Question

·
Height at which magnitude of g will be equal to
that at a depthR/2 is (where Ris radius of earth)
(1) R

(2) √2R
(4) (√2+1)
(3) (√2-1)​

Answer 1

Answer:(√2-1) R

Explanation:

gh=g(1+h/R) ^-2

gd=g(1-R/2/R)

gd=(1/2)g

gh=gd

(R/R+h)^2=1/2

2R^2=R^2+h^2

Negligent 2Rh

Multiple whole eqn by^1/2

√2R-R=h

h=R(√2-1)

Answer 2

(3)$h=(\sqrt{2}-1)R$

Explanation:

Let R be the radius of earth and g be the initial gravity of earth

Gravity at height h is given by

$g_h=g(\frac{R}{R+h})^2$

Gravity at depth d is given by

$g_d=g(1-\frac{d}{R})$

Substitute d=$\frac{R}{2}$

Then , $g_d=g(1-\frac{R}{2R})=\frac{1}{2}g$

We are given that

$g_h=g_d$

$g(\frac{R}{R+h})^2=\frac{1}{2}g$

$\frac{R}{R+h}=\frac{1}{\sqrt 2}$

$\frac{R+h}{R}=\sqrt 2$

$1+\frac{h}{R}==\sqrt 2$

$\frac{h}{R}=\sqrt{2}-1$

$h=(\sqrt{2}-1)R$

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