A) Increase by 4%
B) Increase by 2%
C) Decrease by 2%
D) Decrease by 4%
Correct Answer: D
Solution :
From formula, intensity \[I\propto \frac{1}{{{r}^{2}}}\] For two cases, we have, \[\frac{{{I}_{2}}}{{{I}_{1}}}={{\left( \frac{r}{{{r}_{2}}} \right)}^{2}}\] Here, \[{{r}_{1}}=r,\]\[{{r}_{2}}=4+\frac{2r}{100}\] \[\Rightarrow \]\[\frac{{{I}_{2}}}{{{I}_{1}}}\left[ \frac{r}{\frac{102}{100}r} \right]={{(1+0.02)}^{-2}}\] ?..(i) Expanding equation (i) by binomial theorem, we get \[\frac{{{I}_{2}}}{{{I}_{1}}}=1-0.4\] \[[\because \,\,{{(1+n)}^{-n}}=1-nx]\] or \[{{I}_{2}}={{I}_{1}}(1-0.4)\] \[\frac{{{I}_{1}}-{{I}_{2}}}{{{I}_{1}}}\times 100=4%\] So, intensity decrease by 4%.
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