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  • A rod of mass & length <img alt="\mathrm{ I=R }" src="https://entrancecorner.oncodecogs.com/g

A rod of mass \mathrm{m } & length \mathrm{ I=R } is held very close to the earth so that it is aligned along the radial of earth. Its weight is equal to
 

Option: 1

\mathrm{mg}

 


Option: 2

\mathrm{mg} / 2
 


Option: 3

\mathrm{mg} / 3
 


Option: 4

None of these


Answers (1)

best_answer

The force experted by earth on the differential segment of mass \mathrm{dm} is equal to
\mathrm{ \mathrm{dF}=\frac{\mathrm{GM}(\delta \mathrm{m})}{\mathrm{r}^2} }

The force imparted on te total \mathrm{ \operatorname{rod}=\mathrm{F}=\int \mathrm{\mathrm{dF}} }

\mathrm{ \Rightarrow F = G M \int_R^{2 R} \frac{\delta m}{r^2} }

\mathrm{ =G M \int_R^{2 R}\left(M \cdot d r / r^2\right) }

\mathrm{ =\frac{G M m}{\ell} \int_R^{2 R} \frac{d r}{r^2} }

\mathrm{ =\frac{GMm}{R}\left ( \frac{1}{R}-\frac{1}{2R} \right )=\frac{GMm}{2R^{2}}=\frac{mg}{2} } \mathrm{ \text { where } g=\frac{G M}{R^2} . }

Hence option  2 is correct.


 

Posted by

Gautam harsolia

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