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 ......... $\%$

  • A
    $8$
  • B
    $2$
  • C
    $4$
  • D
    $6$

Explore More

Similar Questions

Time for $20$ oscillations of a pendulum is measured as $t_1 = 39.6\, s$,$t_2 = 39.9\, s$,and $t_3 = 39.5\, s$. What is the precision in the measurements? What is the accuracy of the measurement?

For $z = a^{2} x^{3} y^{1/2}$,where $a$ is a constant. If the percentage error in the measurement of $x$ and $y$ are $4\%$ and $12\%$,respectively,then the percentage error for $z$ will be $........... \%$.

In five successive measurements, the mass of a ball is measured to be $2.61 \,g, 2.58 \,g, 2.40 \,g, 2.73 \,g$ and $2.80 \,g$. The mean absolute error in the measurement is (in $\,g$)

$A$ body travels uniformly a distance of $(13.8 \pm 0.2) \text{ m}$ in a time $(4.0 \pm 0.3) \text{ s}$. Its velocity with error limits and percentage error is

Difficult
View Solution

If $f = x^2$,what is the relative error in $f$?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo