Question

A wire of length \[l = 6 + 0.06\;{\text{cm}}\]and radius \[r = 0.5 + 0.005\;{\text{cm}}\] has mass \[m = 0.3 + 0.003\;{\text{g}}\].
 Maximum percentage error in density is
A. \[{\text{4}}\]

B. \[{\text{2}}\]

C. \[{\text{1}}\]

D. \[{\text{6}}{\text{.8}}\]

Answer 1

Hint: The given components of the wire are length, radius, and mass. By using a basic formula, the percentage error for a variable can be determined-( error)\[ \times 100/\] real reading.
 Obtain the error in volume and mass and then add them to obtain the error in density.

Stepwise solution:

Given,
\[l = 6 + 0.06\;{\text{cm}}\]

\[r = 0.5 + 0.005\;{\text{cm}}\]

\[m = 0.3 + 0.003\;{\text{g}}\]

The percentage error for a variable can be determined by using a basic formula,
Percentage error\[ = \] (error)\[ \times 100/\] real reading.

Now we obtain the percentage error of the length, radius and mass of the wire using the above formula.

Percentage error in length
\[
   = \dfrac{{\left( {0.06} \right) \times 100}}
{6}\% \\
   = 1\% \\
\]

Percentage error in radius
\[
   = \dfrac{{\left( {0.005} \right) \times 100}}
{{0.5}}\% \\
   = 1\% \\
\]

Percentage error in mass
\[
   = \dfrac{{\left( {0.003} \right) \times 100}}
{{0.3}}\% \\
   = 1\% \\
 \]


Now, we need to bear in mind that error is indeed a logarithm while we try to measure the percentage error in a formula-it does not increase in proportion to the reading. So note when multiplying (or splitting) to apply the percentage error then multiplies the strength by the exponentiation error.

For the wire, the density is obtained as, \[\rho = \dfrac{{{\text{mass}}}}
{{{\text{volume}}}}\]

For the wire, the volume is obtained as,\[V = \pi {r^2}l\]
               …… (i)
Therefore, the density is \[\rho = \dfrac{m}
{{\pi {r^2}l}}\]


Now, from equation (i) the error in the volume will be,
Error in volume \[ = 2 \times \] error in radius \[ + \] error in length.
\[
   = 2 \times 1\% + 1\% \\
   = 3\% \\
\]


Now, take the error that we had in the masses into account. And we need to add the error because we are using this mass in division\[/\]
multiplication.

Therefore, the total error in density \[ = \]
error in volume \[ + \]
error in mass
\[
   = 3\% + 1\% \\
   = 4\% \\
\]


Hence, the total error in density is \[4\% \]


Correct option is D.

Note:
In this question we are asked to determine the error in density. The given components along with their errors are length, radius, and mass of the wire. Formula for density is mass per volume. The formula for volume is \[V = \pi {r^2}l\]
Do not get confused when you determine the final error. Do not use this formula, the total error in density \[ = \] error in mass\[ - \] error in volume.
Instead use, the total error in density \[=\] error in volume \[ + \] error in mass