Question

To measure the volume of a cylinderical wire, its diameter is measured by a screw gauge and is found to be 1.25 mm, and its length is measured with a vernier caliper and is found to be 5.50 cm. The least count of the screw gauge is 0.01 mm, and that of the vernier caliper is 0.01 cm.

• A. Percentage error due to area is more than percentage error due to length.
• B. Number of significant figures in volume will be two.
• C. Percentage error in volume will be 1.78%.
• D. Percentage error in length is 1.6%.

Answer 1

Answer: (A), (C)

Percentage error in volume will be 1.78%.

$A=\frac{\pi D^2}{4}\Rightarrow \frac{\delta A}{A}=\frac{2\delta D}{D}=2\times \frac{0.01}{1.25}=\frac{1.6}{100}$
$\frac{\delta l}{l}=\frac{0.01}{5.5}=\frac{0.18}{100}$
$V=A\times l$
$\Rightarrow \frac{\delta V}{V}=\frac{\delta A}{A}+\frac{\delta l}{l}=\frac{1.78}{100}$