In the expression for time period $T$ of simple pendulum $T=2 \pi \sqrt{\frac{l}{g}}$, if the percentage error in time period $T$ and length $l$ are $2 \%$ and $2 \%$ respectively then percentage error in acceleration due to gravity $g$ is equal to ......... $\%$
In the expression for time period $T$ of simple pendulum $T=2 \pi \sqrt{\frac{l}{g}}$, if the percentage error in time period $T$ and length $l$ are $2 \%$ and $2 \%$ respectively then percentage error in acceleration due to gravity $g$ is equal to ......... $\%$
- A
$8$
- B
$2$
- C
$4$
- D
$6$
Similar Questions
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 ($2$)
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 ($2$)
- [IIT 2018]
A public park, in the form of a square, has an area of $(100 \pm 0.2) m ^2$. The side of park is ......... $m$
A public park, in the form of a square, has an area of $(100 \pm 0.2) m ^2$. The side of park is ......... $m$
The period of oscillation of a simple pendulum is $T =2 \pi \sqrt{\frac{ L }{ g }} .$ Measured value of $ L $ is $1.0\, m$ from meter scale having a minimum division of $1 \,mm$ and time of one complete oscillation is $1.95\, s$ measured from stopwatch of $0.01 \,s$ resolution. The percentage error in the determination of $g$ will be ..... $\%.$
The period of oscillation of a simple pendulum is $T =2 \pi \sqrt{\frac{ L }{ g }} .$ Measured value of $ L $ is $1.0\, m$ from meter scale having a minimum division of $1 \,mm$ and time of one complete oscillation is $1.95\, s$ measured from stopwatch of $0.01 \,s$ resolution. The percentage error in the determination of $g$ will be ..... $\%.$
- [JEE MAIN 2021]
Thickness of a pencil measured by using a screw gauge (least count $0.001 \,cm$ ) comes out to be $0.802 \,cm$. The percentage error in the measurement is ........... $\%$
Thickness of a pencil measured by using a screw gauge (least count $0.001 \,cm$ ) comes out to be $0.802 \,cm$. The percentage error in the measurement is ........... $\%$
A force $F$ is applied on a square area of side $L$. If the percentage error in the measurement of $L$ is $2 \%$ and that in $F$ is $4 \%$, what is the maximum percentage error in pressure is .......... $\%$
A force $F$ is applied on a square area of side $L$. If the percentage error in the measurement of $L$ is $2 \%$ and that in $F$ is $4 \%$, what is the maximum percentage error in pressure is .......... $\%$