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Gravitation question

2024 · 29 Jan · Shift 1 · Q66
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  5. /2024 · 29 Jan · Shift 1 · Q66

Gravitation question

2024 · 29 Jan · Shift 1 · Q66

JEE MainPhysicsGravitationMCQ+4 / −1
At what distance above and below the surface of the earth a body will have same weight. (take radius of earth as RRR.)
  1. A
    3R−R2\frac{\sqrt{3} R-R}{2}23​R−R​
  2. B
    R2\frac{R}{2}2R​
  3. C
    5R−R2\frac{\sqrt{5} R-R}{2}25​R−R​
  4. D
    5R−R\sqrt{5} R-R5​R−R
View written solutionFree

Correct answer: C

  1. Weight above the surface

At a height hhh above the Earth's surface, the distance from the Earth's center is R+hR+hR+h.

So the gravitational acceleration there is

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

Hence the weight of a body of mass mmm is

Wabove=mg(RR+h)2W_{\text{above}} = m g\left(\frac{R}{R+h}\right)^2Wabove​=mg(R+hR​)2
  1. Weight below the surface

At a depth ddd below the Earth's surface, the gravitational acceleration is

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

Hence the weight is

Wbelow=mg(1−dR)W_{\text{below}} = m g\left(1-\frac{d}{R}\right)Wbelow​=mg(1−Rd​)
  1. Condition for same weight above and below

We need

Wabove=WbelowW_{\text{above}} = W_{\text{below}}Wabove​=Wbelow​

So,

mg(RR+h)2=mg(1−dR)m g\left(\frac{R}{R+h}\right)^2 = m g\left(1-\frac{d}{R}\right)mg(R+hR​)2=mg(1−Rd​)

Cancelling mgmgmg,

(RR+h)2=1−dR\left(\frac{R}{R+h}\right)^2 = 1-\frac{d}{R}(R+hR​)2=1−Rd​
  1. Interpretation of the question

The wording asks: At what distance above and below the surface of the earth a body will have same weight.

This means the distances above and below are equal. Let that common distance be xxx. Then

h=d=xh=d=xh=d=x

So,

(RR+x)2=1−xR\left(\frac{R}{R+x}\right)^2 = 1-\frac{x}{R}(R+xR​)2=1−Rx​
  1. Solve the equation

Let

y=xRy=\frac{x}{R}y=Rx​

Then

1(1+y)2=1−y\frac{1}{(1+y)^2}=1-y(1+y)21​=1−y

Multiply both sides by (1+y)2(1+y)^2(1+y)2:

1=(1−y)(1+y)21=(1-y)(1+y)^21=(1−y)(1+y)2

Expand:

(1−y)(1+2y+y2)=1(1-y)(1+2y+y^2)=1(1−y)(1+2y+y2)=1 1+2y+y2−y−2y2−y3=11+2y+y^2-y-2y^2-y^3=11+2y+y2−y−2y2−y3=1 1+y−y2−y3=11+y-y^2-y^3=11+y−y2−y3=1

So,

y−y2−y3=0y-y^2-y^3=0y−y2−y3=0 y(1−y−y2)=0y(1-y-y^2)=0y(1−y−y2)=0

Ignoring the trivial solution y=0y=0y=0, we get

1−y−y2=01-y-y^2=01−y−y2=0 y2+y−1=0y^2+y-1=0y2+y−1=0

Thus,

y=−1+52y=\frac{-1+\sqrt{5}}{2}y=2−1+5​​

(positive root only)

Therefore,

x=Ry=R⋅5−12x=Ry=R\cdot \frac{\sqrt{5}-1}{2}x=Ry=R⋅25​−1​ x=5R−R2x=\frac{\sqrt{5}R-R}{2}x=25​R−R​
  1. Match with options
x=5R−R2\boxed{x=\frac{\sqrt{5}R-R}{2}}x=25​R−R​​

This corresponds to Option C.

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