A) \[4.2\times {{10}^{6}}\,\,m\]
B) \[3.19\times {{10}^{6}}\,\,m\]
C) \[1.59\times {{10}^{6}}\,\,m\]
D) none of these
Correct Answer: B
Solution :
Given : Height above the surface of earth \[h=1600\,\,km=1.6\times {{10}^{6}}m\] The acceleration due to gravity at \[a\] depth\[d\], is given by, \[g=g\left( 1-\frac{d}{R} \right)\] Also, the value of \[g\] at a height \[h\] is \[g=g\left( 1-\frac{2h}{R} \right)\] \[=g\left( 1-\frac{2\times 1.6\times {{10}^{6}}}{6.38\times {{10}^{6}}} \right)=0.5g\] Since, \[g\] must be equal to\[\frac{g}{2}\] Hence, \[g\left( 1-\frac{d}{R} \right)=0.5\,g\] \[1-\frac{d}{R}=0.5\] or \[d=0.5R=0.5\times 6.38\times {{10}^{6}}\] \[=3.19\times {{10}^{6}}m\]
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