Question

A student measured the length of a pendulum 1.351 m and time for 30 vibrations is 2 minutes 10 sec using his wrist watch. Find the percent error in g committed.

• A. 1.72%
• B. 1.813%
• C. 1.63%
• D. 1.513%

Answer 1

Answer: (C)

1.63%

Answer 2

Percent error in g

= $\frac{\Delta g}{g} \times 100\left( \frac{\Delta l}{l} + \frac{2\Delta T}{T} \right) \times 100$

= $\left( \frac{.001}{l.351} + \frac{2 \times 1}{130} \right) \times 100 = 0.73 + 1.54 = 1.613\%$