Two slabs with square cross-sections of diffe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The shear stress $\sigma$ is defined as the force $F$ applied per unit area $A$. For a slab of side $l$ and thickness $d$,the area of the face on

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The shear stress $\sigma$ is defined as the force $F$ applied per unit area $A$. For a slab of side $l$ and thickness $d$,the area of the face on which the force is applied is $A = l 	imes d$.Thus,the shear stress for the two slabs is:$\sigma_1 = \frac{F}{l d_1}$ and $\sigma_2 = \frac{F}{l d_2}$.The shear modulus $\eta$ is defined as the ratio of shear stress to the angle of deformation $	heta$ (for small angles,$	heta \approx 	an 	heta$):$\eta = \frac{\sigma}{	heta} \Rightarrow 	heta = \frac{\sigma}{\eta}$.Given $	heta_2 = 2	heta_1$,we substitute the expressions:$\frac{\sigma_2}{\eta_2} = 2 \left( \frac{\sigma_1}{\eta_1} ight)$.Substituting $\sigma_1 = \frac{F}{l d_1}$,$\sigma_2 = \frac{F}{l d_2}$,and $d_2 = 2d_1$:$\frac{F}{l d_2 \eta_2} = 2 \left( \frac{F}{l d_1 \eta_1} ight)$.$\frac{1}{2d_1 \eta_2} = \frac{2}{d_1 \eta_1} \Rightarrow \frac{1}{\eta_2} = \frac{4}{\eta_1} \Rightarrow \eta_2 = \frac{\eta_1}{4}$.Given $\eta_1 = 4 	imes 10^9 \ N/m^2$,we get:$\eta_2 = \frac{4 	imes 10^9}{4} = 1 	imes 10^9 \ N/m^2$.Comparing this with $x 	imes 10^9 \ N/m^2$,we find $x = 1$.

Similar Questions

What is the modulus of rigidity of a liquid?

The dimensions of shear modulus are

The modulus of rigidity of diamond is:

The strain energy stored in a body of volume $V$ due to shear strain $\phi$ is (shear modulus is $\eta$ ).

The dimensional formula for the modulus of rigidity is

Vedclass Test Series

Exam Paper Generator

Online Exam Module