Question

determine the maximum % error in the determination of acceleration due to gravity 'g' using the following observations

length of the thread of the pendulum = 100.1 cm , radius of the bob = 2.52 cm and time period of oscillation = 2.1 sec . The length of the thread is measured with a meter scale of least count 1 mm ,radius of the bob is measured with a vernier callipers of least count 0.01 cm and time period is measured with a stop watch of least count 0.1 sec

Answer 1

we know that

T = 2π √(L/g)

or by rearranging we get

g = 4 Ï€² L / T²

now, here

L = 100 cm = 1.001m

T = 2.1s

g= 4 xπ^{2 x} 1.001/4.41

g= 8.95m/s²

Now percentage error

Δg/g = ΔL/L + 2(ΔT/T)

or

Δg = g.[ΔL/L + 2(ΔT/T)]

where g = 9.8

ΔL = 1mm =0.001m

ΔT= 0.1 sec

Δg = 8.95 [ 0.001/1.001 + 2 x 0.1/2.1]

Δg = 0.853m/s²