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Question

A square shaped slab of lead with side 50 cm and thickness 10.0 cm is subjected to a shearing force (on its narrow face) of magnitude 9.0×104 N. The lower edge is riveted to the floor as shown in figure. How much is the upper edge displaced, if the shear modulus of lead is 5.6×109 Pa?


  1. 4.8×104 m
  2. 3.2×104 m
  3. 1.6×104 m
  4. 0.8×104 m


Solution

The correct option is C 1.6×104 m
Given,
L=50 cm=50×102 m
t=10 cm=10×102 m
G=5.6×109 Pa 
F=9.0×104 N
Area of the face on which force is applied, 
A=50×10=500 cm2=0.05 m2

Let ΔL be the displacement of the upper edge of the slab, due to tangential force F applied.
Then, G=(FA)(ΔLL)
ΔL=FLGA=9×104×50×1025.6×109×0.05
ΔL=1.6×104 m

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