A) 2% decrease
B) 2% increases
C) 1% decrease
D) 1% increases
Correct Answer: A
Solution :
From the relation of g \[g=\frac{GM}{{{R}^{2}}}\]or\[g\propto \frac{1}{{{R}^{2}}}\](at constant mass) \[\therefore \] \[\frac{{{g}_{2}}}{{{g}_{1}}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{2}}\] Where, \[{{R}_{1}}=R\]and\[{{R}_{2}}=R-\frac{R}{100}\] \[\therefore \] \[\frac{{{g}_{2}}}{{{g}_{1}}}={{\left( \frac{2}{R-0.1R} \right)}^{2}}={{(1-0.02)}^{2}}\] \[[{{(1+x)}^{3}}=1+nx\]where \[|x|<<|\] \[\frac{{{g}_{2}}-{{g}_{1}}}{{{g}_{1}}}\times 100=[(1+0.02)-1]\times 100\]=2%decreaes
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