A cylindrical rod of length 1 m and radius 4 | vedclass

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(A) The shear modulus $G$ is defined as the ratio of shear stress to shear strain (angular displacement $	heta$ for small angles).$G = \frac{	ext{Sh

Vedclass · Accepted Answer

$1 / 160 \pi$

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(A) The shear modulus $G$ is defined as the ratio of shear stress to shear strain (angular displacement $	heta$ for small angles).$G = \frac{	ext{Shear Stress}}{	heta}$Shear stress $\sigma = \frac{F}{A}$,where $F = 10^5 \ N$ and $A = \pi r^2$.Given $r = 4 \ cm = 0.04 \ m = 4 	imes 10^{-2} \ m$.Area $A = \pi 	imes (4 	imes 10^{-2})^2 = 16 \pi 	imes 10^{-4} \ m^2$.Now,substitute the values into the formula:$10^{10} = \frac{10^5 / (16 \pi 	imes 10^{-4})}{	heta}$$	heta = \frac{10^5}{16 \pi 	imes 10^{-4} 	imes 10^{10}}$$	heta = \frac{10^5}{16 \pi 	imes 10^6} = \frac{1}{160 \pi} 	ext{ radian}$.

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