A metal cube of side `10 cm` is subjected to a shearing stress of `10^(6)N//m^(2)`. Calculate the modulus of rigidity if the of the cube is displaced by `0.05 cm` with respect to its bottom.
A
`1 xx 10^8 N//m^2`
B
`2 xx 10^8 N//m^2`
C
`3 xx 10^8 N//m^2`
D
`4 xx 10^8 N//m^2`
Text Solution
Verified by Experts
The correct Answer is:
B
Shearing stress, `F/A=10^(6)N//m^(2)` Length of side `L=10cm=0.1m` Shearing displacemet `/_\x=0.005cm` The modulus of rigidity is `eta=F/(Atheta)=(FL)/(A/_\x)` (`:'theta~=tantheta=(/_\x)/L`) `=(10^(6)xx0.1)/0.005=2xx10^(8)N//m^(2)`
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