The percentage error in the measurement of g is : (Given that g= $\frac{4\pi^2L}{T^2}$, L= 10$\pm$0.1 cm, T = 100$\pm$1 s)
Answer & explanation
Correct answer: option 4
Given g= $\frac{4\pi^2L}{T^2}$
Fractional error in value of g = $\frac{\Delta g}{g}\times$100
=$\frac{\Delta L}{L}\times$100+$\frac{2\Delta T}{T}\times$100=$\frac{0.1}{10}\times$100+$\frac{2}{100}\times$100 = 3%