A clock regulated by u seconds pendulum keeps correct time. During summer the length of the pendulum increases to 1.01 m How much will the clock gain or lose in one day ?

A clock regulated by u seconds pendulum keeps correct time. During sum

1.	A clock regulated by u seconds pendulum keeps correct time. During summer the length of the pendulum increases to 1.01 m How much will the clock gain or lose in one day ?
Answer» Solution :The time PERIOD of a simple pendulum is `T= 2pi sqrt((l)/(g))= (2pi)/(sqrt(g))(l^(1//2))` differentiating T w.r.t. l and dividing T on both slides we GET `(dT)/(T)= (1)/(2)(dl)/(l)` In CASE of a seconds pendulum , T= 2s and `l= (8)/(pi^(2))= 0.9927` Also dl= 1.01-0.9927= 0.0173m. `:. (dT)/(2)= (1)/(2)((0.0173)/(0.9927))` i.e the change in time period or change in time for oscillation , `dT= (0.0173)/(0.9927)s` `dT= (1)/(2)(dl)/(l)T=(1)/(2)(dl)/(l)86,400` per day `= (1)/(2) (0.0173)/(0.9927) xx 86400= 752.9s` When the lenght INCREASE the time period of the pendulum increase. So, the clock loses time or runs slow. (Note: when the lenght decrease the time period of the pendulum decrease and there will be gain of time or the clock FAST.)