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Gravitation question

2020 · 5 Sep · Shift 1 · Q60
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  5. /2020 · 5 Sep · Shift 1 · Q60

Gravitation question

2020 · 5 Sep · Shift 1 · Q60

JEE MainPhysicsGravitationMCQ+4 / −1
The value of the acceleration due to gravity is g1 at a height h = R2{R \over 2}2R​(R = radius of the earth) from the surface of the earth. It is again equal to g1 at a depth d below the surface of the earth. The ratio (dR)\left( {{d \over R}} \right)(Rd​) equals :
  1. A
    59{5 \over 9}95​
  2. B
    19{1 \over 9}91​
  3. C
    79{7 \over 9}97​
  4. D
    49{4 \over 9}94​
View written solutionFree

Correct answer: A

  1. Acceleration due to gravity at height

At a height hhh above the earth's surface,

gh=g(RR+h)2g_h = g\left(\frac{R}{R+h}\right)^2gh​=g(R+hR​)2

Given h=R2h = \frac{R}{2}h=2R​,

g1=g(RR+R/2)2=g(R3R/2)2=g(23)2=4g9g_1 = g\left(\frac{R}{R+R/2}\right)^2 = g\left(\frac{R}{3R/2}\right)^2 = g\left(\frac{2}{3}\right)^2 = \frac{4g}{9}g1​=g(R+R/2R​)2=g(3R/2R​)2=g(32​)2=94g​

So,

g1=4g9g_1 = \frac{4g}{9}g1​=94g​
  1. Acceleration due to gravity at depth

At a depth ddd below the earth's surface, assuming uniform density,

gd=g(1−dR)g_d = g\left(1-\frac{d}{R}\right)gd​=g(1−Rd​)

It is given that this is again equal to g1g_1g1​, so

g(1−dR)=4g9g\left(1-\frac{d}{R}\right) = \frac{4g}{9}g(1−Rd​)=94g​

Cancel ggg:

1−dR=491-\frac{d}{R} = \frac{4}{9}1−Rd​=94​

Hence,

dR=1−49=59\frac{d}{R} = 1-\frac{4}{9} = \frac{5}{9}Rd​=1−94​=95​
  1. Match with options
dR=59\frac{d}{R} = \frac{5}{9}Rd​=95​

So the correct option is A.

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