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

Two rods of different materials and identical cross sectional area, are joined face to face at one end and their free ends are fixed to the rigid walls. If the temperature of the surroundings is increased by 30oC, the magnitude of the displacement of the joint of the rod is (length of rods l1=l2=1 unit, ratio of their young's moduli Y1/Y2=2, coefficient of linear expansion are α1 and α2)

A
5(α2α1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
10(α1α21)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
10(α22α1)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
5(2α1α2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 10(α22α1)
l1α2θ+l2α2θ=F1l2A1y2+f2l2A2y2
1+α1×30+1×α2×30=F×1A×2y+F×1A×y
[where , y1 & y2 are young's modulus y1y2=2]
30(α1+α2)=3F2Ay
10(α1+α2)=F2Ay
Thus, displacement of joint is
F1l1Ay1l2α1θ=F×l1A1y21×α1×30=10(α1+α2)30α1
=10(α22α1)

1176567_790162_ans_280ae7528b0e469ea7e2dfdbe40eb9c3.jpg

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