Question

In a simple pendulum experiment, length is
measured as 31.4 cm with an accuracy of
1mm. The time for 100 oscillations of
pendulum is 112.0s with an accuracy of 0.01s.
The percentage accuracy in g is
11 223 ) 1.3 4) 2.3.

Answer 1

Hey Dear,

◆ Answer -

∆g/g % = 2.1 %

● Explaination -

# Given -

l = 31.4 cm

∆l = 0.1 cm

∆T = 0.01 s

# Solution -

Mean time period of pendulum is -

T = 112/100

T = 1.12 s

Time period of pendulum is given by -

T = 2π√(l/g)

T² = 4π²l/g

g = 4π²l/T²

Relative error in g is thus -

∆g/g = ∆l/l + 2∆T/T

∆g/g = 0.1/31.4 + 2×0.01/1.12

∆g/g = 0.021

Percentage error in measurement of g is -

∆g/g % = 0.021 × 100 %

∆g/g % = 2.1 %

Therefore, percentage error in measurement of g is 2.1 %.

Thanks dear...