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A second's clock with a iron pendulum is constructed so a to keep correct time at 10^(@)C. Give alpha_("iron")=12xx10^(-6) per""^(@)C. What will be the change in the rate when the temperature rises to 25^(@)C? |
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Answer» Solution :When the pendulum keeps correct TIME, its period of vibration is 2 sec and so it makes `(24xx60xx60)/2=43200` Vibratiions /day If lenth of the pendulum at `10^(@)C` is `l_(10)` and at `25^(@)C` is `l_(25)` `:.l_(25)=l_(10)[1+alpha(25-10)]=l_(10)[1+15alpha]` as `T=2pi sqrt(L//g)` i.e. `T prop sqrt(l)` i.e. `n prop 1/(sqrt(l))`, n is no of vibrations per sec `:.(n_(25))/(n_(10))=sqrt((l_(10))/(l_(24)))=[1+15alpha]^(-1//2)~~1=15/2alpha` `:.n_(25)=n_(10)(1-15/2xx12xx10^(-6))` `=43200[1-0.00009]=43119.2` That is the clock makes (3200-43196.12=3.88 vibratiion loss per day. That is clock losses `3.88xx2=7.76` sec per day). |
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