Question

A wire has a mass $(0.3 \pm 0.003)g$, radius $(0.5 \pm 0.005)mm$ and length $(6 \pm 0.06)cm$. The maximum percentage error in the measurement of its density is -
A). 1
B). 2
C). 3
D). 4

Answer 1

Hint: In order to deal with this question we will use the formula of density which is the ratio of the mass of wire to the mass of volume, then we will proceed further by writing it in the form of error percentage.

Formula used: $V = \pi {r^2}L,\rho = \dfrac{m}{V} = \dfrac{m}{{\pi {r^2}L}},\log \dfrac{a}{b} = \log a - \log b$

Complete step-by-step solution:
Given that
$
  m + \Delta m = (0.3 \pm 0.003)g \\
  r + \Delta r = (0.5 \pm 0.005)mm \\
  L + \Delta L = (6 \pm 0.06)cm $
We know that the volume of the wire is given by, $V = \pi {r^2}L$
Thus, the density of the wire, $\rho = \dfrac{m}{V} = \dfrac{m}{{\pi {r^2}L}}$
Taking log on both the sides, we get
$
  \because \rho = \dfrac{m}{{\pi {r^2}L}} \\
   \Rightarrow \log (\rho ) = \log \left( {\dfrac{m}{{\pi {r^2}L}}} \right) \\
   \Rightarrow \log (\rho ) = \log \left( m \right) - \log (\pi {r^2}L){\text{ }}\left[ {\because \log \dfrac{a}{b} = \log a - \log b} \right] $
Now after differentiating both the side, we get:
$
   \Rightarrow \dfrac{1}{\rho }d\rho = \dfrac{1}{m}dm + 2\dfrac{1}{r}dr + \dfrac{1}{L}dL \\
   \Rightarrow \dfrac{{\Delta \rho }}{\rho } = \dfrac{{\Delta m}}{m} + 2\dfrac{{\Delta r}}{r} + \dfrac{{\Delta L}}{L} \\
  \therefore \dfrac{{\Delta \rho }}{\rho } \times 100 = \dfrac{{\Delta m}}{m} \times 100 + 2\dfrac{{\Delta r}}{r} \times 100 + \dfrac{{\Delta L}}{L} \times 100 $
Notice that we had a negative sign which changed to positive after differentiating , this is because we are trying to compute the error in the measurement or the deviation of the result so it is cumulative i.e. irrespective of sign it will be positive.
Now, Let us substitute all the values known to us in the above-found formula, we get
$
  \dfrac{{\Delta \rho }}{\rho } \times 100 = \dfrac{{\Delta m}}{m} \times 100 + 2\dfrac{{\Delta r}}{r} \times 100 + \dfrac{{\Delta L}}{L} \times 100 \\
   \Rightarrow \dfrac{{\Delta \rho }}{\rho } \times 100 = \dfrac{{0.003}}{{0.3}} \times 100 + 2\dfrac{{0.005}}{{0.5}} \times 100 + \dfrac{{0.06}}{6} \times 100 \\
   \Rightarrow \dfrac{{\Delta \rho }}{\rho } \times 100 = 1 + 2 + 1 \\
   \Rightarrow \dfrac{{\Delta \rho }}{\rho } \times 100 = 4\% $
Hence, the maximum percentage error in the measurement of its density is 4%
So, the correct answer is option D.

Note: Error is the difference between the actual value of any physical quantity and its calculated value. In physics, there are essentially three types of mistakes, spontaneous mistakes, blunders, and systematic errors. In some data, the approximation error occurs between an exact value of the quantity under consideration and the approximation to it. An approximation error may occur because the data measured by the instrument is not accurate Or we use approximations instead of the real details.