$Assertion$ : When percentage errors in the measurement of mass and velocity are $1\%$ and $2\%$ respectively, the percentage error in $K.E.$ is $5\%$.
$Reason$ : $\frac{{\Delta E}}{E} = \frac{{\Delta m}}{m} + \frac{{2\Delta v}}{v}$
$Assertion$ : When percentage errors in the measurement of mass and velocity are $1\%$ and $2\%$ respectively, the percentage error in $K.E.$ is $5\%$.
$Reason$ : $\frac{{\Delta E}}{E} = \frac{{\Delta m}}{m} + \frac{{2\Delta v}}{v}$
- [AIIMS 2010]
- A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
- B
If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
- C
If the Assertion is correct but Reason is incorrect.
- D
If both the Assertion and Reason are incorrect.
Similar Questions
Write rule for error in result due to multiplication and division.
Write rule for error in result due to multiplication and division.
The internal and external radii of a hollow cylinder are measured with the help of a vernier callipers.Their values are $(4.23 \pm 0.01)\,\,cm$ and $(3.87 \pm 0.01)\,\,cm,$ respectively. The thickness of the wall of the cylinder is
The internal and external radii of a hollow cylinder are measured with the help of a vernier callipers.Their values are $(4.23 \pm 0.01)\,\,cm$ and $(3.87 \pm 0.01)\,\,cm,$ respectively. The thickness of the wall of the cylinder is
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]
A physical quantity $P$ is given by $P= \frac{{{A^3}{B^{\frac{1}{2}}}}}{{{C^{ - 4}}{D^{\frac{3}{2}}}}}$. The quantity which brings in the maximum percentage error in $P$ is
A physical quantity $P$ is given by $P= \frac{{{A^3}{B^{\frac{1}{2}}}}}{{{C^{ - 4}}{D^{\frac{3}{2}}}}}$. The quantity which brings in the maximum percentage error in $P$ 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 $\frac{\Delta R^{\prime}}{R^{\prime 2}}=\frac{\Delta R_{1}}{R_{1}^{2}}+\frac{\Delta R_{2}}{R_{2}^{2}}$
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 $\frac{\Delta R^{\prime}}{R^{\prime 2}}=\frac{\Delta R_{1}}{R_{1}^{2}}+\frac{\Delta R_{2}}{R_{2}^{2}}$