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Home » The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm respectively. The area of the sheet inappropriate significant figures and error is:a) 164 ± 3 b) 163.62 ± 2.6 c) 163.6 ± 2.6 d) 163.62 ± 3
The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm respectively. The area of the sheet inappropriate significant figures and error is:
a) 164 ± 3 cm^{2}
b) 163.62 ± 2.6 cm^{2}
c) 163.6 ± 2.6 cm^{2}
d) 163.62 ± 3 cm^{2}
CBSE | Class 11 | NCERT Exemplar | Physics | Units and Measurements
The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm respectively. The area of the sheet inappropriate significant figures and error is:
a) 164 ± 3 cm^{2}
b) 163.62 ± 2.6 cm^{2}
c) 163.6 ± 2.6 cm^{2}
d) 163.62 ± 3 cm^{2}

Correct answer is a) 164 ± 3 cm^{2}

Error in product of quantities: Suppose x=a \times b
Let,

\Delta a be the absolute error in measurement of a,
\Delta b be the absolute error in measurement of b,
\Delta x be the absolute error in calculation of x, i.e. product of a and b.

The maximum fractional error in \mathrm{x} can be calculated as,

\frac{\Delta \mathrm{x}}{\mathrm{x}}=\pm\left(\frac{\Delta \mathrm{a}}{\mathrm{a}}+\frac{\Delta \mathrm{b}}{\mathrm{b}}\right).

Percentage error in the value of \mathrm{x} is given as the sum of Percentage error in value of a and Percentage error in value of b

According to the given question, length is l=(16.2 \pm 0.1) \mathrm{cm}

Breadth is \mathrm{b}=(10.1 \pm 0.1) \mathrm{cm}

Area will be \mathrm{A}=1 \times \mathrm{b}=(16.2 \mathrm{~cm}) \times(10.1 \mathrm{~cm})=163.62 \mathrm{~cm}^{2}

According to the norm, the area will only have three significant figures, while the error will only have one significant figure. We receive area as a result of rounding off.

\mathrm{A}=164 \mathrm{~cm}^{2}
If \Delta \mathrm{A} is error in the area, then relative error is calculated as \frac{\delta 4}{\mathrm{~A}}.

\frac{\Delta 4}{\mathrm{~A}}=\frac{\Delta \mathrm{l}}{\mathrm{l}}+\frac{\Delta \mathrm{b}}{\mathrm{b}}=\frac{0.1 \mathrm{~cm}}{16.2 \mathrm{~cm}}+\frac{0.1 \mathrm{~cm}}{10.1 \mathrm{~cm}}

=\frac{1.01+1.62}{16.2 \times 10.1}=\frac{2.63}{163.62}

\Rightarrow \Delta \mathrm{A}=\mathrm{A} \times \frac{2.63}{163.62} \mathrm{~cm}^{2}=162.62 \times \frac{2.63}{163.62}=2.63 \mathrm{~cm}^{2}
\Delta \mathrm{A}=3 \mathrm{~cm}^{2} (By rounding off to one significant figure)
Area, \mathrm{A}=\mathrm{A} \pm \Delta \mathrm{A}(164 \pm 3) \mathrm{cm}^{2}

i

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