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 \[=0.2\text{ }kcal\text{/}kg\text{ }{}^\circ C,\] final temperature of the body)\[=100{}^\circ C\] \[\Delta E=m\times 0.2\times 100\] \[=20\,m\,kcal\] \[=20\,m\times 4.2\times {{10}^{3}}J\] Now, gain in mass is given by \[\Delta m=\frac{\Delta E}{{{c}^{2}}}\] \[=\frac{20\,m\times 4.2\times {{10}^{3}}}{{{(3\times {{10}^{8}})}^{2}}}\] Therefore percentage increase in mass is given by \[=\frac{\Delta m}{m}\times 100\] \[=\frac{20\,m\times 4.2\times {{10}^{3}}\times 100}{{{(3\times {{10}^{8}})}^{2}}\times m}\] \[=9.3\times {{10}^{-11}}%\]
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