Assertion : When percentage errors in the mea | vedclass

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(A) The kinetic energy $E$ is given by the formula $E = \frac{1}{2}mv^2$.Taking the relative error,we use the formula for propagation of errors:$\frac

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If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

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(A) The kinetic energy $E$ is given by the formula $E = \frac{1}{2}mv^2$.Taking the relative error,we use the formula for propagation of errors:$\frac{\Delta E}{E} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}$.Given that the percentage error in mass $\frac{\Delta m}{m} 	imes 100 = 1\%$ and the percentage error in velocity $\frac{\Delta v}{v} 	imes 100 = 2\%$.Substituting these values into the error equation:$\frac{\Delta E}{E} 	imes 100 = (\frac{\Delta m}{m} 	imes 100) + 2 	imes (\frac{\Delta v}{v} 	imes 100)$$\frac{\Delta E}{E} 	imes 100 = 1\% + 2 	imes 2\% = 1\% + 4\% = 5\%$.Since both the Assertion and the Reason are correct and the Reason correctly explains the propagation of error formula used in the Assertion,the correct option is $A$.

$Assertion$ : When percentage errors in the measurement of mass and velocity are $1\%$ and $2\%$ respectively,the percentage error in $K.E.$ is $5\%$.
$Reason$ : $\frac{{\Delta E}}{E} = \frac{{\Delta m}}{m} + \frac{{2\Delta v}}{v}$

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$Assertion$ : When percentage errors in the measurement of mass and velocity are $1\%$ and $2\%$ respectively,the percentage error in $K.E.$ is $5\%$.$Reason$ : $\frac{{\Delta E}}{E} = \frac{{\Delta m}}{m} + \frac{{2\Delta v}}{v}$