A) Decreases by 19%
B) Increases by 19%
C) Decreases by more than 19%
D) Increase by more than 19%
Correct Answer: D
Solution :
[d] \[g=\frac{GM}{{{R}^{2}}}\] (i) \[g'=\frac{Gm}{{{\left( \frac{90}{100}R \right)}^{2}}}=\frac{100}{81}-\frac{GM}{{{R}^{2}}}\] (ii) From Eqs. (i) and (ii), \[g'=\frac{100}{81}g\Rightarrow \frac{g'}{g}=\frac{100}{81}\] \[\frac{g'}{g}-1=\frac{100}{81}-1\] \[\therefore \Delta g=\frac{19}{81}g=23%\]of g. So increase is more then 19% of g.
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