Question

While measuring acceleration due to gravity by a simple pendulum a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of the time period. His percentage error in the measurement of the value of $g$ will be:

• A. $2$%
• B. $4$%
• C. $7$%
• D. $10$%

Answer 1

The correct option is C $7$%

$T=2\pi \sqrt{\frac{l}{g}}\Rightarrow T^2=\frac{4{\pi }^{2}l}{g}\Rightarrow g=\frac{4{\pi }^{2}l}{T^2}$

$\frac{△g}{g}\times 100=\frac{\Delta l}{l}\times 100+2\frac{△T}{T}\times 100=1+2\times 3=7\%$