A metal cube of side 10 cm is subjected to a shearing stress of 106N/m2. Calculate the modulus of rigidity in ×108Nm−2 if the top of the cube is displaced by 0.05 cm with respect to its bottom.
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is 2.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Shearing stress , F/A=106N/m2Length of side, L=10cm=0.1mShearing displacement , Δx=0.05cm=0.0005mThe modulus of rigidity is η=FAθ=FLAΔx∵θ≅tanθ=ΔxL=106×0.10.0005=2×108N/m2