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

A square lead slab of side 50cm and thickness 10cm is subjected to a shearing force (on its narrow face) of 9×104N. The lower edge is riveted to the floor. How much will the upper edge be displaced?
(Shear modulus of lead =5.6×109Nm2)

A
0.16mm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1.6mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.16cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.6cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 0.16mm
The lead slab is fixed and force is applied parallel to the narrow face as shown in the figure. Area of the face parallel to which this force is applied is
A=50cm×10cm=0.5m×0.1m=0.05m2
If ΔL is the displacement of the upper edge of the slab due to tangential force, F, then
η=F/AΔL/LorΔL=FLηA
Substituting the given values, we get
ΔL=(9×104N)(0.5m)5.6×109Nm2×0.05m2=1.6×104m=0.16mm
1031890_937025_ans_4ec8ed4ece9544e1a1a900b221bfd040.PNG

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Equilibrium 2.0
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon