A) 6 m
B) 9 m
C) 12 m
D) 15 m
Correct Answer: C
Solution :
\[F\propto \frac{1}{{{d}^{2}}},\]given \[F'=\frac{64F}{100}F,d'=(d+3)\] \[\therefore \frac{F}{F'}=\frac{{{(d+3)}^{2}}}{{{d}^{2}}}\] \[\Rightarrow \frac{64F}{100F}=\frac{{{d}^{2}}}{{{(d+3)}^{3}}}\Rightarrow \frac{8}{10}=\frac{d}{d+3}\] \[8d+24=10d\] \[2d=24\Rightarrow d=12m\]You need to login to perform this action.
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