Question:
A certain body weighs $22.42\,g$ and has a measured volume of $4.7\,cc$. The possible error in the measurement of mass and volume are $0.01\,gm$ and $0.1\,cc$. Then maximum error in the density will be
A certain body weighs $22.42\,g$ and has a measured volume of $4.7\,cc$. The possible error in the measurement of mass and volume are $0.01\,gm$ and $0.1\,cc$. Then maximum error in the density will be
Updated On: Jun 25, 2024
- $22\%$
- $2\%$
- $0.2\%$
- $0.02\%$
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The Correct Option is B
Solution and Explanation
Density $\rho=\frac{mass \, m}{volume \, V} .....(i)$
Take logarithm to take base $e$ on the both sides of eqn $(i)$, we get
$ ln\rho=lnm-lnV .....(ii)$
Differentiate eqn $(ii)$, on both sides, we get
$ \frac{? \rho}{\rho}=\frac{? m}{m}-\frac{? V}{V}$
Errors are always added, Error in the density $\rho$ will be
$ =\big[\frac{? m}{m}+\frac{? V}{V}\big]\times100 \%$
$=\big[\frac{0.01}{22.42}+\frac{0.1}{4.7}\big]\times100 \%$
$=2 \%$
Take logarithm to take base $e$ on the both sides of eqn $(i)$, we get
$ ln\rho=lnm-lnV .....(ii)$
Differentiate eqn $(ii)$, on both sides, we get
$ \frac{? \rho}{\rho}=\frac{? m}{m}-\frac{? V}{V}$
Errors are always added, Error in the density $\rho$ will be
$ =\big[\frac{? m}{m}+\frac{? V}{V}\big]\times100 \%$
$=\big[\frac{0.01}{22.42}+\frac{0.1}{4.7}\big]\times100 \%$
$=2 \%$
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Concepts Used:
Physical World
The physical world includes the complications of the natural world around us. It is a type of analysis of the physical world around us to understand how it works. The fundamental forces that control nature are:
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