Question

The acceleration due to gravity is measured by using simple pendulum . If 'a ' and 'b' are relative errors in me?surement of length and time period respectively then percentage error in the measurement of acceleration due to gravity will be??

Answer 1

$$\begin{gathered} DearStudent, \\ T=2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}} \\ {\mathrm{T}}^2=\frac{4{\mathrm{\pi }}^2\mathrm{L}}{\mathrm{g}} \\ \mathrm{g}=\frac{4{\mathrm{\pi }}^2\mathrm{L}}{{\mathrm{T}}^2} \\ \frac{∆\mathrm{g}}{\mathrm{g}}=\frac{∆\mathrm{L}}{\mathrm{L}}+2\frac{∆\mathrm{T}}{\mathrm{T}} \\ \frac{∆\mathrm{g}}{\mathrm{g}}\times 100=\frac{∆\mathrm{L}}{\mathrm{L}}\times 100+2\frac{∆\mathrm{T}}{\mathrm{T}}\times 100 \\ \%\mathrm{g}=\%\mathrm{L}+2*\%\mathrm{T} \\ \mathrm{percentage}\mathrm{error}\mathrm{in}\mathrm{g}=\mathrm{a}+2\mathrm{b} \\ \mathrm{Regards} \end{gathered}$$