Verified Solution • 2025–26 Session moderate

QJEEM20Pha3b8: JEE Main 2020 Physics - Class 11 JEEM20Pha3b8 Solution: The height ‘h’ at which the weight of a body will be the same…

The height ‘h’ at which the weight of a body will be the same as that at the same depth ‘h’ from the surface of the earth is (Radius of the earth is R and effect of the rotation of the earth is neglected)

Multiple Choice Options

  • A R2{R \over 2}
  • B 5R−R2{{\sqrt 5 R - R} \over 2}
  • C 3R−R2{{\sqrt 3 R - R} \over 2}
  • D 52R−R{{\sqrt 5 } \over 2}R - R

Step-by-Step Verified Solution

JEE Main 2020 (Online) 2nd September Evening Slot Physics - Gravitation Question 139 English Explanation

M = mass of earth

M1 = mass of shaded portion

Re = Radius of earth

M1 = M43πR3.43π(R−h)3{M \over {{4 \over 3}\pi {R^3}}}.{4 \over 3}\pi {\left( {R - h} \right)^3}

= M(R−h)3R3{{M{{\left( {R - h} \right)}^3}} \over R^3}

Given, Weight of body is same at P and Q

∴\therefore mgP = mgQ

⇒\Rightarrow gP = gQ

⇒\Rightarrow GM1(R−h)2=GM(R+h)2{{G{M_1}} \over {{{\left( {R - h} \right)}^2}}} = {{GM} \over {{{\left( {R + h} \right)}^2}}}

⇒\Rightarrow GM(R−h)3R3(R−h)2=GM(R+h)2{{GM{{\left( {R - h} \right)}^3}} \over {{R^3}{{\left( {R - h} \right)}^2}}} = {{GM} \over {{{\left( {R + h} \right)}^2}}}

⇒\Rightarrow (R – h) (R + h)2 = R3

⇒\Rightarrow R3 – hR2 + h2R – h3 + 2R2h – 2Rh2 = R3

⇒\Rightarrow R2h – Rh2 – h3 = 0

⇒\Rightarrow R2 – Rh – h2 = 0

⇒\Rightarrow h2 + Rh – R2 = 0

⇒\Rightarrow h = −R±R2+4R22{{ - R \pm \sqrt {{R^2} + 4{R^2}} } \over 2}

⇒\Rightarrow h = −R+5R2{{ - R + \sqrt 5 R} \over 2}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring