Question

The measurement value of length of a simple pendulum is 20 cm known with 2 mm accuracy. The time for 50 oscillations was measured to be 40 s within 1 s resolution. Calculate the percentage accuracy in the determination of acceleration due to gravity ‘g’ from the above measurement.

Answer 1

mark me brainliest please

Answer 2

Answer:

Explanation:

∵ 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 %