A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm, 1.23 mm, 1.19 mm and 1.20 mm. The percentage error is x \over 121\% . The value of x is \underline\hspace2cm .

Units and Measurements PYQ — JEE Main 2022 · 28 Jun · Shift 2 · Q68 with Solution

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Correct answer: 150

The thickness readings are: $1.22\text{ mm},\;1.23\text{ mm},\;1.19\text{ mm},\;1.20\text{ mm}$

$\bar{x}=\frac{1.22+1.23+1.19+1.20}{4}$

$\bar{x}=\frac{4.84}{4}=1.21\text{ mm}$

$|1.22-1.21|=0.01$ $|1.23-1.21|=0.02$ $|1.19-1.21|=0.02$ $|1.20-1.21|=0.01$

$\Delta x=\frac{0.01+0.02+0.02+0.01}{4}$

$\Delta x=\frac{0.06}{4}=0.015\text{ mm}$

$\%\text{ error}=\frac{\Delta x}{\bar{x}}\times 100$

$\%\text{ error}=\frac{0.015}{1.21}\times 100$

Since $1.21=\frac{121}{100}$ ,

$\frac{0.015}{1.21}\times 100=\frac{0.015\times 100}{1.21}$

$=\frac{1.5}{1.21}=\frac{150}{121}\%$

So this is of the form: $\frac{x}{121}\%$

Hence, $x=150$

Stored correct answer = $150$

This matches our derived answer.