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 \[mg=mg\left( 1-\frac{2h}{{{R}_{e}}} \right)\] \[i.e.,\] \[w=w\left( 1-\frac{2h}{{{R}_{e}}} \right)\] \[\frac{99}{100}w=w\left( 1-\frac{2h}{{{R}_{e}}} \right)\] \[i.e.,\] \[h=0.005\,{{R}_{e}}\] At a point below the surface of earth at depth A. The weight of body is given by \[w=w\left( 1-\frac{h}{{{R}_{e}}} \right)\] \[\frac{w}{w}=0.995\] \[%\,\Delta w=\frac{(1-0.995)w}{w}\times 100\] \[%\,\Delta w=0.5%\] (decreases)You need to login to perform this action.
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