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 \(....\%\)
Pick an option to check your answer.
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 \(=36 \%\)
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