A) 1 : 1
B) 1 : 2
C) 2 : 1
D) 1 : 4
Correct Answer: A
Solution :
Key Idea Acceleration due to gravity decreases both at altitude and depth. If g is the acceleration due to gravity at a point, at a height h above the surface of earth, and g be acceleration due to gravity on earths surface, then \[g=g\left( 1-\frac{2h}{R} \right)\] ? (i) If g be the acceleration due to gravity at a point at depth d, below the surface of earth, then \[g=g\left( 1-\frac{d}{R} \right)\] ... (ii) But \[d=2h\] (given) \[\therefore \] \[g=g\left( 1-\frac{2h}{R} \right)\] \[\therefore \] \[\frac{g}{g}=\frac{g\left( 1-\frac{2h}{R} \right)}{g\left( 1-\frac{2h}{R} \right)}=1\] \[\therefore \] \[g:g=1:1\] Note The value of acceleration due to gravity, also decreases due to rotation of earth.You need to login to perform this action.
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