Question

Two slabs with square cross section of different materials $(1,2)$ with equal sides $(l)$ and thickness $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ such that $\mathrm{d}_{2}=2 \mathrm{~d}_{1}$ and $l>\mathrm{d}_{2}$. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is $\theta_{2}=2 \theta_{1}$. If the shear moduli of material 1 is $4 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$, then shear moduli of material 2 is $\mathrm{x} \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$, where value of x is _______ .

Hint:

Use the deformation angle to find the shear modulus of the material.

Answer 1

Correct Answer: 1

1. Deformation angle: \[ 2 \theta_{1} = \theta_{2} \] \[ 2 \frac{\sigma_{1}}{\eta_{1}} = \frac{\sigma_{2}}{\eta_{2}} \] \[ 2 \left( \frac{F}{l d_{1} \eta_{1}} \right) = \frac{F}{l d_{2} \eta_{2}} \] \[ \eta_{2} = \frac{\eta_{1}}{4} = 1 \times 10^{9} \] \[ x = 1 \] The calculated value is 1, corresponding to option (1).