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(B) Step $1$: Calculate the mean value of the readings.$X_{mean} = \frac{1.22 + 1.23 + 1.19 + 1.20}{4} = \frac{4.84}{4} = 1.21\,mm$.Step $2$: Calculat

Vedclass · Accepted Answer

$150$

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(B) Step $1$: Calculate the mean value of the readings.$X_{mean} = \frac{1.22 + 1.23 + 1.19 + 1.20}{4} = \frac{4.84}{4} = 1.21\,mm$.Step $2$: Calculate the absolute errors for each reading.$|\Delta X_1| = |1.22 - 1.21| = 0.01\,mm$$|\Delta X_2| = |1.23 - 1.21| = 0.02\,mm$$|\Delta X_3| = |1.19 - 1.21| = 0.02\,mm$$|\Delta X_4| = |1.20 - 1.21| = 0.01\,mm$Step $3$: Calculate the mean absolute error.$\Delta X_{mean} = \frac{0.01 + 0.02 + 0.02 + 0.01}{4} = \frac{0.06}{4} = 0.015\,mm$.Step $4$: Calculate the percentage error.$	ext{Percentage error} = \left( \frac{\Delta X_{mean}}{X_{mean}} ight) 	imes 100\% = \left( \frac{0.015}{1.21} ight) 	imes 100\% = \frac{1.5}{1.21}\% = \frac{150}{121}\%$.Comparing this with $\frac{x}{121}\%$,we get $x = 150$.

$A$ student in the laboratory measures the thickness of a wire using a screw gauge. The readings are $1.22\,mm$,$1.23\,mm$,$1.19\,mm$,and $1.20\,mm$. The percentage error is $\frac{x}{121}\%$. The value of $x$ is ..............

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