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

Answer: (C)

$7\%$

Answer 2

$T=2\pi\sqrt{\frac{l}{g}}\,\Rightarrow T^{2}=\frac{4\pi^{2}l}{g}\,\Rightarrow g=\frac{4\pi^{2}l}{T^{2}}$ $\frac{\Delta g}{g}\times100=\frac{\Delta l}{l}\times 100+2 \frac{\Delta T}{T}\times100$ $=1\%+2\times3\%=7\%$