We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be $2.63 \;s , 2.56 \;s , 2.42\; s , 2.71 \;s$ and $2.80 \;s$. Calculate the absolute errors, relative error or percentage error.
We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be $2.63 \;s , 2.56 \;s , 2.42\; s , 2.71 \;s$ and $2.80 \;s$. Calculate the absolute errors, relative error or percentage error.
Answer The mean perlod of oscillation of the pendulum
$T \;=\frac{(2.63+2.56+2.42+2.71+2.80) \,s}{5}$
$\quad=\frac{13.12}{5} \;s$
$=2.624\, s $
$=2.62 \,s$
As the periods are measured to a resolution of $0.01 \,s ,$ all times are to the second decimal; it is proper to put this mean perlod also to the second decimal.
The errors in the measurements are
$2.63 \,s -2.62 \,s =0.01 \,s$
$2.56 \,s-2.62 \,s=-0.06 \,s$
$2.42\, s -2.62\, s =-0.20 \,s$
$2.71 \,s -2.62\, s =0.09 \,s$
$2.80\, s-2.62\, s=0.18\, s$
The arthmetic mean of all the absolute errors (for arithmetic mean, we take only the magnitudes) is
$ \Delta T_{\text {mean}} =[(0.01+0.06+0.20+0.09+0.18) \,s ] / 5 $
$=0.54 \,s / 5 $
$=0.11 \,s $
$T=2.6 \pm 0.1 \,s$
$\delta a=\frac{0.1}{2.6} \times 100=4 \%$
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- [NEET 2023]
The percentage error in measurement of a physical quantity $m$ given by $m = \pi \tan \theta $ is minimum when $\theta $ $=$ .......... $^o$ (Assume that error in $\theta $ remain constant)
The percentage error in measurement of a physical quantity $m$ given by $m = \pi \tan \theta $ is minimum when $\theta $ $=$ .......... $^o$ (Assume that error in $\theta $ remain constant)
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 ($2$)
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]
Two resistance are measured in $Ohm$ and is given as
$R_1 = 3 \Omega \pm 1\%$ and $R_2 = 6 \Omega \pm 2\%$ When they are connected in parallel, the percentage error in equivalent resistance is.......... $\%$
Two resistance are measured in $Ohm$ and is given as
$R_1 = 3 \Omega \pm 1\%$ and $R_2 = 6 \Omega \pm 2\%$ When they are connected in parallel, the percentage error in equivalent resistance is.......... $\%$