If the time period of oscillation of a pendulum is measured as 2.5 second using a stop watch with the least count $\frac{1}{2}$ second, then the permissible error in the measurement is

Answer (5)

Given,

$\displaystyle T=2.5s, \Delta T=\frac {1} {2} s$

Therefore, the permissible error in measurement of time period is

$\displaystyle \frac {\Delta T} {T} \times 100=\frac {(\frac {1} {2})} {(2.5)} =20\%$