The following data is given for calculating density of cylinder

Mass=$6.7\pm 0.1 g$

Radius=$0.087\pm 0.001 cm$

Length=$3.28\pm 0.01 cm$


Let density by d

$$\frac{\Delta d}{d}=\frac{\Delta m}{m}+\frac{\Delta l}{l} + \frac{2\Delta r}{r}$$

Plugging in the values, we end up with something around 4.69%. The answer is 2.4%.

I have seen the solution for this question, and in that, they took the original equation as

$$\frac{\Delta d}{d}=\frac{\Delta m}{m}+ \frac{\Delta r}{2r} + \frac{\Delta l}{l}$$

The main difference in the both solutions is the denominator of the relative error in radius. As far as I know, we multiple the powers in the numerator. So why did they take it like that?


1 Answer 1


Your equation for the error in the density is correct. The $2$ should be in the numerator of the radius term because the factor in the volume is $r^2$. I get about $4.096\%$ error (though I would report fewer decimals, I show them for comparison). The solution manual is wrong in the equation, then evaluates it correctly.

  • $\begingroup$ I even uploaded this question on a third part website, and they did it the same as the manual and achieved the right answer. $\endgroup$
    – Aditya
    Commented Aug 17, 2019 at 15:20
  • $\begingroup$ Just try it. Take the mass and length as exact. If the radius is exact as well, the density is $\frac {6.7}{\pi 0.087^2\cdot 3.28} \approx 85.90$. If the radius is high the density is $\frac {6.7}{\pi 0.088^2\cdot 3.28} \approx 83.96$. A shift of $\frac 1{87}$ in the radius made a shift about $\frac 2{87}$ in the density $\endgroup$ Commented Aug 17, 2019 at 15:28
  • $\begingroup$ That’s true, and then we get 2.29%(approx), but why am I not getting the answer by my calculations? $\endgroup$
    – Aditya
    Commented Aug 17, 2019 at 15:52
  • $\begingroup$ The $2.29\%$ is just the effect of $\Delta r$ because I kept the other two exact. I don't know how you got $4.69\%$ because you didn't show the work. $\endgroup$ Commented Aug 17, 2019 at 18:24
  • $\begingroup$ We get 4.69 by just inputting the values. I don’t see why I need to show it. I used a calc and checked $\endgroup$
    – Aditya
    Commented Aug 18, 2019 at 0:37

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