Test Index

Gravitation Part 1

Section: Physics

© examsnet.com

Question : 45 of 100

Marks: +1, -0

The height 'h' at which the weight of a bodywill be the same as that at the same depth 'h'from the surface of the earth is (Radius of theearth is

R

and effect of the rotation of the earthis neglected):

[2 Sep 2020 Shift 2]

$\frac{\sqrt{5} R - R}{2}$
$\frac{\sqrt{5}}{2} R - R$
$\frac{R}{2}$
$\frac{\sqrt{3}R - R}{2}$

Solution:

\blacklozenge M =

mass of earth

M_{1} =

mass of shaded portion

R =

Radius of earth

\blacklozenge M_1 = \frac{M}{\frac{4}{3} \pi R^{3}} \cdot \frac{4}{3} \pi (R-h)^{3}

= \frac{M(R-h)^{3}}{R}

◆ Weight of body is same at

P

and

Q

i.e.

mg_P = mg_Q

g_P = g_Q

\frac{GM_1}{(R-h)^{2}} = \frac{GM}{(R+h)^{2}}

\frac{GM (R-h)^{3}}{(R-h)^{2} R^{3}} = \frac{GM}{(R+h)^{2}}

(R-h)(R+h)^{2} = R^{3}

R^{3} - h R^{2} - h^{2} R - h^{3}

+ 2 R^{2} h - 2 R h^{2} = R^{3}

R^{2} - (Rh)^{2} - h^{3} = 0

R^{2} - Rh - h^{2} = 0

h^{2} + R h - R^{2} = 0

\Rightarrow h = \frac{-R \pm \sqrt{R^{2} + 4 R^{2}}}{2}

ie

h = \frac{-R + \sqrt{5} R}{2}

= \left(\frac{\sqrt{5} - 1}{2}\right) R

© examsnet.com

Go to Question:

Prev Question

Next Question