The period of oscillation of a simple pendulum is T = 2 π l g .   Measured value of l is 20 . 0…

The period of oscillation of a simple pendulum is T=2πlg.  Measured value of l is 20.0 cm, known to 1 mm  accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wristwatch of 1 s resolution. The accuracy in the determination of g is 
  1. 5%
  2. 4%
  3. 3%
  4. 1%

Solution

∴T=2πlg⇒g=4π2lT2

∴ Error in g can be calculated as

Δgg=Δll+2ΔTT. 

∴ Total time for n oscillation is t=nT where T= time for oscillation.

⇒Δtt=ΔTT

⇒Δgg=Δll+2Δtt

Given that Δl=1 mm=10-3 m, l=20×10-2 m

Δt=1 s, t=90 s. 

% error in g is

Δgg×100=Δll+2Δtt×100

=10-320×10-2+2×190×100=12+209=0.5+2.22=2.72%

≅3%

Asked in: JEE Main 2015 (04 Apr)

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