A) Rs = \[\sqrt{2}\]RB
B) RS = \[\frac{{{R}_{B}}}{2}\]
C) Rs = 4RB
D) Rs = \[\frac{{{R}_{B}}}{2}\]
Correct Answer: B
Solution :
Increase in length,\[\Delta L=\frac{FL}{Y.A}=\frac{FL}{Y.\pi {{R}^{2}}}\]As F, L and \[\Delta L\]are same hence,\[Y.{{R}^{2}}=a\,cons\tan t\] \[\therefore \] \[2.0\times {{10}^{11}}R_{s}^{2}=1.0\times {{10}^{11}}R_{B}^{2}\]\[\Rightarrow \] \[{{R}_{s}}=\frac{{{R}_{B}}}{\sqrt{2}}\]You need to login to perform this action.
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