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Home » Measure of two quantities along with the precision of the respective measuring instrument is:A = 2.5 m/s ± 0.5 m/sB = 0.10 s ± 0.01 sThe value of AB will bea) (0.25 ± 0.08) mb) (0.25 ± 0.5) mc) (0.25 ± 0.05) md) (0.25 ± 0.135) m
Measure of two quantities along with the precision of the respective measuring instrument is:
A = 2.5 m/s ± 0.5 m/s
B = 0.10 s ± 0.01 s
The value of AB will be
a) (0.25 ± 0.08) m
b) (0.25 ± 0.5) m
c) (0.25 ± 0.05) m
d) (0.25 ± 0.135) m
CBSE | Class 11 | NCERT Exemplar | Physics | Units and Measurements
Measure of two quantities along with the precision of the respective measuring instrument is:
A = 2.5 m/s ± 0.5 m/s
B = 0.10 s ± 0.01 s
The value of AB will be
a) (0.25 ± 0.08) m
b) (0.25 ± 0.5) m
c) (0.25 ± 0.05) m
d) (0.25 ± 0.135) m

Correct answer is a) (0.25 ± 0.08) m

Here,

\mathrm{A}=2.5 \mathrm{~ms}^{-1} \pm 0.5 \mathrm{~ms}^{-1}, \mathrm{~B}=0.10 \mathrm{~s} \pm 0.01 \mathrm{~s}

\mathrm{AB}=\left(2.5 \mathrm{~ms}^{-1}\right)(0.10 \mathrm{~s})=0.25 \mathrm{~m} \frac{\Delta \mathrm{AB}}{\mathrm{AB}}=\left(\frac{\Delta \mathrm{A}}{\mathrm{A}}+\frac{\Delta \mathrm{B}}{\mathrm{B}}\right)=\left(\frac{0.5}{2.5}+\frac{0.01}{0.10}\right)=0.3

\Delta \mathrm{AB}=0.3 \times 0.25 \mathrm{~m}=0.075 \mathrm{~m}=0.08 \mathrm{~m}

The value of \mathrm{AB}(0.25 \pm 0.8) \mathrm{m}

i

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