We can use the formula for the time period of a simple pendulum:
T = 2π√(L/g)
Taking the derivative of this formula with respect to 'g', we get:
(dT/dg) = -πL/(g^2√(L/g))
Now, we can calculate the relative error in g as follows:
(Δg/g) = (ΔT/T) / (dT/dg/g)
where ΔT/T is the relative error in T and Δg/g is the relative error in g.
We can calculate ΔT/T as follows:
ΔT/T = (Δt/t) + (ΔL/L)
where Δt is the error in time measurement and ΔL is the error in length measurement.
Δt = 0.01 s (the resolution of the stopwatch)
ΔL = 0.5 mm (half of the minimum division of the meter scale)
ΔT/T = (0.01/1.95) + (0.5/1000) = 0.0051282
Now, we can substitute the values in the expression for (Δg/g):
(Δg/g) = 0.0051282 / [-π(1.0)/(g^2√(1.0/g))] = -0.0051282g^2√g/π
Taking the absolute value and multiplying by 100 to get the percentage error, we get:
|(Δg/g)| × 100 = 0.0051282g^2√g/π × 100 = 1.03%
Therefore, the answer is option (2) 1.03%.
