 in area of cross section. If it takes 200 N to stretch this wire by 1 mm, how much force will be required to stretch a wire of the same material as well as diameter from its normal length of 10 m to a length of 1002 cm?

A steel wire is 1 m long and 1 mm^2 in area of cross section. If it takes 200 N to stretch this wire by 1 mm, how much force will be required to stretch a wire of the same material as well as diameter from its normal length of 10 m to a length of 1002 cm?

Grade:12

3 Answers

Aziz Alam IIT Roorkee
askIITians Faculty 233 Points
7 years ago
Length(l0)=10m Area(A)=1 mm2
l’=10.001m F=200N
Y=F*l_0/((l'-l_0)*A) \Rightarrow Y=2*10^{12} N/m^2
\Rightarrow F=Y*A*(l'-l_0)/l_0 \Rightarrow F=4000N
Payal
17 Points
2 years ago
Caution:The above ans is wrong. 
Correct  ans is 400 N
Soln:
Y=(F×l)/l'×A where l' signifies change in length .
In 1st case l'=1mm i.e. 10^(-3)m 
Y= 2×10^12
Thus for 2nd case,
F=(Y l'A)/l 
F= 400N
 
Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the solution to your problem.
 
We have, F1 = 200 N, L1 = 1 m, A1 = 1 mm2 = 10-6 m2, ΔL1 = 1 mm = 10-3 m
Now, we know, Y = F1L1/A1ΔL1 = (200 x 1)/(10-6 x 10-3) = 2 x 1011 N/m2
 
We now have, L2 = 10 m, ΔL2 = 10.02 – 10 = 0.02 m, A2 = 1mm2 = 10-6 m2, Y = 2 x 1011 N/m2
Hence, F = YΔL2A2/L2 = (2 x 1011) x (0.02) x (10–6)/10 = 400 N
 
Thanks and regards,
Kushagra

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