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