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If the earth were a homogeneous sphere of radius R and a straight hole bored in it through its centre, show that a particle dropped into the hole will execute shm and find out its period.

Answer» <html><body><p></p>Solution :Acceleration due to gravity at the surface of the earth <br/> `g= (GM )/R^2` <br/> M-Mass of the earth and R-radius<br/> G-Gravitational constant <br/>If `rho`is the <a href="https://interviewquestions.tuteehub.com/tag/density-17451" style="font-weight:bold;" target="_blank" title="Click to know more about DENSITY">DENSITY</a> of earth then<br/>`M = 4/3piR^2rho` <br/> `:. g = 4/3 piRrhoG ""...(1)` <br/> At a depth h below the surface of the earth <br/> `g.=4/3 pi(R-h)rhoG""...(2)` <br/>Dividing eq. (2) by eq. (1) <br/> `(g.)/g=(4/3pi(R-h)rhoG)/(4/3piRrhoG)` <br/> `g.=g""((R-h))/R`<br/> Let <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> be the distance of the particle from the centre of the earth. Then R-h = x<br/> Then acceleration `=(d^2x)/(dt^2) =g.` <br/> i.e.`(d^2x)/(dt^2) = -(gx)/R` <br/> Negative sign is put to show that acceleration acts towards the centre of the earth which is opposite to the direction of increasing x. <br/> `(d^2x)/(dt^2)+g/Rx=0` <br/> `omega^2=g/R,omega=sqrt(g/R)` <br/> Hence the particle <a href="https://interviewquestions.tuteehub.com/tag/dropped-2594665" style="font-weight:bold;" target="_blank" title="Click to know more about DROPPED">DROPPED</a> into the hole execute <a href="https://interviewquestions.tuteehub.com/tag/shm-630759" style="font-weight:bold;" target="_blank" title="Click to know more about SHM">SHM</a> <br/> Priod `T=(2pi)/omega=2pisqrt(R/g)`</body></html>


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