On the surface of earth, the time period of a simple pendulum is T_(e). When the pendulum is taken to a height of 2R above the surface of earth, the time period becomes T_(h). (R is radius of earth.) The value of (T_(e))/(T_(h)) is

On the surface of earth, the time period of a simple pendulum is T_(e)

1.	On the surface of earth, the time period of a simple pendulum is T_(e). When the pendulum is taken to a height of 2R above the surface of earth, the time period becomes T_(h). (R is radius of earth.) The value of (T_(e))/(T_(h)) is
Answer» <html><body><p>`1 : 9`<br/>`9 : 1`<br/>`1 : <a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>`<br/>`3 : 1`</p>Solution :As time <a href="https://interviewquestions.tuteehub.com/tag/period-1151023" style="font-weight:bold;" target="_blank" title="Click to know more about PERIOD">PERIOD</a> of a <a href="https://interviewquestions.tuteehub.com/tag/simple-1208262" style="font-weight:bold;" target="_blank" title="Click to know more about SIMPLE">SIMPLE</a> <a href="https://interviewquestions.tuteehub.com/tag/pendulum-1149901" style="font-weight:bold;" target="_blank" title="Click to know more about PENDULUM">PENDULUM</a> <br/> `T = 2pisqrt((l)/(g))` <br/> On earth.s surface `g_(e) = (GM)/(R^(2))` <br/> At a height `h = <a href="https://interviewquestions.tuteehub.com/tag/2r-300472" style="font-weight:bold;" target="_blank" title="Click to know more about 2R">2R</a>, g_(h) = (GM)/((R + 2R)^(2)) = (GM)/(9R^(2))` <br/> `(T_(e))/(T_(h)) = sqrt((g_(h))/(g_(e))) = sqrt((GM.R^(2))/(9R^(2).GM))` <br/> `(T_(e))/(T_(h)) = sqrt((1)/(9)) = 1:3`</body></html>