Question

time of 10 oscillations measured by stop clock of a simple pendulum of length of 0.1m is 20 seconds. least count of the stopwatch is 0.1 second. in calculation of g=4π^2×L/T^2 the percentage error of g will be?

Answer 1

Answer:

Delta g / g = 2 Delta T / T + Delta L / L

Explanation:

= 0.1 / 20 + 0.1 / 1

= 0.05 + 0.1

= 0.15 ×100

= 15%