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(A) The stress $\sigma$ at a distance $x$ from the free end is given by $\sigma = \frac{F}{A}$.For a small element of length $dx$ and mass $dm = \rho

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) The stress $\sigma$ at a distance $x$ from the free end is given by $\sigma = \frac{F}{A}$.For a small element of length $dx$ and mass $dm = ho A dx$,the force $dF$ required to accelerate it is $dF = (dm)a = ho A a dx$.Thus,the rate of change of stress with respect to $x$ is $\frac{d\sigma}{dx} = \frac{1}{A} \frac{dF}{dx} = \frac{ho A a}{A} = ho a$.This represents the slope of the stress-distance graph,so $	an 	heta = ho a$.Since the acceleration $a$ is constant for the entire rod,we have $\frac{	an 	heta_1}{	an 	heta_2} = \frac{ho_1}{ho_2}$.Given $	heta_1 = 37^\circ$ and $	heta_2 = 53^\circ$,and $ho_1 = 9 	ext{ g/cm}^3$:$\frac{	an 37^\circ}{	an 53^\circ} = \frac{9}{ho_2} \Rightarrow \frac{3/4}{4/3} = \frac{9}{ho_2} \Rightarrow \frac{9}{16} = \frac{9}{ho_2}$.Therefore,$ho_2 = 16 	ext{ g/cm}^3$.

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