Question

In an experiment of simple pendulum, the timemeasured was 50 s for 25 vibrations, when thelength of the simple pendulum was taken 100 cm.if the least count of the stopwatch is 0.2s and that of meter scale is 0.1cm then find the maximum percentage error in the measurement of g​

Answer 1

Given: The time measured is 50 s for 25 vibrations, the length of the simple pendulum was taken 100 cm.

To find: The maximum percentage error in the measurement of g​?

Solution:

• Now we know that T = 2π x √ (l/g)

            g = 2πl / T²

           Δg/g x 100 = Δl/l x 100 + 2ΔT/T x 100

• We have given that the least count of the stopwatch is 0.2 s and that of meter scale is 0.1 cm.

• So putting values in formula, we get:

           Δg/g x 100 = 0.1/100 x 100 + 2 x 0.2/50 x 100

           Δg/g = 0.1 + (2 x 0.2 x 2)

           Δg/g = 0.1 + 0.8

           Δg/g = 0.9

Answer:

           So the maximum percentage error in the measurement of g​ is  0.9%