Question

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

• A. $\pm 0.01c{m}^2$
• B. $\pm 0.1c{m}^2$
• C. $\pm 0.11c{m}^2$
• D. $\pm 0.2c{m}^2$

Answer 1

The correct option is D $\pm 0.2c{m}^2$

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