Gravity at poles exceeds gravity at the equatore in the ration `201:200`. A pendulum regulated for poles is taken to the equator. Calculate how many seconds a day it willl gain or lose?

Explanation

Let g and `g^(')` be the values of acceleration due to gravity at poles and equatore respectively.
Then `(g)/(g^('))=(201)/(200)`
Let l be the length of pendulum and T and `T^(')` be the periods of the pendulum at the poles and equator respectively. Then
`T=2pisqrt((l)/(g)) and T^(')=2pisqrt((l)/(g^('))`
`:. (T^('))/(T)=2pisqrt((l)/(g^(')))xx(1)/(2pi)sqrt((g)/(l))`
`=sqrt((g)/(g^(')))=sqrt((201)/(200))=(1+(1)/(200))^(1/2)`
`=(1+(1)/(2)xx(1)/(200)+...)=1.0025`
`T^(')=1.0025T`
`:. (T^(')-T)/(T)=(1.0025T-T)/(T)=0.0025`
Thus increase in time period per second at the equatore is 0.0025 s, i.e., at the equatore, the penulum will lose 0.0025s per second. Therefore loss in time per day `=0.0025 xx 24xx60xx60`
`=216s`