Question

The length of a pendulum is measured as $1.01$ m and time for $30$ oscillations is measured as $1$ minute $3$ seconds. Error in length is $0.01$ m and error in time is $3$ seconds. The percentage error in the measurement of acceleration due to gravity is:

• A. $1$
• B. $5$
• C. $10$
• D. $15$

Answer 1

The correct option is C $10$

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

$⟹g=\frac{4{\pi }^{2}l}{T^2}$

Thus $\frac{\Delta g}{g}=\frac{\Delta l}{l}+2\frac{\Delta T}{T}$

Percentage error in measurement of:

$g=\frac{\Delta g}{g}\times 100=(\frac{\Delta l}{l}+2\frac{\Delta T}{T})\times 100$

$=(\frac{0.01}{1.01}+2\frac{3}{63})\times 100$% $=10$%