Question

The acceleration due to gravity(g) is measured on earth’s surface by using a simple pendulum. If time period of the pendulum T = (4.00 ± 0.01) s and length of pendulum l = (4.00 ± 0.01) m were measured, then value of g upto correct significant figures, is (use 2 = 10.0) Choose answer: (9.8 ± 0.2) m/s2 (10.0 ± 0.1) m/s2 (9.8 ± 0.8) m/s2 (10.0 ± 0.8) m/s2

Answer 1

answer : (10 ± 0.8) m/s²

explanation : given, time period T = (4.00 ± 0.01)s

length of pendulum , l = (4.00 ± 0.01)m

using formula,

T = 2π√{l/g}

squaring both sides,

T² = 4π²l/g

⇒g = 4π²l/T²

here, l = 4.00, T = 4.00 and π² = 10.0 [given]

so, g = 4 × 10 × 4/4² = 10 m/s²

now find error in g,

∆g/g = ∆l/l + 2 × ∆T/T

⇒∆g/10 = 0.01/4 + 2 × 0.01/4

⇒∆g = 10(0.01/4 + 0.02/4)

⇒ ∆g = 0.75 ≈ 0.8

hence , g = (10.0 ± 0.8) m/s²

Answer 2

$\huge\bold\purple{Answer:-}$

(10.0 ± 0.8) m/s2