A student in the laboratory measures thickness of a wire using screw gauge. The readings are $1.22\,mm , 1.23\,mm , 1.19\,mm$ and $1.20\,mm$. The percentage error is $\frac{ x }{121} \%$. The value of $x$ is ..............
A student in the laboratory measures thickness of a wire using screw gauge. The readings are $1.22\,mm , 1.23\,mm , 1.19\,mm$ and $1.20\,mm$. The percentage error is $\frac{ x }{121} \%$. The value of $x$ is ..............
- [JEE MAIN 2022]
- A
$100$
- B
$150$
- C
$98$
- D
$140$
Similar Questions
Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$
Assertion $A$ : A spherical body of radius $(5 \pm 0.1)$ $mm$ having a particular density is falling through a liquid of constant density. The percentage error in the calculation of its terminal velocity is $4\,\%$.
Reason $R$ : The terminal velocity of the spherical body falling through the liquid is inversely proportional to its radius.
In the light of the above statements, choose the correct answer from the options given below on :
Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$
Assertion $A$ : A spherical body of radius $(5 \pm 0.1)$ $mm$ having a particular density is falling through a liquid of constant density. The percentage error in the calculation of its terminal velocity is $4\,\%$.
Reason $R$ : The terminal velocity of the spherical body falling through the liquid is inversely proportional to its radius.
In the light of the above statements, choose the correct answer from the options given below on :
- [JEE MAIN 2023]
The mean time period of second's pendulum is $2.00s$ and mean absolute error in the time period is $0.05s$. To express maximum estimate of error, the time period should be written as
The mean time period of second's pendulum is $2.00s$ and mean absolute error in the time period is $0.05s$. To express maximum estimate of error, the time period should be written as
$Assertion$: In the measurement of physical quantities direct and indirect methods are used.
$Reason$ : The accuracy and precision of measuring instruments along with errors in measurements should be taken into account, while expressing the result.
$Assertion$: In the measurement of physical quantities direct and indirect methods are used.
$Reason$ : The accuracy and precision of measuring instruments along with errors in measurements should be taken into account, while expressing the result.
- [AIIMS 2017]
The measured value of the length of a simple pendulum is $20 \mathrm{~cm}$ with $2 \mathrm{~mm}$ accuracy. The time for $50$ oscillations was measured to be $40$ seconds with $1$ second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $\mathrm{N} \%$. The value of $\mathrm{N}$ is:
The measured value of the length of a simple pendulum is $20 \mathrm{~cm}$ with $2 \mathrm{~mm}$ accuracy. The time for $50$ oscillations was measured to be $40$ seconds with $1$ second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $\mathrm{N} \%$. The value of $\mathrm{N}$ is:
- [JEE MAIN 2024]
A student performs an experiment for determination of $g \left(=\frac{4 \pi^{2} l }{ T ^{2}}\right), \ell =1 m$ and he commits an error of $\Delta \ell$. For $T$ he takes the time of $n$ oscillations with the stop watch of least count $\Delta T$ and he commits a human error of $0.1 s$ For which of the following data, the measurement of $g$ will be most accurate?
A student performs an experiment for determination of $g \left(=\frac{4 \pi^{2} l }{ T ^{2}}\right), \ell =1 m$ and he commits an error of $\Delta \ell$. For $T$ he takes the time of $n$ oscillations with the stop watch of least count $\Delta T$ and he commits a human error of $0.1 s$ For which of the following data, the measurement of $g$ will be most accurate?