A rubber cube of side 5 ~cm has one face fixe | vedclass

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(C) Given: Side of the cube $L = 5 ~cm = 0.05 ~m$. Area $A = L^2 = (0.05)^2 = 25 	imes 10^{-4} ~m^2$. Tangential force $F = 1800 ~N$. Modulus of rigi

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$15$

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(C) Given: Side of the cube $L = 5 ~cm = 0.05 ~m$. Area $A = L^2 = (0.05)^2 = 25 	imes 10^{-4} ~m^2$. Tangential force $F = 1800 ~N$. Modulus of rigidity $\eta = 2.4 	imes 10^6 ~N/m^2$.Shear stress is defined as $\sigma = F/A = 1800 / (25 	imes 10^{-4}) ~N/m^2$.Shear strain is defined as $	heta = x/L$,where $x$ is the lateral displacement.Using the formula $\eta = 	ext{Shear stress} / 	ext{Shear strain}$,we have $\eta = (F/A) / (x/L)$.Rearranging for $x$: $x = (F \cdot L) / (A \cdot \eta)$.Substituting the values: $x = (1800 	imes 0.05) / (25 	imes 10^{-4} 	imes 2.4 	imes 10^6)$.$x = 90 / (25 	imes 2.4 	imes 10^2) = 90 / 6000 = 0.015 ~m$.Converting to millimeters: $x = 0.015 	imes 1000 ~mm = 15 ~mm$.

$A$ rubber cube of side $5 ~cm$ has one face fixed,while a tangential force $1800 ~N$ is applied on its opposite face. If the modulus of rigidity of rubber is $2.4 \times 10^6 ~N/m^2$,then the lateral displacement of the strained face is . . . . . . (in $mm$)

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