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: Value 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, \[h=0.41\,r\]
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