How to rectify "time lost " by pendulum clock by alteration in its length? question:
If a clock loses $5$ seconds per day ,what is the alteration required in the length 
of pendulum in order that the clock keeps correct time 
$(a)\dfrac{4}{86400} $times its original length be shortened
$(b)\dfrac{1}{86400}$ times its original length be shortened
$(c)\dfrac{1}{8640}$ times its original length be shortened
$(d)\dfrac{4}{8640} $times its original length be shortened
my attempt:
there are $86400$ seconds in a day 
but clock is slow so it only counts $86395$ seconds 
so, the factor by which clock is slow is $\dfrac{86395}{86400}=0.99994212$
so, new pendulum's length should be $\left(\dfrac{86395}{86400}\right)^2\times$(original length)
but none of the options are matching my answer 
please tell me right approach to solve this problem 
thank you!

I think none of the answers are right.  Based on this description of pendulum motion, the period $T$ is related to length $L$ as $$T = 2\pi \sqrt{\frac{L}{g}}$$ Let $L_s$ be the length of the currently slow pendulum and $L$ the desired length of the pendulum.  Since you want the difference as a ratio to apply to $L_s$, that is ${L}/{L_s}$, then set this up as a ratio on $T$: $$\frac{\sqrt{L}}{\sqrt{L_s}} =  \frac{T}{T_s}$$Note that the $2\pi$ and $g$ terms divide out.  Then,$$\\ \frac{{L}}{{L_s}} = \bigg(\frac{T}{T_s}\bigg)^2 = \bigg(\frac{86400}{86400+5}\bigg)^2$$I don't think any of those choices reduces to this.
