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Physics Question on elastic moduli

Two parallel and opposite forces each 5000 N5000\, N are applied tangentially to the upper and lower faces of a cubical metal block of side 25 cm25 \,cm. The angle of shear is (The shear modulus of the metal is 80 GPa80 \,GPa.)

A

10−4 rad10^{-4}\, rad

B

10−5 rad10^{-5}\, rad

C

10−6 rad10^{-6}\, rad

D

10−7 rad10^{-7}\, rad

Answer

10−6 rad10^{-6}\, rad

Explanation

Solution

Here, F=5000 NF= 5000 \,N L=25 cmL=25\, cm =25×10−2 m=25\times10^{-2}\,m AA =(25×10−2m)2=\left(25\times10^{-2}m\right)^{2} =625×10−4m2=625\times10^{-4}m^{2} η=80Gpa\eta=80 Gpa =80×109pa=80\times10^{9}pa Angle of shear, θ\theta =FAη=\frac{F}{A\eta} =5000625×10−4×80×109=\frac{5000}{625\times10 ^{-4}\times80\times10^{9}} =10−6 rad=10^{-6} \, rad