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A metal cube of side 20cm is subjected t...

A metal cube of side 20cm is subjected to a shearing force of 4000N. The top face is displaced through 0.50 cm with respect to the bottom. Find the rigidity modulus of the metal.

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Length of one side `= L = 0.20m`
Shearing force `= F = 4000N`
Displacement of the top face `= l = 0.005m`
Area `= A = L^(2) = 0.04m^(2)`
Shearing strain `= theta = ( l)/( L)`
Rigidity moludus `=G = ( F)/( A theta ) = ( Fl )/(Al )`
`= ( 4000 xx 0 .20)/( 0.04 xx 0.005 )`
`= 4 xx 10^(6) N //m^(2)`
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