Question
The breadth of the meter stick,b= 2.75±0.05
Thickness of the meter stick,t=0.7±0.05
The diameter each of ten identical nails,d=5.280, 3.715, 4.020, 3.715, 2.690, 3.640, 3.610, 3.605, 2.660, 3.650, ±0.05
a) determine the volume of the meter rule,v
b) what is the absolute E or maximum error in v?
c) determine the main diameter of the nails, and the standard deviation (Sv) of the nails.
d) what is the precise value of the radius of the nails?
Thickness of the meter stick,t=0.7±0.05
The diameter each of ten identical nails,d=5.280, 3.715, 4.020, 3.715, 2.690, 3.640, 3.610, 3.605, 2.660, 3.650, ±0.05
a) determine the volume of the meter rule,v
b) what is the absolute E or maximum error in v?
c) determine the main diameter of the nails, and the standard deviation (Sv) of the nails.
d) what is the precise value of the radius of the nails?
Answers
GPT-4o mini
Let's break down the problem step by step.
### a) Determine the Volume of the Meter Rule (v)
The volume \( v \) of a rectangular prism is given by the formula:
\[
v = b \times t \times L
\]
where:
- \( b \) is the breadth,
- \( t \) is the thickness,
- \( L \) is the length.
Assuming the length of the meter stick is \( L = 1 \, \text{meter} \) (which is standard for a meter stick), we have:
\[
b = 2.75 \, \text{m} \quad ( \text{with uncertainty} \pm 0.05\, \text{m})
\]
\[
t = 0.7 \, \text{m} \quad ( \text{with uncertainty} \pm 0.05 \, \text{m})
\]
\[
L = 1 \, \text{m}
\]
Now substituting the values into the volume formula:
\[
v = 2.75 \times 0.7 \times 1 = 1.925 \, \text{m}^3
\]
### b) What is the Absolute Error or Maximum Error in v?
The absolute error in volume can be calculated using the formula for propagation of uncertainty with multiplication. For \( v = b \cdot t \cdot L \):
\[
\left( \frac{\Delta v}{v} \right) = \left( \frac{\Delta b}{b} \right) + \left( \frac{\Delta t}{t} \right)
\]
where:
- \( \Delta b = 0.05 \, \text{m} \)
- \( \Delta t = 0.05 \, \text{m} \)
Calculating the relative uncertainties:
For \( b \):
\[
\frac{\Delta b}{b} = \frac{0.05}{2.75} \approx 0.01818
\]
For \( t \):
\[
\frac{\Delta t}{t} = \frac{0.05}{0.7} \approx 0.07143
\]
So summing these:
\[
\left( \frac{\Delta v}{v} \right) \approx 0.01818 + 0.07143 \approx 0.08961
\]
Now, find the absolute error \( \Delta v \):
\[
\Delta v = v \times \left( \frac{\Delta v}{v} \right) = 1.925 \times 0.08961 \approx 0.172
\]
Thus, the volume and its uncertainty:
\[
v = 1.925 \pm 0.172 \, \text{m}^3
\]
### c) Determine the Main Diameter of the Nails and the Standard Deviation (Sv) of the Nails
The diameters of the nails are:
- \( d_1 = 5.280 \)
- \( d_2 = 3.715 \)
- \( d_3 = 4.020 \)
- \( d_4 = 3.715 \)
- \( d_5 = 2.690 \)
- \( d_6 = 3.640 \)
- \( d_7 = 3.610 \)
- \( d_8 = 3.605 \)
- \( d_9 = 2.660 \)
- \( d_{10} = 3.650 \)
Now, let's find the main (mean) diameter (\(\bar{d}\)):
\[
\bar{d} = \frac{\sum d_i}{n} = \frac{5.280 + 3.715 + 4.020 + 3.715 + 2.690 + 3.640 + 3.610 + 3.605 + 2.660 + 3.650}{10}
\]
Calculating the sum:
\[
\sum d_i \approx 5.280 + 3.715 + 4.020 + 3.715 + 2.690 + 3.640 + 3.610 + 3.605 + 2.660 + 3.650 \approx 37.090
\]
Calculating the mean:
\[
\bar{d} = \frac{37.090}{10} = 3.709 \, \text{(approximately)}
\]
Now, let's calculate the standard deviation \( S_d \):
\[
S_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n}}
\]
Now calculating each \( (d_i - \bar{d})^2 \):
1. \((5.280 - 3.709)^2 \approx 2.46376\)
2. \((3.715 - 3.709)^2 \approx 0.000036\)
3. \((4.020 - 3.709)^2 \approx 0.096961\)
4. \((3.715 - 3.709)^2 \approx 0.000036\)
5. \((2.690 - 3.709)^2 \approx 1.038576\)
6. \((3.640 - 3.709)^2 \approx 0.004761\)
7. \((3.610 - 3.709)^2 \approx 0.009801\)
8. \((3.605 - 3.709)^2 \approx 0.011556\)
9. \((2.660 - 3.709)^2 \approx 1.094929\)
10. \((3.650 - 3.709)^2 \approx 0.003601\)
Sum of squared differences:
\[
\sum (d_i - \bar{d})^2 \approx 2.46376 + 0.000036 + 0.096961 + 0.000036 + 1.038576 + 0.004761 + 0.009801 + 0.011556 + 1.094929 + 0.003601 \approx 4.617
\]
Now, calculate standard deviation \( S_d \):
\[
S_d = \sqrt{\frac{4.617}{10}} \approx 0.680
\]
### d) What is the Precise Value of the Radius of the Nails?
The radius \( r \) of each nail is given by:
\[
r = \frac{d}{2}
\]
So, using the mean diameter:
\[
r = \frac{\bar{d}}{2} = \frac{3.709}{2} \approx 1.8545 \, \text{(approximately)}
\]
### Summary of Results:
- **Volume of the Meter Rule:** \( v = 1.925 \pm 0.172 \, \text{m}^3 \)
- **Mean Diameter of the Nails:** \( \bar{d} \approx 3.709 \, \text{mm} \)
- **Standard Deviation of the Nails:** \( S_d \approx 0.680 \, \text{mm} \)
- **Radius of the Nails:** \( r \approx 1.8545 \, \text{mm} \)
### a) Determine the Volume of the Meter Rule (v)
The volume \( v \) of a rectangular prism is given by the formula:
\[
v = b \times t \times L
\]
where:
- \( b \) is the breadth,
- \( t \) is the thickness,
- \( L \) is the length.
Assuming the length of the meter stick is \( L = 1 \, \text{meter} \) (which is standard for a meter stick), we have:
\[
b = 2.75 \, \text{m} \quad ( \text{with uncertainty} \pm 0.05\, \text{m})
\]
\[
t = 0.7 \, \text{m} \quad ( \text{with uncertainty} \pm 0.05 \, \text{m})
\]
\[
L = 1 \, \text{m}
\]
Now substituting the values into the volume formula:
\[
v = 2.75 \times 0.7 \times 1 = 1.925 \, \text{m}^3
\]
### b) What is the Absolute Error or Maximum Error in v?
The absolute error in volume can be calculated using the formula for propagation of uncertainty with multiplication. For \( v = b \cdot t \cdot L \):
\[
\left( \frac{\Delta v}{v} \right) = \left( \frac{\Delta b}{b} \right) + \left( \frac{\Delta t}{t} \right)
\]
where:
- \( \Delta b = 0.05 \, \text{m} \)
- \( \Delta t = 0.05 \, \text{m} \)
Calculating the relative uncertainties:
For \( b \):
\[
\frac{\Delta b}{b} = \frac{0.05}{2.75} \approx 0.01818
\]
For \( t \):
\[
\frac{\Delta t}{t} = \frac{0.05}{0.7} \approx 0.07143
\]
So summing these:
\[
\left( \frac{\Delta v}{v} \right) \approx 0.01818 + 0.07143 \approx 0.08961
\]
Now, find the absolute error \( \Delta v \):
\[
\Delta v = v \times \left( \frac{\Delta v}{v} \right) = 1.925 \times 0.08961 \approx 0.172
\]
Thus, the volume and its uncertainty:
\[
v = 1.925 \pm 0.172 \, \text{m}^3
\]
### c) Determine the Main Diameter of the Nails and the Standard Deviation (Sv) of the Nails
The diameters of the nails are:
- \( d_1 = 5.280 \)
- \( d_2 = 3.715 \)
- \( d_3 = 4.020 \)
- \( d_4 = 3.715 \)
- \( d_5 = 2.690 \)
- \( d_6 = 3.640 \)
- \( d_7 = 3.610 \)
- \( d_8 = 3.605 \)
- \( d_9 = 2.660 \)
- \( d_{10} = 3.650 \)
Now, let's find the main (mean) diameter (\(\bar{d}\)):
\[
\bar{d} = \frac{\sum d_i}{n} = \frac{5.280 + 3.715 + 4.020 + 3.715 + 2.690 + 3.640 + 3.610 + 3.605 + 2.660 + 3.650}{10}
\]
Calculating the sum:
\[
\sum d_i \approx 5.280 + 3.715 + 4.020 + 3.715 + 2.690 + 3.640 + 3.610 + 3.605 + 2.660 + 3.650 \approx 37.090
\]
Calculating the mean:
\[
\bar{d} = \frac{37.090}{10} = 3.709 \, \text{(approximately)}
\]
Now, let's calculate the standard deviation \( S_d \):
\[
S_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n}}
\]
Now calculating each \( (d_i - \bar{d})^2 \):
1. \((5.280 - 3.709)^2 \approx 2.46376\)
2. \((3.715 - 3.709)^2 \approx 0.000036\)
3. \((4.020 - 3.709)^2 \approx 0.096961\)
4. \((3.715 - 3.709)^2 \approx 0.000036\)
5. \((2.690 - 3.709)^2 \approx 1.038576\)
6. \((3.640 - 3.709)^2 \approx 0.004761\)
7. \((3.610 - 3.709)^2 \approx 0.009801\)
8. \((3.605 - 3.709)^2 \approx 0.011556\)
9. \((2.660 - 3.709)^2 \approx 1.094929\)
10. \((3.650 - 3.709)^2 \approx 0.003601\)
Sum of squared differences:
\[
\sum (d_i - \bar{d})^2 \approx 2.46376 + 0.000036 + 0.096961 + 0.000036 + 1.038576 + 0.004761 + 0.009801 + 0.011556 + 1.094929 + 0.003601 \approx 4.617
\]
Now, calculate standard deviation \( S_d \):
\[
S_d = \sqrt{\frac{4.617}{10}} \approx 0.680
\]
### d) What is the Precise Value of the Radius of the Nails?
The radius \( r \) of each nail is given by:
\[
r = \frac{d}{2}
\]
So, using the mean diameter:
\[
r = \frac{\bar{d}}{2} = \frac{3.709}{2} \approx 1.8545 \, \text{(approximately)}
\]
### Summary of Results:
- **Volume of the Meter Rule:** \( v = 1.925 \pm 0.172 \, \text{m}^3 \)
- **Mean Diameter of the Nails:** \( \bar{d} \approx 3.709 \, \text{mm} \)
- **Standard Deviation of the Nails:** \( S_d \approx 0.680 \, \text{mm} \)
- **Radius of the Nails:** \( r \approx 1.8545 \, \text{mm} \)