Question

If the length of second's pendulum is increased by 2% ,how many seconds it will lose per day.

• A. 3927s
• B. 3727s
• C. 3427s
• D. 864s

Answer 1

The correct option is D 864s

Time period of seconds pendulum=2s

We have $T\propto \sqrt{l}\therefore \frac{T^1}{T}=\sqrt{\frac{l^1}{l}}$

Given that $l^1=1.02l\therefore \frac{l^1}{l}=1.02\Rightarrow \frac{T^1}{T}=\sqrt{1.02}\approx 1.01\Rightarrow T^1=2.02s$

ie, it will complete one oscillation in 2.02s but shows as 2 s.

Therefore, clock loses 0.02s for every 2s or otherwise, 0.01s for every second.

$\therefore$ in one day it will lose$=24\times 60\times 60\times 0.01s=864s$