wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A copper wire of length 1.0 m and a steel wire of length 0.5 m having equal cross-sectional areas are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by 1 mm. If the Young's moduli of copper and steel are 1.0×1011 Nm2 and 2.0×1011 Nm2, the total extension of the composite wire is


A

1.25 mm

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

1.50 mm

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

1.75 mm

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

2.00 mm

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

1.25 mm


When a wire of length L, cross-sectional area A and Young's modulus Y is stretched with a force F, the extension l in the wire is given by
l=FLAY
Since F and A are the same for the two wires, we have
For copper wire lc=FLcAYc
For steel wire ls=FLsAYs
ls=lc(LsLc)(YcYs)
=1 mm×(0.5 m1.0 m)×(1.0×1011 Nm22.0×1011 Nm2)
=0.25 mm
Total extension = 1 + 0.25 = 1.25 mm.
Hence, the correct choice is (a).


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Hooke's Law
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon