The period of oscillation of a simple pendulum is $T = 2 \pi \sqrt{\frac{L}{g}}$. The measured value of $L$ is $1.0 \text{ m}$ using a meter scale with a minimum division of $1 \text{ mm}$,and the time for one complete oscillation is $1.95 \text{ s}$ measured using a stopwatch with a resolution of $0.01 \text{ s}$. The percentage error in the determination of $g$ will be ..... $\%$.

  • A
    $1.13$
  • B
    $1.03$
  • C
    $1.33$
  • D
    $1.30$

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Similar Questions

In an experiment of a simple pendulum,the errors in the measurement of the length of the pendulum $(L)$ and time period $(T)$ are $3 \%$ and $2 \%$,respectively. The maximum percentage error in the value of $\frac{L}{T^2}$ is (in $\%$)

$A$ student determined Young's Modulus of elasticity using the formula $Y = \frac{M g L^{3}}{4 b d^{3} \delta}$. The value of $g$ is taken to be $9.8 \, m/s^2$, without any significant error. His observations are as follows:
Physical QuantityLeast count and Observed value
Mass $(M)$$1 \, g$ and $2 \, kg$
Length of bar $(L)$$1 \, mm$ and $1 \, m$
Breadth of bar $(b)$$0.1 \, mm$ and $4 \, cm$
Thickness of bar $(d)$$0.01 \, mm$ and $0.4 \, cm$
Depression $(\delta)$$0.01 \, mm$ and $5 \, mm$

Then the fractional error in the measurement of $Y$ is:

Consider a physical quantity $Z$ expressed as $Z = \frac{A B^{1/2}}{C^2}$. If the relative error in the magnitudes of $A, B,$ and $C$ is $1\%$, then the relative error in $Z$ will be: (in $\%$)

In an experiment,the following observations were recorded: $L = 2.820 \, m, M = 3.00 \, kg, l = 0.087 \, cm$,Diameter $D = 0.041 \, cm$. Taking $g = 9.81 \, m/s^2$ and using the formula $Y = \frac{4MgL}{\pi D^2 l}$,the maximum permissible error in $Y$ is ......... $\%$.

Time intervals measured by a clock give the following readings: $1.25 \; s, 1.24 \; s, 1.27 \; s, 1.21 \; s$,and $1.28 \; s$. What is the percentage relative error of the observations?

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