Question

In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for $$20$$ oscillations is measured by using a watch of $$1$$ second least count. The mean value of time taken comes out to be $$30 s$$. The length of pendulum is measured by using a meter scale of least count $$1 mm$$ and the value obtained is $$55.0 cm$$. The percentage error in the determination of g is close to :-

Answer 1

Correct option is D. $$6.8 \%$$
$$T = \dfrac{30 sec}{20} $$ $$\Delta T = \dfrac{1}{20} sec$$
$$L = 55 cm$$ $$\Delta L = 1 \, mm = 0.1 \, cm$$
$$g = \dfrac{4 \pi^2 L}{T^2}$$
percentage error in g is
$$\dfrac{\Delta g}{g} \times 100 \% = \left(\dfrac{\Delta L}{L} + \dfrac{2 \Delta T}{T} \right) 100\%$$
$$= \left(\dfrac{0.1}{55} + \dfrac{2 \left(\dfrac{1}{20} \right)}{\dfrac{30}{20}} \right) 100 \% = 6.8 \%$$