Question

A student makes an error of 1% in measuring length of pendulum and negative error of 3% in value of time periods. The percentage error in the measurement of ‘g’ will be

• A. 5%
• B. 1%
• C. 7%
• D. 2%

Answer 1

The correct option is C 7%

$g=4{\pi }^2\frac{1}{T^2}$
$\frac{\Delta 9}{g}\times 100=\frac{\Delta l}{l}\times 100+2(\frac{\Delta T}{T}\times 100)$
$=1\%+2\left(3\%\right)$
$=7\%$