A system has internal energy equal to $E_1,450\mathrm{ }\mathrm{J}$ of heat is taken out of it and $500\mathrm{ }\mathrm{J}$ of work is done on it. The final energy of the system will be

 Work is done on the system,

$∆\mathrm{E}={\mathrm{E}}_2-{\mathrm{E}}_1=-\mathrm{Q}+\mathrm{W}$

${\mathrm{E}}_2-{\mathrm{E}}_1=-450+500=50 \mathrm{J}$

${\mathrm{E}}_2={\mathrm{E}}_1+50 \mathrm{J}$

$\therefore$ The final energy of the system is ${\mathrm{E}}_1+50\mathrm{ }\mathrm{J}$.

Hence the correct answer is (A) $\left({\mathrm{E}}_1+50\right)$.