The time period of oscillation of simple pendulum is given by $$T=2$$$$π$$lg where l is $$100$$ cm and is known to have $$0.1$$ cm accuracy. The time period is $$2s$$. The time (t) of $$100$$ oscillations is measured by a stop watch of least count $$0.1s$$. The maximum percentage error in measurement of g is

Question

The time period of oscillation of simple pendulum is given by $$T=2$$$$π$$lg where  l is $$100$$ cm  and is known to have  $$0.1$$ cm accuracy. The time period is  $$2s$$. The time (t) of $$100$$ oscillations is measured by a stop watch of least count $$0.1s$$. The maximum percentage error in measurement of g is

Infinity Learn · Accepted Answer

GivenThe time period of simple pendulum $$T=2$$$$π$$lg⇒$$g=4$$$$π$$$$2lT2$$⇒Δ$$gg=$$Δ$$ll+2$$Δ$$TTt=100T=200s$$⇒Δ$$tt=$$ΔTT⇒Δ$$gg=$$Δ$$ll+2$$⋅Δtt⇒Δ$$gg=0.1100+20.11002$$⇒Δgg×$$100=0.2$$%

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The time period of oscillation of simple pendulum is given by T=2πlg where l is 100 cm and is known to have 0.1 cm accuracy. The time period is 2s. The time (t) of 100 oscillations is measured by a stop watch of least count 0.1s. The maximum percentage error in measurement of g is

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