Question

In an experiment to determine acceleration due to gravity, the length of the pendulum is measured as 98 cm be a scale of least count of 1 cm. The period of swing/oscillations is measured with the help of a stop watch having a least count of 1s. The time period of 50 oscillation is found to be 98 s. Express value of g with proper error limits.

Answer 1

Answer:

(10.1±0.3)

Explanation:

T = 2π √l/g

T = 98/50 = 1.96

T² = 4π² × l/g

g= 4× 3.14² × 0.98/1.96²

g=10.06 m/s²

error in g = 1/98+ 1/98×2

= 3/98

delta g = 10* 3/98

= 0.3 m/s²

g=(10.1±0.6) m/s²