The ratio of the weight of a body at a height of (R/10) from the surface of the earth to that at a depth of (R/10) is (R is radius of earth)

Question

The ratio of the weight of a body at a height of (R/10) from the surface of the earth to that at a depth of (R/10) is (R is radius of earth) (A) 4: 5 (B) 1: 1 (C) 9: 8 (D) 2: 3

Tardigrade Admin · Accepted Answer

ωh=(m g/(1+(h)R)2) Given h=(R/10) =(100/121) m g ωd=m g(1-(d/R)) =m g(1-(1/10)) (∴ d=(R/10)) =(9/10) m g ∴ (ωh/ωd)=(100 m g/121) × (10/9) m g =(1000/121 × 9) ≈ (8/9)

Q. The ratio of the weight of a body at a height of $\frac{R}{10}$ from the surface of the earth to that at a depth of $\frac{R}{10}$ is ( $R$ is radius of earth)

$4 : 5$

$1 : 1$

$9 : 8$

$2 : 3$

$8 : 9$

Q. The ratio of the weight of a body at a height of 10R​ from the surface of the earth to that at a depth of 10R​ is (R is radius of earth)

4:5

1:1

9:8

2:3

8:9