Question

In case of measurement of g, if error in measurement of length of pendulum is 2%, the percentage error in time period is1 %. The maximum error in measurement of g is

• A. 1 %
• B. 2 %
• C. 4 %
• D. no error.

Answer 1

The correct option is C 4 %

Error in length $\frac{\Delta l}{l}\times 100=2$%
Error in time period $\frac{\Delta T}{T}\times 100=1$%
Acceleration due to gravity $g=\frac{4{\pi }^{2}l}{T^2}$
Error in measurement of $g$ $\frac{\Delta g}{g}\times 100=\frac{\Delta l}{l}\times 100+2\times \frac{\Delta T}{T}\times 100$
$⟹$ $\frac{\Delta g}{g}\times 100=2+2\times 1=4$%