Question

In an experiment of simple pendulum time period measured was 50 sec for 25 vibration when the length of the simple pendulum was taken 100 cm. If the least count of stop watch is 0.1 s and that of meter scale is 0.1 cm. Calculate the maximum percentage error in the measurement of value of g.

• A.
• B.
• C.
• D. none of these

Answer 1

The correct option is C

The time period of a simple pendulum is given by T = $2\pi \sqrt{\frac{L}{g}}$ or $T^2=\frac{4{\pi }^{2}l}{g}$ or $g=\frac{4{\pi }^{2}l}{T^2}$ As 4 and $\pi$ are constant, maximum permissible error in g is given by Now $\frac{△g}{g}=\frac{△l}{l}+\frac{2△T}{T}$ Here $△l=0.1$ cm, L = 1m = 100 cm, $△$T = 0.1s, T = 50s.

$\therefore \frac{△g}{g}=\frac{0.1}{100}+2\left(\frac{0.1}{50}\right)$ = $\frac{0.1}{100}+\left(\frac{0.1}{25}\right);\frac{△g}{g}\times 100=[\frac{0.1}{100}+\frac{0.1}{25}]\times 100=0.1+0.4=0.5\%$