An experiment is performed to obtain the valu | vedclass

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

Question

(B) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$.Squaring both sides,we get $T^2 = 4\pi^2 \frac{L}{g}$,which

Vedclass · Accepted Answer

$2.7$

Vedclass · Answer

(B) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$.Squaring both sides,we get $T^2 = 4\pi^2 \frac{L}{g}$,which implies $g = 4\pi^2 \frac{L}{T^2}$.The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.Given: $L = 20.0$ cm,$\Delta L = 1$ mm = $0.1$ cm.$T_{total} = 90.0$ s,$\Delta T_{total} = 1$ s. Since $T = \frac{T_{total}}{100}$,$\Delta T = \frac{\Delta T_{total}}{100} = \frac{1}{100} = 0.01$ s.Substituting the values:$\frac{\Delta g}{g} = \frac{0.1}{20.0} + 2 	imes \frac{1}{90} = 0.005 + 0.0222 = 0.0272$.Percentage error = $0.0272 	imes 100 = 2.72 \% \approx 2.7 \%$.

Similar Questions

The resistance $R = V / I$ where $V = (100 \pm 5) \; V$ and $I = (10 \pm 0.2) \; A$. Find the percentage error in $R$. (in $\%$)

Thickness of a pencil measured by using a screw gauge (least count $0.001 \,cm$) comes out to be $0.802 \,cm$. The percentage error in the measurement is ........... $\%$

In the expression for time period $T$ of a simple pendulum $T = 2 \pi \sqrt{\frac{l}{g}}$,if the percentage error in time period $T$ and length $l$ are $2 \%$ and $2 \%$ respectively,then the percentage error in acceleration due to gravity $g$ is equal to ......... $\%$

What is error in measurement? What is mistake in measurement?

The least count of a stopwatch is $0.2\, s$. The time for $20\, oscillations$ of a pendulum is measured to be $25\, s$. The percentage error in the measurement of time will be ........ $\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module