Question

The period of oscillation of a simple pendulum is T = 2π√l/g . measured value of L is 20.0cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using wrist watch of 1s revolution. The accuracy in the determination of g is .

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Answer 1

Here, $\sf{T=2\pi\sqrt{\dfrac{L}{g}}}$

Squaring both sides, we get,

$\sf:\implies\:T^2=\dfrac{4\pi^2L}{g}$

$\sf:\implies\:g=\dfrac{4\pi^2L}{T^2}$

The relative error in g is,

$\sf:\implies\:\dfrac{\Delta g}{g}=\dfrac{\Delta L}{L}+2\dfrac{\Delta T}{T}$

Here, $\sf{T=\dfrac{t}{n}}$ and $\sf{\Delta T=\dfrac{\Delta t}{n}}$

Therefore, $\sf\dfrac{\Delta T}{T}=\dfrac{\Delta t}{t}$

The errors in both L and t are the least count errors.

$\sf:\implies\:\dfrac{\Delta g}{g}=\dfrac{0.1}{20}+2\bigg(\dfrac{1}{90}\bigg)$

$\sf:\implies\:\dfrac{\Delta g}{g}=0.005+0.022=0.027$

The percentage error in g is

$\sf:\implies\:\dfrac{\Delta g}{g}\times 100=0.027\times 100$

$:\implies\:\underline{\boxed{\bf{\red{\dfrac{\Delta g}{g}\times 100=2.7\%\approx3\%}}}}$

Answer 2

Answer:

Accuracy in the determination of g is approax 3%

Explanation:

Central idea :-Given, time period T = 2π√L/g this, changes can be expressed as

±2∆T/T = ± ∆L/L ± ∆g/g

According to the Question, we can write

∆L/L = 0.1cm / 20.0cm = 1/200

Again time period

• T = 90/100s

• and ∆T = 1/100s

• => ∆T/T = 1/90

• Now, T = 2π√L/g

• ∵ g = 4π²L/T²

• ∴ Δg/g = ΔL/L + 2ΔT/T

or, Δg/g × 100% = (ΔL/L) × 100% + (2 ΔT/T)×100%

• = (1/200×100)% + 2×1/90×100%

• = 2.72% ≈ 3%

Thus accuracy in the determination of g is approax 3% .