Explain least count and least count error. Write a note on least count error.
Explain least count and least count error. Write a note on least count error.
The smallest value that can be measured by the measuring instrument is called its least count.
- All the readings or measured values are good only up to this value.
- Error associated with resolution of the instrument is called least count error.
- Least count of vernier is $0.01 \mathrm{~cm}$ and least count of spherometer is $0.001 \mathrm{~cm}$.
- Least count error belong to category of random error but within a limited size.
- It occur with both systematic and random error.
- Least count of meter scale is $1 \mathrm{~mm}$.
- Using instrument of higher precision, improving experimental technique we can reduce least count error.
Repeating the observation several times and taking arithmetic mean of all observation the mean value would be very close to the true value of the measured quantity.
Similar Questions
A physical quantity $X$ is related to four measurable quantities $a,\, b,\, c$ and $d$ as follows $X = a^2b^3c^{\frac {5}{2}}d^{-2}$. The percentange error in the measurement of $a,\, b,\, c$ and $d$ are $1\,\%$, $2\,\%$, $3\,\%$ and $4\,\%$ respectively. What is the percentage error in quantity $X$ ? If the value of $X$ calculated on the basis of the above relation is $2.763$, to what value should you round off the result.
A physical quantity $X$ is related to four measurable quantities $a,\, b,\, c$ and $d$ as follows $X = a^2b^3c^{\frac {5}{2}}d^{-2}$. The percentange error in the measurement of $a,\, b,\, c$ and $d$ are $1\,\%$, $2\,\%$, $3\,\%$ and $4\,\%$ respectively. What is the percentage error in quantity $X$ ? If the value of $X$ calculated on the basis of the above relation is $2.763$, to what value should you round off the result.
If there is a positive error of $50\%$ in the measurement of velocity of a body, then the error in the measurement of kinetic energy is .............. $\%$
If there is a positive error of $50\%$ in the measurement of velocity of a body, then the error in the measurement of kinetic energy is .............. $\%$
Calculate the mean $\%$ error in five observation
$80.0,80.5,81.0,81.5,82$
Calculate the mean $\%$ error in five observation
$80.0,80.5,81.0,81.5,82$
- [AIIMS 2019]
A certain body weighs $22.42\;g$ and has a measured volume of $4.7 \;cc .$ The possible error in the measurement of mass and volume are $0.01\; gm$ and $0.1 \;cc .$
Then maximum error in the density will be
A certain body weighs $22.42\;g$ and has a measured volume of $4.7 \;cc .$ The possible error in the measurement of mass and volume are $0.01\; gm$ and $0.1 \;cc .$
Then maximum error in the density will be
- [AIPMT 1991]
Durring Searle's experiment, zero of the Vernier scale lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale. The $20^{\text {th }}$ division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of $2 \ kg$ is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale but now the $45^{\text {th }}$ division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is $2 m$. and its cross-sectional area is $8 \times 10^{-7} m ^2$. The least count of the Vernier scale is $1.0 \times 10^{-5} m$. The maximum percentage error in the Young's modulus of the wire is
Durring Searle's experiment, zero of the Vernier scale lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale. The $20^{\text {th }}$ division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of $2 \ kg$ is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale but now the $45^{\text {th }}$ division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is $2 m$. and its cross-sectional area is $8 \times 10^{-7} m ^2$. The least count of the Vernier scale is $1.0 \times 10^{-5} m$. The maximum percentage error in the Young's modulus of the wire is
- [IIT 2014]