(C) The surface area of a sphere is given by $A = 4\pi r^2$. Using the rules of error propagation,the relative error in $A$ is given by $\frac{\Delta

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(C) The surface area of a sphere is given by $A = 4\pi r^2$. Using the rules of error propagation,the relative error in $A$ is given by $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$. Given that the percentage error in the radius is $\frac{\Delta r}{r} \times 100 = 0.3\%$. Therefore,the percentage error in the surface area is $\frac{\Delta A}{A} \times 100 = 2 \times (0.3\%) = 0.6\%$. Thus,the Assertion is correct. The Reason states the formula $\frac{\Delta A}{A} = \frac{4\Delta r}{r}$,which is incorrect because the correct relation is $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$. Hence,the Assertion is correct but the Reason is incorrect.

Class 11 Physics Units, Dimensions and Measurement Errors of Measurement

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$Assertion$ : The error in the measurement of radius of the sphere is $0.3\%$. The permissible error in its surface area is $0.6\%$.
$Reason$ : The permissible error is calculated by the formula $\frac{\Delta A}{A} = \frac{4\Delta r}{r}$.

AIIMS 2008 All AIIMS

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.

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Units, Dimensions and Measurement

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Errors of Measurement

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$A$ physical quantity $A$ is related to four observables $a, b, c,$ and $d$ as follows: $A = \frac{a^2 b^3}{c \sqrt{d}}$. The percentage errors of measurement in $a, b, c,$ and $d$ are $1\%, 3\%, 2\%,$ and $2\%$ respectively. What is the percentage error in the quantity $A$?

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$A$ metal wire has mass $(0.4 \pm 0.002) \, g$,radius $(0.3 \pm 0.001) \, mm$ and length $(5 \pm 0.02) \, cm$. The maximum possible percentage error in the measurement of density will nearly be $.......\%$

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$A$ body travels uniformly a distance of $(13.8 \pm 0.2) \ m$ in a time $(4.0 \pm 0.3) \ s$. The percentage error in velocity is ......... $\%$

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Three students $S_{1}, S_{2}$ and $S_{3}$ perform an experiment for determining the acceleration due to gravity $(g)$ using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.

Student No. Length of pendulum $(cm)$ No. of oscillations $(n)$ Total time for oscillations $(s)$ Time period $(s)$

$1.$ $64.0$ $8$ $128.0$ $16.0$

$2.$ $64.0$ $4$ $64.0$ $16.0$

$3.$ $20.0$ $4$ $36.0$ $9.0$

(Least count of length $= 0.1 \, cm$,least count for time $= 0.1 \, s$)
If $E_{1}, E_{2}$ and $E_{3}$ are the percentage errors in $g$ for students $1, 2$ and $3$ respectively,then the minimum percentage error is obtained by student no. ....... .

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Student No.	Length of pendulum $(cm)$	No. of oscillations $(n)$	Total time for oscillations $(s)$	Time period $(s)$
$1.$	$64.0$	$8$	$128.0$	$16.0$
$2.$	$64.0$	$4$	$64.0$	$16.0$
$3.$	$20.0$	$4$	$36.0$	$9.0$

$Assertion$ : The error in the measurement of radius of the sphere is $0.3\%$. The permissible error in its surface area is $0.6\%$. $Reason$ : The permissible error is calculated by the formula $\frac{\Delta A}{A} = \frac{4\Delta r}{r}$.

Similar Questions

$A$ physical quantity $A$ is related to four observables $a, b, c,$ and $d$ as follows: $A = \frac{a^2 b^3}{c \sqrt{d}}$. The percentage errors of measurement in $a, b, c,$ and $d$ are $1\%, 3\%, 2\%,$ and $2\%$ respectively. What is the percentage error in the quantity $A$?

$A$ metal wire has mass $(0.4 \pm 0.002) \, g$,radius $(0.3 \pm 0.001) \, mm$ and length $(5 \pm 0.02) \, cm$. The maximum possible percentage error in the measurement of density will nearly be $.......\%$

We can reduce random errors by

$A$ body travels uniformly a distance of $(13.8 \pm 0.2) \ m$ in a time $(4.0 \pm 0.3) \ s$. The percentage error in velocity is ......... $\%$

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$Assertion$ : The error in the measurement of radius of the sphere is $0.3\%$. The permissible error in its surface area is $0.6\%$.
$Reason$ : The permissible error is calculated by the formula $\frac{\Delta A}{A} = \frac{4\Delta r}{r}$.