Question

A certain body weight $22.42g$ and has a measured volume of $4.7cc$. The possible error in the measurement of mass and volume is $0.01gm$ and $0.1cc$. The maximum error in the density will

• A. $22$ %
• B. $2$ %
• C. $0.2$ %
• D. $0.02$%

Answer 1

The correct option is B $2$ %

$m=22.42gm,\Delta m=0.01gm\Rightarrow \frac{\Delta m}{m}=\frac{0.01}{22.42}=\frac{1}{22.42}$

$v=4.7cc,\Delta v=0.1cc\Rightarrow \frac{\Delta v}{v}=\frac{0.1}{4.7}=\frac{1}{4.7}$

density $d=\frac{m}{v}$

$\Rightarrow$ error $\frac{\Delta d}{d}=\frac{\Delta m}{m}-\frac{\Delta v}{v}$

$ma{x}^m\frac{\Delta d}{d}{|}_{max}=\frac{\Delta m}{m}+\frac{\Delta v}{v}$

$\frac{\Delta d}{d}{|}_{max}\times 100\%=(\frac{\Delta m}{m}+\frac{\Delta v}{v})\times 100\%$

$\Rightarrow \frac{\Delta d}{d}{|}_{max}\times 100\%=\frac{1}{2242}\times 100\%+\frac{1}{47}\times 100\%$

$=2.17\%$

nearly 2%