We know that shearing stress=
F/A=10^
4
Nm
^−2
Length of side of cube=10cm=10/100
m=0.1m
Shearing displacement=△n=0.05cm
=0.05
/100m=0.0005m
We know that modulus of rigidity is η=
F/Aθ
= FL/
A△x [tanθ≈θ= △x/L
] = =
10^4x0.1/5x10^-4
⇒10^7/5^1⇒2x10^6Nm^-2
Hence, modulus of rigidity is equal to 2x10^6Nm^-2
The answer is D , i hope it helps !
Question 1 FALSO
Question 2 CIERTO
Question 5 FALSO
Question 6 CIERTO
Question 7 CIERTO
