Radius ( mathrm{r}) , length (l) and resistance (mathrm{R}) of a metal wire was measured in laborato

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1.Units, Dimensions and Measurement

hard

The radius ( $\mathrm{r})$, length $(l)$ and resistance $(\mathrm{R})$ of a metal wire was measured in the laboratory as
$\mathrm{r}=(0.35 \pm 0.05) \mathrm{cm}$
$\mathrm{R}=(100 \pm 10) \mathrm{ohm}$
$l=(15 \pm 0.2) \mathrm{cm}$
The percentage error in resistivity of the material of the wire is :

A$25.6 \%$

B$39.9 \%$

C$37.3 \%$

D$35.6 \%$

(JEE MAIN-2024)

Solution

$\rho={R} \frac{A}{\ell}$
$\frac{\Delta \rho}{\rho}=\frac{\Delta \mathrm{R}}{\mathrm{R}}+2 \frac{\Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \ell}{\ell}$
$=\frac{10}{100}+2 \times \frac{0.05}{0.35}+\frac{0.2}{15}$
$=\frac{1}{10}+\frac{2}{7}+\frac{1}{75}$
$\frac{\Delta \rho}{\rho}=39.9 \%$

Standard 11

Physics

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medium

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$Reason$ : The accuracy and precision of measuring instruments along with errors in measurements should be taken into account, while expressing the result.

easy

(AIIMS-2017)

If the random error in the arithmetic mean of $50$ observations is $\alpha$, then the random error in the arithmetic mean of $150$ observations would be

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medium

The resistance $R =\frac{V}{I}$ where $V= 100 \pm 5 \,volts$ and $ I = 10 \pm 0.2$ amperes. What is the total error in $R$ ……… $\%$

medium

The radius ( $\mathrm{r})$, length $(l)$ and resistance $(\mathrm{R})$ of a metal wire was measured in the laboratory as $\mathrm{r}=(0.35 \pm 0.05) \mathrm{cm}$ $\mathrm{R}=(100 \pm 10) \mathrm{ohm}$ $l=(15 \pm 0.2) \mathrm{cm}$ The percentage error in resistivity of the material of the wire is :

Solution

Similar Questions

In an experiment, the values of refractive indices of glass were found to be $1.54, 1.53,$$1.44,1.54,1.56$ and $1.45$ in successive measurements, then Mean absolute error is

$Assertion$: 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.

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The radius ( $\mathrm{r})$, length $(l)$ and resistance $(\mathrm{R})$ of a metal wire was measured in the laboratory as
$\mathrm{r}=(0.35 \pm 0.05) \mathrm{cm}$
$\mathrm{R}=(100 \pm 10) \mathrm{ohm}$
$l=(15 \pm 0.2) \mathrm{cm}$
The percentage error in resistivity of the material of the wire is :

$Assertion$: 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.