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 0.1 cm and that of vernier callipers is 0.01 cm. What will be the error in its area ?

\(\pm\) 0.2 cm²

\(\Delta A = (\frac{\Delta l}{l} + \frac{\Delta b}{b})A\)

\(\Delta A = \pm (\frac{0.1}{10.0} + \frac{0.01}{1.00})(10.0 * 1.00)cm^2\)

\(\Delta A = \pm 0.2 cm^2\)