Question

The mass of a body is \[20.000\] g and its volume is \[10.00c{m^3}\] . If the measured values are expressed to the correct significant figures, the maximum error in the value of density is
A. \[0.001g.c{m^{ - 3}}\]
B. \[0.010g.c{m^{ - 3}}\]
C. \[0.100g.c{m^{ - 3}}\]
D. None of these

Answer 1

Hint: Here we have to know the formula of density. We all know that density is equal to mass upon volume. i.e. \[\dfrac{m}{v}\] here $m$ is equal to mass and $v$ is equal to volume. We also should know the formula of error calculation which we are going to use here to solve this question. There is nothing which is error free. So we should know the calculation.

Complete step by step answer:
Here the formula of density is \[\dfrac{m}{v}\]. Here $m$ is the mass which is given in this question is \[20.000\] g and in this question volume is also given which is equal to \[10.00c{m^3}\]. Now, the density would be \[20.000/10.00 = 2gm/c{m^3}\].
Now according to the formula the error in density is
\[\Delta \rho /\rho = (\Delta m/m) + (\Delta v/v)\] ,
\[ \Rightarrow\Delta \rho /\rho= (0.001/20.000) + (0.01/10) \\
\Rightarrow\Delta \rho /\rho = 1.05 \times {10^{ - 3}}\]
Here, \[\Delta \rho \] is equal to the change in density. Now,
\[\Delta \rho \]\[ = \]\[1.05 \times {10^{ - 3}}\]\[ \times \]\[2\]
\[\Delta \rho = 2.1 \times {10^{ - 3}}
\therefore\Delta \rho = 0.002g/c{m^3}\]

So the right option is D.

Note:We can get confused by the formula of the error calculation. Because this is a tricky one. One should know that nothing is errorless. And also we should calculate the error in the same unit either in the S.I unit or in the C.G.S unit. Many students forget to convert all the units in the same form. Which can cause mark deduction. Here all the units in the question are given in the same category so we need not change it.