Two slabs with square cross section of different materials (1, 2) with equal sides ($l$) and thickness $d_1$ and $d_2$ such that $d_2 = 2d_1$ and $l > d_2$. Considering that the 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 modulus of material 1 is $4\times10^9\ \text{N/m}^2$, then the shear modulus of material 2 is $x\times10^9\ \text{N/m}^2$, where the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 1
The force acts on the narrow top face of area $l\,d$, so the shear stress is $\dfrac{F}{ld}$ and the shear angle is
$$\theta = \frac{F}{l\,d\,\eta}$$
$$\frac{\theta_2}{\theta_1} = \frac{d_1\eta_1}{d_2\eta_2} = \frac{\eta_1}{2\eta_2} = 2 \Rightarrow \eta_2 = \frac{\eta_1}{4} = 1\times10^9\ \text{N/m}^2$$
So $x = 1$.
Solution by Sreeraj P, M.Sc Physics