In an experiment, on the measurement of g using a simple pendulum, the time period was measured with an accuracy of 0.2% while the length was meausred with an accuracy of 0.5%. The percentage accuracy in the value of g thus obtained is

Question

In an experiment, on the measurement of g using a simple pendulum, the time period was measured with an accuracy of 0.2% while the length was meausred with an accuracy of 0.5%. The percentage accuracy in the value of g thus obtained is (A) 0.25% (B) 0.1% (C) 0.7% (D) 0.9%

Tardigrade Admin · Accepted Answer

Given (Δ T/T) = 0.2 % , (Δ L/L) = 0.5 % Since T = 2 π √(L/g) ⇒ : :: :: g = (4 π2 L/T2) ∴ : :: : (Δ g/g) = (Δ L/L) + (2 Δ T/T) = 0.5 % + 2(0.2 %) = 0.9 %

Q. In an experiment, on the measurement of g using a simple pendulum, the time period was measured with an accuracy of 0.2% while the length was meausred with an accuracy of 0.5%. The percentage accuracy in the value of g thus obtained is

0.25%

0.1%

0.7%

0.9%

Given $\frac{Δ T}{T} = 0.2%, \frac{Δ L}{L} = 0.5%$
Since $T = 2 π \frac{L}{g}$
$\Rightarrow g = \frac{4 π ^{2} L}{T ^{2}}$
$∴ \frac{Δ g}{g} = \frac{Δ L}{L} + \frac{2Δ T}{T}$
$= 0.5% + 2 (0.2%)$
$= 0.9%$

Q. In an experiment, on the measurement of g using a simple pendulum, the time period was measured with an accuracy of 0.2% while the length was meausred with an accuracy of 0.5%. The percentage accuracy in the value of g thus obtained is

0.25%

0.1%

0.7%

0.9%

Given TΔT​=0.2%,LΔL​=0.5% Since T=2πgL​​ ⇒g=T24π2L​ ∴gΔg​=LΔL​+T2ΔT​ =0.5%+2(0.2%) =0.9%

Given $\frac{Δ T}{T} = 0.2%, \frac{Δ L}{L} = 0.5%$
Since $T = 2 π \frac{L}{g}$
$\Rightarrow g = \frac{4 π ^{2} L}{T ^{2}}$
$∴ \frac{Δ g}{g} = \frac{Δ L}{L} + \frac{2Δ T}{T}$
$= 0.5% + 2 (0.2%)$
$= 0.9%$