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Gravitation question

2009 · Shift 0 · Q73
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Gravitation question

2009 · Shift 0 · Q73

JEE MainPhysicsGravitationMCQ+4 / −1
The height at which the acceleration due to gravity becomes g9{g \over 9}9g​(where g=g=g= the acceleration due to gravity on the surface of the earth) in terms of R,R,R, the radius of the earth, is:
  1. A
    R2{R \over {\sqrt 2 }}2​R​
  2. B
    R/2R/2R/2
  3. C
    2  R\sqrt 2 \,\,R2​R
  4. D
    2 R2\,R2R
View written solutionFree

Correct answer: D

  1. Acceleration due to gravity at height hhh

    At a height hhh above the earth's surface, the acceleration due to gravity is gh=g(RR+h)2g_h = g\left(\frac{R}{R+h}\right)^2gh​=g(R+hR​)2 where RRR is the radius of the earth.

  2. Given condition

    We are told that gh=g9g_h = \frac{g}{9}gh​=9g​

    Substituting into the formula: g(RR+h)2=g9g\left(\frac{R}{R+h}\right)^2 = \frac{g}{9}g(R+hR​)2=9g​

  3. Cancel ggg from both sides

    (RR+h)2=19\left(\frac{R}{R+h}\right)^2 = \frac{1}{9}(R+hR​)2=91​

  4. Take square root

    RR+h=13\frac{R}{R+h} = \frac{1}{3}R+hR​=31​

  5. Solve for hhh

    3R=R+h3R = R + h3R=R+h h=2Rh = 2Rh=2R

  6. Match with options

    The required height is 2R\boxed{2R}2R​

    So the correct option is D.

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