# Finding the percentage error in the measurement of density.

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=$$3.28\pm 0.01 cm$$

MY SOLUTION-

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?

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.
• 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 Commented Aug 17, 2019 at 15:28
• 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. Commented Aug 17, 2019 at 18:24
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