Question

To find the value of 'g' using simple pendulum T = 2.00 sec; $l$ = 1.00 m was measured. Estimate maximum permissible error in 'g'. Also find the value of 'g'. (use $u^2$ = 10)

• A. 'g' = (10 $\pm$ 2) $m/s^2$
• B. 'g' = (10.0 $\pm$ 0.2) $m/s^2$
• C. 'g' = (10 $\pm$ 0.002) $m/s^2$
• D. 'g' = (10.0 $\pm$ 0.02) $m/s^2$

Answer 1

Answer: (B)

'g' = (10.0 $\pm$ 0.2) $m/s^2$

$T=2\pi \sqrt{\frac{l}{g}}\Rightarrow g=\frac{4{\pi }^{2}l}{T^2}$
${\left(\frac{\Delta g}{g}\right)}_{max}=\frac{\Delta l}{l}+2\frac{\Delta T}{T}\left(\frac{0.01}{1.00}+2\frac{0.01}{2.00}\right)\times 100\%$ = 2 %
value of g = $\frac{4{\pi }^{2}l}{T^2}=\frac{4\times 10\times 1.00}{(2.00{)}^2}=10.l0m/s^2$
${\left(\frac{\Delta g}{g}\right)}_{max}=2/100$
So $\frac{\Delta g_{max}}{10.0}=\frac{2}{100}$ so $(\Delta g{)}_{max}=0.2$ = max error in 'g'
so 'g' = (10.0 $\pm$ 0.2) $m/s^2$