Q. The length of a strip measured with a meter rod is $10.0\, cm$. Its width measured with a vernier callipers is $1.00\, cm$. The least count of the meter rod is $0.1\, cm$ and that of vernier callipers is $0.01\, cm$. What will be the error in its area?

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A

$\pm 0.01\, cm ^{2}$

B

$\pm 0.1\, cm ^{2}$

C

$\pm 0.11\, cm ^{2}$

D

$\pm 0.2\, cm ^{2}$

Solution:

$\Delta A =\left(\frac{\Delta l}{l}+\frac{\Delta b}{b}\right) A=\pm\left(\frac{0.1}{10.0}+\frac{0.01}{1.00}\right) \times(10.0 \times 1.00) cm ^{2}$
$=\pm 0.02 \times 10=\pm 0.2\, cm ^{2}$

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