A) \[6.28\times {{10}^{5}}\,\,N\]
B) \[6.28\times {{10}^{4}}\,\,N\]
C) \[6.28\times {{10}^{4}}\text{ }dyne\]
D) \[6.28\times {{10}^{5}}\text{ }dyne\]
Correct Answer: B
Solution :
Young's modulus of a rope \[Y=\frac{FL}{A\Delta l}\] Given, \[L=10m,\,\,A=\pi {{r}^{2}}=\pi {{(1)}^{2}}=\pi ;\] \[Y=20\times {{10}^{11}}\,dyne/c{{m}^{2}},\,\Delta l=1cm,\] \[F=\frac{Y.A.\Delta l}{L}\] \[F=\frac{20\times {{10}^{11}}\times 1\times 1}{10\times {{10}^{2}}}\] \[F=6.28\times {{10}^{9}}dyne\] \[F=6.28\times {{10}^{4}}N\]
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