A) 66.6m
B) 60.0m
C) 33,3m
D) 30.0m
Correct Answer: C
Solution :
The internal restoring force acting per unit area of cross-section of the deformed body is called stress. If an external force F is applied to the cross-sectional area A of the body, then stress \[=\frac{F}{A}=\frac{mg}{A}\] where mass m = volume \[\times \] density =\[Al\rho \] \[\therefore \] Stress \[=\frac{Al\rho g}{A}=\rho \lg \] \[\therefore \] \[l=\frac{breaking\text{ }stress}{\rho g}\] Given, breaking stress \[={{10}^{6}}\,N/{{m}^{2}}\], \[\rho =3\times {{10}^{3}}kg/{{m}^{3}},g=10\,m/{{s}^{2}}\], \[\therefore \] \[l=\frac{{{10}^{6}}}{(3\times {{10}^{3}})\times 10}\] \[l=33.3\,m\]You need to login to perform this action.
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