A pendulum clock runs 20s slow per day. What should be the change in length of the clock so that it records correct time? Take the pendulum as a simple pendulum.

Explanation

Half time period of a simple pendulum
`t=T/2=pisqrt(L/g)`
For a perfect seconds pendulum t = 1 s
`therefore" "1=pisqrt(L/g)or,L=g/pi^(2)" "...(1)`
Also, `1d=24xx60xx60=86400s`. Number of half oscillations executed in a day by a pendulum which runs 20 s slow per day = 86400 - 20 = 86380.
Hence, half time period of that pendulum `t_(1)=86400/86380s`.
If the length of this simple pendulum is `L_(1)`.
then `t_(1)=pisqrt(L_(1)/g)or,L_(1)=(g t_(1)^(2))/pi^(2)" "...(2)`
Subtracting equation (1) from equation (2),
`L_(1)-L=g/pi^(2)(t_(1)^(2)-1)=980/pi^(2)[(86400/86380)^(2)-1]`
= 0.046 cm = 0.46 mm
Hence, to get correct time, length of the pendulum is to be decreased by 0.46 mm.
Alternative Method :
Half time period of a perfect seconds pendulum = 1 s.
Number of half oscillations of a clock that runs slow by `t_(0)`s in a day = `86400-t_(0)`.
Hence, half time period = `86400/(86400-t_(0))s`
`therefore` Increase in the value of half time period,
`dt=86400/(86400-t_(0))-1=t_(0)/(86400-t_(0))s`
If the value of `t_(0)` is negligibly smaller than 86400,
`dt=t_(0)/86400s`
Now, `t=pisqrt(L/g)or,logt=logpi+1/2logL-1/2logg`
Differentiating, `(dt)/t=1/2(dL)/L-1/2(dg)/g" "...(1)`
Here, dL = increase in length and dg = increase in acceleration due to gravity.
Putting values of t and dt in (1),
`t_(0)/86400=1/2((dL)/L-(dg)/g)`
or, `t_(0)=43200((dL)/L-(dg)/g)" "...(2)`
This equation can be used as a rule for a seconds pendulum. For a seconds pendulum `L=g/pi^(2)` and on the surface of the earth `g=980cm*s^(-2)`. If the clock runs fast, value of `t_(0)` is negative.
In the given problem, `t_(0)=20s` ,
If there is no change in the value of g, dg = 0
`therefore" "t_(0)=43200xx(dL)/L`
or, `dL=(20xxL)/43200=20/43200xx980/pi^(2)=0.046cm=0.46mm`
Hence, length of the defective clock has increased by 0.46mm. Thus to get correct time, its length needs to be decreased by 0.46 mm.