A) 9%
B) \[9.3\times {{10}^{-11}}%\]
C) 10%
D) None of these
Correct Answer: B
Solution :
Heat taken by the given body \[=mc\,\Delta \theta \] [Specific heat of body\[=0.2\text{ }kcal/kg{}^\circ C,\]final temperature of the body\[=100{}^\circ C\]] \[\Delta E=m\times 0.2\times 100=20\text{ }m\text{ }kcal\] \[=20m\times 4.2\times {{10}^{3}}J\] Now, gain in mass is given by \[\Delta m=\frac{\Delta E}{{{c}^{2}}}\] \[(\because E=m{{c}^{2}})\] \[=\frac{20m\times 4.2\times {{10}^{3}}}{{{(3\times {{10}^{8}})}^{2}}}\] Therefore, percentage increase in mass is given by \[=\frac{\Delta m}{m}\times 100=\frac{20m\times 4.2\times {{10}^{3}}}{{{(3\times {{10}^{8}})}^{2}}\times m}\times 100\] \[=9.3\times {{10}^{-11}}%\]
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