A) 0.5% decrease
B) 2% decrease
C) 0.5% increase
D) 1% increase
Correct Answer: A
Solution :
At an altitude h the acceleration due to gravity is \[g=g\left( 1-\frac{2h}{{{R}_{e}}} \right)\] or \[mg;=mg\left( 1-\frac{2h}{{{R}_{e}}} \right)\] i.e., \[\omega =\omega \left( 1-\frac{2h}{{{R}_{e}}} \right)\] \[\frac{99}{100}\omega =\omega \left( 1-\frac{2h}{{{R}_{e}}} \right)\] i.e., \[h=0.005{{R}_{e}}\] At a point below the surface of earth at depth h. The weight of body is given by \[\omega =\omega \left( 1-\frac{2h}{{{R}_{e}}} \right)\] \[\frac{\omega }{\omega }=0.995\] \[%\Delta \omega =\frac{(1-0.995)\omega }{\omega }\times 100\] \[%\Delta \omega =0.5%\] (decreases)
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