In order to determine the Young's Modulus of a wire of radius $0.2\, cm$ (measured using a scale of least count $=0.001\, cm )$ and length $1 \,m$ (measured using a scale of least count $=1\, mm$ ), a weight of mass $1\, kg$ (measured using a scale of least count $=1 \,g$ ) was hanged to get the elongation of $0.5\, cm$ (measured using a scale of least count $0.001\, cm$ ). What will be the fractional error in the value of Young's Modulus determined by this experiment? (in $\%$)

  • [JEE MAIN 2021]
  • A

    $0.14$

  • B

    $0.9$

  • C

    $9$

  • D

    $1.4$

Similar Questions

The density of a cube is measured by measuring its mass and length of its sides. If the maximum error in the measurement of mass and length are $4\%$ and $3\%$ respectively, the maximum error in the measurement of density will be   ........ $\%$

  • [AIIMS 2013]

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 $R$ ......... $\%$

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]

The percentage error in the measurement of $g$ is $.....\%$ (Given that $g =\frac{4 \pi^2 L }{ T ^2}, L =(10 \pm 0.1)\,cm$, $T =(100 \pm 1)\,s )$

  • [NEET 2022]

If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation $z=x / y$. If the errors in $x, y$ and $z$ are $\Delta x, \Delta y$ and $\Delta z$, respectively, then

$\mathrm{z} \pm \Delta \mathrm{z}=\frac{\mathrm{x} \pm \Delta \mathrm{x}}{\mathrm{y} \pm \Delta \mathrm{y}}=\frac{\mathrm{x}}{\mathrm{y}}\left(1 \pm \frac{\Delta \mathrm{x}}{\mathrm{x}}\right)\left(1 \pm \frac{\Delta \mathrm{y}}{\mathrm{y}}\right)^{-1} .$

The series expansion for $\left(1 \pm \frac{\Delta y}{y}\right)^{-1}$, to first power in $\Delta y / y$, is $1 \mp(\Delta y / y)$. The relative errors in independent variables are always added. So the error in $\mathrm{z}$ will be $\Delta \mathrm{z}=\mathrm{z}\left(\frac{\Delta \mathrm{x}}{\mathrm{x}}+\frac{\Delta \mathrm{y}}{\mathrm{y}}\right)$.

The above derivation makes the assumption that $\Delta x / x<<1, \Delta \mathrm{y} / \mathrm{y} \ll<1$. Therefore, the higher powers of these quantities are neglected.

($1$) Consider the ratio $\mathrm{r}=\frac{(1-\mathrm{a})}{(1+\mathrm{a})}$ to be determined by measuring a dimensionless quantity a.

If the error in the measurement of $\mathrm{a}$ is $\Delta \mathrm{a}(\Delta \mathrm{a} / \mathrm{a} \ll<1)$, then what is the error $\Delta \mathrm{r}$ in

$(A)$ $\frac{\Delta \mathrm{a}}{(1+\mathrm{a})^2}$ $(B)$ $\frac{2 \Delta \mathrm{a}}{(1+\mathrm{a})^2}$ $(C)$ $\frac{2 \Delta \mathrm{a}}{\left(1-\mathrm{a}^2\right)}$ $(D)$ $\frac{2 \mathrm{a} \Delta \mathrm{a}}{\left(1-\mathrm{a}^2\right)}$

($2$) In an experiment the initial number of radioactive nuclei is $3000$ . It is found that $1000 \pm$ $40$ nuclei decayed in the first $1.0 \mathrm{~s}$. For $|\mathrm{x}| \ll 1$, In $(1+\mathrm{x})=\mathrm{x}$ up to first power in $x$. The error $\Delta \lambda$, in the determination of the decay constant $\lambda$, in $\mathrm{s}^{-1}$, is

$(A) 0.04$    $(B) 0.03$    $(C) 0.02$   $(D) 0.01$

Give the answer quetion ($1$) and ($2$)

  • [IIT 2018]