A) 9%
B) \[9.3\times {{10}^{-11}}%\]
C) 10%
D) None of these
Correct Answer: B
Solution :
\[E=mc\Delta \theta \] \[\Delta m=\frac{mc\Delta \theta }{{{c}^{2}}}\] \[=\frac{m\times 0.2\times 100\times 4.2\times {{10}^{3}}}{{{(3\times {{10}^{8}})}^{2}}}\] \[\frac{\Delta m}{m}=\frac{20\times 4.2\times {{10}^{3}}}{{{(3\times {{10}^{8}})}^{2}}}\] % increase in mass\[=\frac{\Delta m}{m}\times 100\] \[=\frac{20\times 4.2\times {{10}^{3}}}{{{(3\times {{10}^{8}})}^{2}}}\times 100\] \[=9.3\times {{10}^{-11}}\]
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