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Alchen [17]
3 years ago
14

A student measures the diameter of a small cylindrical object and gets the following readings: 4.32, 4.35, 4.31, 4.36, 4.37, 4.3

4 cm (a) What is the average diameter from these readings? (b) What is the standard deviation of these measurements? The student also measured the length of the object to be (0.126 ± 0.005) m and the mass to be object to be (1.66 ± 0.05) kg. Using the method from this week's lab, determine (c) the density and (d) the proportion of error in the density calculation.
Physics
1 answer:
Zinaida [17]3 years ago
3 0

Answer:

a. \bar{d}=4.34 cm

b. \sigma=0.023 cm

c. \rho=(0.0089\pm 0.00058) kg/cm^{3}

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}

  • x_{i} is values of each diameter
  • N is the total number of values. N=6

\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}

\bar{d}=4.34 cm

b) The standard deviation equations is:

\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}

If we put all this values in that equation we will get:

\sigma=0.023 cm

Then the mean diameter will be:

\bar{d}=(4.34\pm 0.023)cm

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: V=\pi R^{2}h

Then:

\rho=\frac{m}{\pi R^{2}h}

Using the values that we have, we can calculate the value of density:

\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}

We need to use propagation of error to find the error of the density.

\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}  

  • δm is the error of the mass value.
  • δd is the error of the diameter value.
  • δh is the error of the length value.

Let's find each partial derivative:

1. \frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089

2.  \frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004

3. \frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071

Therefore:

\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}

\delta\rho=0.00058

So the density is:

\rho=(0.0089\pm 0.00058) kg/cm^{3}

I hope it helps you!

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