Question

The measurement value of length of a simple pendulum is 20cn know with 2mm accuracy the time for 50 oscillations was measured to be 40s within 1s resolution calculate the percentage accuracy in the determination of acceleration due to gravity 'g' from the above measurements.

Answer 1

Answer:

Ello✌️

gud mrng

∵ time period is time taken to complete a revolution .

e.g., T = t/n , here n is number of revolution and t is time.

so, ∆T = ∆t/n, because n is Constant term.

Hence, ∆T/T = ∆t/n × n/t = ∆t/t -----(1)

Now, T = 2π√(l/g)

Taking square both sides,

T² = 4π²l/g ⇒ g = 4π²l/T²

∴ ∆g/g = ∆l/l + 2∆T/T

⇒ ∆g/g = ∆l/l + 2∆t/t [ from equation (1) ]

∴ % error in g = 100( ∆l/l + 2∆t/t)

Here, ∆l = 1mm = 0.1 cm , l = 20cm

∆t = 1 s and t = 90 s

∴ % error in g = 100(0.1/20 + 2 × 1/90)

= 100( 0.005 + 0.022)

= 100( 0.027)

= 2.7 %

Hence, % error in g = 2.7 %

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