Question

If length of a pendulum of wall clock is increased by 0.1% then find error in time per day. Could someone answer this??

Answer 1

Let initially , Length of pendulum is L
Then, time period ,$\bold{T=\sqrt{\frac{L}{g}}}$ = 1s { Let }
time in a day , t = 1 × 60 × 60 × 24 s

After increasing length 0.1% ,
Length will be L'=L + 0.001L= 1.001L
Now, new time period
$\bold{T'=\sqrt{\frac{L'}{g}}}=\bold{\sqrt{\frac{1.001L}{g}}}=\bold{\sqrt{1.001}\sqrt{\frac{L}{g}}}\\\\\bold{T'=\sqrt{1.001}s}$
New time in a day ,t' = √{1.001} × 60 × 60 × 24s

Error in a day = t' - t
= √{1.001} × 60 × 60 × 24 - 1 × 60 × 60 × 24
= 60 × 60 × 24( √{1.001} - 1)
= 60 × 60 × 24 ( 1.0005 - 1)
= 60 × 60 × 24 × 0.0005
= 43.2 s

Answer 2

T = √l/g

∴ ΔT/T ×100 =1/2.Δl/l

∴Δt/t×100=1/2×0.1=0.05%

error = 0.05

so for one day , 43.20 sec

3 MARKs