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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2022
  • Medium
  • Numerical

An object is taken to a height above the surface of earth at a distance \(\frac{5}{4} R\) from the centre of the earth. Where radius of earth, \(R=6400\,km\). The percentage decrease in the weight of the object will be \(....\%\)

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Answer & Solution

Correct Answer

(A) \(36\)

Step-by-step Solution

Detailed explanation

\(g _{\text {eff }}=\frac{ g }{\left(1+\frac{ h }{ R }\right)^{2}} ; g _{ eff }=\frac{ g }{\left(1+\frac{1}{4}\right)^{2}}=\frac{16\,g }{25}\)

\(\text { change }=\frac{g_{\text {off }}-g}{g} \times 100=\frac{\frac{16}{25}-1}{1} \times 100\)

\(=\frac{-9}{25} \times 100=-36 \%\)

Hence \(\%\) decrease in the weight \(=36 \%\)