In a simple pendulum experiment for determina | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The formula for acceleration due to gravity is $g = \frac{4\pi^2 L}{T^2}$.Taking the relative error, we have $\frac{\Delta g}{g} = \frac{\Delta L}

Vedclass · Accepted Answer

$6.8$

Vedclass · Answer

(C) The formula for acceleration due to gravity is $g = \frac{4\pi^2 L}{T^2}$.Taking the relative error, we have $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$.Given:$L = 55.0\,cm$, $\Delta L = 1\,mm = 0.1\,cm$.Time for $20$ oscillations is $t = 30\,s$, so $T = \frac{30}{20} = 1.5\,s$.The least count of the watch is $1\,s$, so the error in total time is $\Delta t = 1\,s$. Thus, $\Delta T = \frac{\Delta t}{20} = \frac{1}{20} = 0.05\,s$.Now, calculate the percentage error:$\frac{\Delta g}{g} 	imes 100 = \left( \frac{\Delta L}{L} + 2\frac{\Delta T}{T} ight) 	imes 100\%$$= \left( \frac{0.1}{55.0} + 2 	imes \frac{0.05}{1.5} ight) 	imes 100\%$$= \left( 0.001818 + 0.06666 ight) 	imes 100\%$$= 0.06848 	imes 100\% \approx 6.8\%$.

Similar Questions

$A$ physical quantity $S$ is related to four observables $a, b, c$,and $d$ as $S = \frac{\sqrt{a} b}{c^3 d^4}$. If the percentage errors of measurement in $a, b, c$,and $d$ are $2 \%$,$1 \%$,$1 \%$,and $1 \%$ respectively,then the percentage error in the quantity $S$ is: (in $\%$)

The relative error in the determination of the surface area of a sphere is $\alpha$. Then the relative error in the determination of its volume is

In an experiment,the values of refractive indices of glass were found to be $1.54, 1.53, 1.44, 1.54, 1.56$ and $1.45$ in successive measurements. The mean absolute error is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module