(A) Let the maximum temperature be $T_1 = 44^{\circ} C \pm 0.5^{\circ} C$ and the minimum temperature be $T_2 = 22^{\circ} C \pm 0.5^{\circ} C$.The te

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(A) Let the maximum temperature be $T_1 = 44^{\circ} C \pm 0.5^{\circ} C$ and the minimum temperature be $T_2 = 22^{\circ} C \pm 0.5^{\circ} C$.The temperature difference is given by $\Delta T = T_1 - T_2$.The value of the difference is $44^{\circ} C - 22^{\circ} C = 22^{\circ} C$.According to the rule of error propagation for subtraction,when two quantities are subtracted,their absolute errors are added.Therefore,the error in the difference is $\Delta(\Delta T) = \Delta T_1 + \Delta T_2 = 0.5^{\circ} C + 0.5^{\circ} C = 1.0^{\circ} C$.Thus,the temperature difference is $22^{\circ} C \pm 1^{\circ} C$.

Class 11 Physics Units, Dimensions and Measurement Errors of Measurement

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If the maximum and minimum temperatures at a place on a day are measured as $44^{\circ} C \pm 0.5^{\circ} C$ and $22^{\circ} C \pm 0.5^{\circ} C$ respectively,then the temperature difference is

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$22^{\circ} C \pm 1^{\circ} C$
B
$22^{\circ} C \pm 0.5^{\circ} C$
C
$22^{\circ} C \pm 0.25^{\circ} C$
D
$22^{\circ} C \pm 1.5^{\circ} C$

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If the maximum and minimum temperatures at a place on a day are measured as $44^{\circ} C \pm 0.5^{\circ} C$ and $22^{\circ} C \pm 0.5^{\circ} C$ respectively,then the temperature difference is

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$A$ physical quantity $x$ is calculated from the relation $x = \frac{a^2 b^3}{c \sqrt{d}}$. If the percentage errors in $a, b, c,$ and $d$ are $2\%, 1\%, 3\%,$ and $4\%$ respectively,what is the percentage error in $x$?

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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 $\frac{\Delta A}{A} = \frac{4\Delta r}{r}$.