Measure of two quantities along with the precision of respective measuring instrument $A = 2.5\,m{s^{ - 1}} \pm 0.5\,m{s^{ - 1}}$, $B = 0.10\,s \pm 0.01\,s$ The value of $AB$ will be
$\left( {0.25 \pm 0.08} \right)\,m$
$\left( {0.25 \pm 0.5} \right)\,m$
$\left( {0.25 \pm 0.05} \right)\,m$
$\left( {0.25 \pm 0.135} \right)\,m$
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
$z \pm \Delta z=\frac{x \pm \Delta x}{y \pm \Delta y}=\frac{x}{y}\left(1 \pm \frac{\Delta x}{x}\right)\left(1 \pm \frac{\Delta y}{y}\right)^{-1} .$
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 $z$ will be $\Delta z=z\left(\frac{\Delta x}{x}+\frac{\Delta y}{y}\right)$.
The above derivation makes the assumption that $\Delta x / x \ll<1, \Delta y / y \ll<1$. Therefore, the higher powers of these quantities are neglected.
($1$) Consider the ratio $r =\frac{(1- a )}{(1+ a )}$ to be determined by measuring a dimensionless quantity a.
If the error in the measurement of $a$ is $\Delta a (\Delta a / a \ll<1)$, then what is the error $\Delta r$ in
$(A)$ $\frac{\Delta a }{(1+ a )^2}$ $(B)$ $\frac{2 \Delta a }{(1+ a )^2}$ $(C)$ $\frac{2 \Delta a}{\left(1-a^2\right)}$ $(D)$ $\frac{2 a \Delta a}{\left(1-a^2\right)}$
($2$) In an experiment the initial number of radioactive nuclei is $3000$ . It is found that $1000 \pm$ 40 nuclei decayed in the first $1.0 s$. For $|x|<1$, In $(1+x)=x$ up to first power in $x$. The error $\Delta \lambda$, in the determination of the decay constant $\lambda$, in $s ^{-1}$, is
$(A) 0.04$ $(B) 0.03$ $(C) 0.02$ $(D) 0.01$
Give the answer or quetion ($1$) and ($2$)
The resistance $R =\frac{ V }{ I },$ where $V =(50 \pm 2) \;V$ and $I=(20 \pm 0.2)\;A.$ The percentage error in $R$ is $x\%$. The value of $x$ to the nearest integer is .........
Two resistors of resistances $R_{1}=100 \pm 3$ $ohm$ and $R_{2}=200 \pm 4$ $ohm$ are connected $(a)$ in series, $(b)$ in parallel. Find the equivalent resistance of the $(a)$ series combination, $(b)$ parallel combination. Use for $(a)$ the relation $R=R_{1}+R_{2}$ and for $(b)$ $\frac{1}{R^{\prime}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$ and $\frac{\Delta R^{\prime}}{R^{\prime 2}}=\frac{\Delta R_{1}}{R_{1}^{2}}+\frac{\Delta R_{2}}{R_{2}^{2}}$
The percentage error in measurement of mass and speed are $3\%$ and $2\%$ then error in kinetic energy will be .......... $\%$
What is accuracy in measurement ? Accuracy depend on which factors ?