A) \[2\,\,m\,\,{{l}^{3}}{{\omega }^{2}}\sin \theta .\cos \theta \]
B) \[\,m\,\,{{l}^{2}}{{\omega }^{2}}\sin 2\theta \]
C) \[\,m\,\,{{l}^{2}}\sin 2\theta \]
D) \[\,{{m}^{1/2}}\,\,{{l}^{1/2}}\,\omega \,\,\sin \theta .\cos \theta \]
Correct Answer: B
Solution :
Energy absorbed in the ionisation of 1 mole of \[Mg\,to\,M{{g}^{+}}(g)=750KJ\] Energy left unused \[=\text{12}00-\text{75}0=\text{45}0\text{ kJ}\] % of \[M{{g}^{+}}(g)\]converted into \[M{{g}^{2+}}(g)\] \[=\frac{450}{1450}\times 100=31%\] Hence, the % of \[\text{M}{{\text{g}}^{+}}\left( \text{g} \right)=\text{1}00-\text{31}=\text{69}%\]You need to login to perform this action.
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