A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He uses a stopwatch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statements is(are) true?

We know that, ΔT/T = Δt/t = 1/40
T = 2 sec/Oscillation
ΔT = 0.05 sec
The time period of a simple pendulum is given by T = 2π√(l/g), where l is the length of the pendulum and g is the acceleration due to gravity.
Squaring both sides, we get T² = 4π²(l/g).
Therefore, g = 4π²l/T².
Taking the logarithm of both sides, we get:
ln(g) = ln(4π²l) - 2ln(T)
Differentiating, we get:
Δg/g = 0 - 2(ΔT/T) = -2(ΔT/T)
Since ΔT/T = 1/40, we have Δg/g = -2(1/40) = -1/20 = -0.05
The percentage error in g is given by |Δg/g| × 100% = |-0.05| × 100% = 5%
Therefore, the percentage error in the determination of g is 5%.