How to rectify "time lost " by pendulum clock by alteration in its length?

Question

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!

Answer 1

$$1-\left(\dfrac{86395}{86400}\right)^2 = 1-\left(1-\dfrac{5}{86400}\right)^2 = 1-\left(1-\dfrac{10}{86400}+\dfrac{5^2}{86400^2}\right) = \dfrac{1}{8640}-\dfrac{1}{298598400}$$ which suggests to me that you might be expected to give answer $(b)$

Answer 2

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.