What is the % error in g?

Question

in an experiment to determine g, a student measures the length of the thread used in the pendulum as 0.49 m but forgets to add the radius of the bob which is of diameter 2x 10 ^ -2 m. he also commits an error of 0.5s in measuring the time of 75 s 50 oscillations. the maximum error in value of g that he will commit in this observation will be:

1) 3% 2) 2% 3) 3.3% 4) 3.33%

(pls provide required solution)

Answer 1

Pendulum period in seconds

T ≈ 2π√(L/G)

L is length of pendulum in meters

G is gravitational acceleration = 9.8 m/s²

Since G is the result, solve for G

L/G = (T/2π)²

G = L / (T/2π)²

First he has a length error of .02/(.49+.02) = 3.92%

Since G is proportional to L, the error directly relates.

"time of 75 s 50 oscillations" not sure what this means. I'll assume it as "time of 75 sec for 50 oscillations"

His time error is:

0.5/75 = 0.67%

But T is under the √, which reduces the error in half, so error is 0.33%

We know the length error will cause G to read low, but the time error is unknown in direction, so worse case, we have to add errors.

Adding the errors, 3.92% + 0.33% = 4.25%

which does not match any of the answers.

Perhaps you typed one of the numbers wrong? I double checked the numbers.