A) \[3.6\,m/{{s}^{2}}\]
B) \[6.6\,m/{{s}^{2}}\]
C) \[7.66\,m/{{s}^{2}}\]
D) \[9.65\,m/{{s}^{2}}\]
Correct Answer: D
Solution :
The value of acceleration due to gravity at a depth h, below the surface of earth of radius R is \[g=g\left( 1-\frac{h}{R} \right)\] Given, \[g=9.8\,m/{{s}^{2}},h=100\,km=100\times {{10}^{3}}m\], \[R=6380\times {{10}^{3}}m\]. \[\therefore \] \[g=9.8\left( 1-\frac{100\times {{10}^{3}}}{6380\times {{10}^{3}}} \right)\] \[=9.8\left( 1-\frac{1}{63.8} \right)=9.8\times \frac{62.8}{63.8}=9.65\,m/{{s}^{2}}\] Note: The value of acceleration due to gravity goes on decreasing below the surface of earth and becomes zero at the centre of the earth.
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