A) \[0.31\] times \[r\]
B) \[0.41\] times \[r\]
C) \[0.51\] times \[r\]
D) \[0.61\] times \[r\]
Correct Answer: B
Solution :
Key Idea: Valued of acceleration due to gravity decreases on going above the earth's surface. The weight of a body is given by \[w=mg\] (on earth's surface) At a height \[h\] above the earth's surface \[w'=\frac{mg}{{{\left( 1+\frac{h}{R} \right)}^{2}}}\] \[\therefore \] \[\frac{w}{w'}={{\left( 1+\frac{h}{R} \right)}^{2}}\] Given, \[w=80\,\,kg,\,\,w'=40\,\,kg\] \[\therefore \] \[\frac{80}{40}={{\left( 1+\frac{h}{R} \right)}^{2}}\] \[\Rightarrow \] \[2={{\left( 1+\frac{h}{R} \right)}^{2}}\] \[\Rightarrow \] \[\sqrt{2}=1.41=1+\frac{h}{R}\] \[\Rightarrow \] \[h=0.41\,\,R\] Given, \[R=r\] \[\therefore \] \[h=0.41\,\,r\]
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