A) \[1.6\times {{10}^{-2}}m\]
B) \[2.9\times {{10}^{-3}}m\]
C) \[5.6\times {{10}^{-4}}m\]
D) \[6.0\times {{10}^{-5}}m\]
Correct Answer: D
Solution :
The given, \[T=7.2\times {{10}^{-2}}N/m\] \[P=2.4\times {{10}^{3}}N/{{m}^{2}}\] or Pa \[R=?\] We know that, \[P=\frac{2T}{R}\] or \[R=\frac{2T}{P}\] \[=\frac{2\times 7.2\times {{10}^{-2}}}{2.4\times {{10}^{3}}}=6.0\times {{10}^{-5}}m\]
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