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Gravitation question

2021 · 31 Aug · Shift 2 · Q62
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  5. /2021 · 31 Aug · Shift 2 · Q62

Gravitation question

2021 · 31 Aug · Shift 2 · Q62

JEE MainPhysicsGravitationMCQ+4 / −1
If RE be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is : (Given : r < RE)
  1. A
    1−rRE−r2RE2−r3RE31 - {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} - {{{r^3}} \over {R_E^3}}1−RE​r​−RE2​r2​−RE3​r3​
  2. B
    1+rRE+r2RE2+r3RE31 + {r \over {{R_E}}} + {{{r^2}} \over {R_E^2}} + {{{r^3}} \over {R_E^3}}1+RE​r​+RE2​r2​+RE3​r3​
  3. C
    1+rRE−r2RE2+r3RE31 + {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} + {{{r^3}} \over {R_E^3}}1+RE​r​−RE2​r2​+RE3​r3​
  4. D
    1+rRE−r2RE2−r3RE31 + {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} - {{{r^3}} \over {R_E^3}}1+RE​r​−RE2​r2​−RE3​r3​
View written solutionFree

Correct answer: D

  1. Acceleration due to gravity at depth rrr below the surface

If the depth below Earth’s surface is rrr, then distance from Earth’s center is RE−rR_E-rRE​−r For a uniformly dense Earth, gravity inside varies linearly with distance from center: gd=g(1−rRE)g_d = g\left(1-\frac{r}{R_E}\right)gd​=g(1−RE​r​)

  1. Acceleration due to gravity at height rrr above the surface

If the height above Earth’s surface is rrr, then distance from Earth’s center is RE+rR_E+rRE​+r So, gh=g(RERE+r)2g_h = g\left(\frac{R_E}{R_E+r}\right)^2gh​=g(RE​+rRE​​)2

Thus the required ratio is gdgh=g(1−rRE)g(RERE+r)2\frac{g_d}{g_h} = \frac{g\left(1-\frac{r}{R_E}\right)}{g\left(\frac{R_E}{R_E+r}\right)^2}gh​gd​​=g(RE​+rRE​​)2g(1−RE​r​)​

Cancelling ggg: gdgh=(1−rRE)(1+rRE)2\frac{g_d}{g_h} = \left(1-\frac{r}{R_E}\right)\left(1+\frac{r}{R_E}\right)^2gh​gd​​=(1−RE​r​)(1+RE​r​)2

Let x=rREx=\frac{r}{R_E}x=RE​r​ Then, gdgh=(1−x)(1+x)2\frac{g_d}{g_h}=(1-x)(1+x)^2gh​gd​​=(1−x)(1+x)2

  1. Expand the expression

First, (1+x)2=1+2x+x2(1+x)^2=1+2x+x^2(1+x)2=1+2x+x2

Now multiply: (1−x)(1+2x+x2)=1+2x+x2−x−2x2−x3(1-x)(1+2x+x^2) = 1+2x+x^2 - x - 2x^2 - x^3(1−x)(1+2x+x2)=1+2x+x2−x−2x2−x3 =1+x−x2−x3= 1+x-x^2-x^3=1+x−x2−x3

Replacing x=rREx=\frac{r}{R_E}x=RE​r​, gdgh=1+rRE−r2RE2−r3RE3\frac{g_d}{g_h}=1+\frac{r}{R_E}-\frac{r^2}{R_E^2}-\frac{r^3}{R_E^3}gh​gd​​=1+RE​r​−RE2​r2​−RE3​r3​

  1. Match with options

This matches Option D: 1+rRE−r2RE2−r3RE31 + \frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3}1+RE​r​−RE2​r2​−RE3​r3​

Therefore, the correct answer is D.

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