A) 2% decrease
B) 2% increase
C) 1% decrease
D) 1% increase
Correct Answer: A
Solution :
From the relation of g \[g=\frac{GM}{{{R}^{2}}}\] or \[g\propto \frac{1}{{{R}^{2}}}\](at constant mass) As \[{{R}_{1}}=R,\] \[{{R}_{2}}=R-\frac{R}{100}\] Hence, \[\frac{{{g}_{2}}}{{{g}_{1}}}={{\left( \frac{{{R}_{r}}}{{{R}_{r}}} \right)}^{2}}\] \[\frac{{{g}_{2}}}{{{g}_{1}}}={{\left( \frac{R}{R-0.1R} \right)}^{2}}={{(1-0.01)}^{2}}\] As \[[{{(1+x)}^{n}}=1+nx\,\,\,\,\,\,\,where\,\,\left| x \right|<<1]\] or \[\frac{{{g}_{2}}}{{{g}_{1}}}=1+0.02\] or \[\frac{{{g}_{2}}-{{g}_{1}}}{{{g}_{1}}}\times 100=[(1+0.02)-1]\times 100=2%\]
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