Question

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E. Questions Consider the cylinder case. The uncertainty in any one measurement of the density can be determined in the following way. (a) Take the instrumental limit of error associated with the Vernier calipers used to be one-half the least count or smallest measureable increment 8lVernier = 0.001 cm Assume the uncertainty in all linear measurements to be the same and equal to the result above, i.e., 81 = 8D = 8lvernier. The propagated error formulas for volume and density can be combined so that (5ppтор 18m) ² - 11 - (9) + + T / 72 m ' P If the uncertainty in the mass measurement (i.e., the mass balance's limit of error) is taken to be 8m = 0.05 g, then use the following set of trial values me = 27.50 g 1 = 7.690 cm D = 1.260 cm to determine peyl, and use Eq. (9) to determine Spprop: Pey] =- 8Pprop =- _[kg/m²) _[kg/m) (b) This value calculated for 8 pprop should be small and should not account for any difference between your experimental determination, Peyl, and the accepted value. i.e., Ap= Peyi - Pail = - _[kg/m] Therefore the length and mass measurements are not responsible for the disc ancy. A possible source for the discrepancy is a deviation from perfe cylinder. The diameter varies along the length, and the length varies a cross-section. If the measurements were done carefully, these variati ancy is a deviation from perfection in the h, and the length varies across the arefully, these variations would

Answer 1

2. Suppose that one measures the area of a small rectangular plate very accurately by making five measurements at different locations along its length and width with the following results in cm :

(a) The mean length of a small rectangular plate which will be given by -

$\bar{L}$_avg = (sum of all data points) / (total number of data points)

$\bar{L}$_avg = [(5.425 cm) + (5.422 cm) + (5.422 cm) + (5.423 cm) + (5.424 cm)] / (5)

$\bar{L}$_avg = (27.116 cm) / (5)

$\bar{L}$_avg = 5.4232 cm

(b) The standard error in the length which will be given by -

$\delta$L = _$\sqrt{}$(1 / N) (L_i - $\bar{L}$_avg)²

where, (L_i - $\bar{L}$_avg)² = [(5.425 cm) - (5.4232 cm)]² = 3.24 x 10^-6 cm²

[(5.422 cm) - (5.4232 cm)]² = 1.44 x 10^-6 cm²

[(5.422 cm) - (5.4232 cm)]² = 1.44 x 10^-6 cm²

[(5.423 cm) - (5.4232 cm)]² = 4 x 10^-8 cm²

[(5.424 cm) - (5.4232 cm)]² = 6.4 x 10^-7 cm²

then, we get

$\delta$L = _$\sqrt{}$(1 / 5) [(3.24 x 10^-6 cm²) + (1.44 x 10^-6 cm²) + (1.44 x 10^-6 cm²) + (4 x 10^-8 cm²) + (6.4 x 10^-7 cm²)]

$\delta$L = _$\sqrt{}$(1 / 5) (6.8 x 10^-6 cm²)

$\delta$L = _$\sqrt{}$1.36 x 10^-6 cm²

$\delta$L = 0.00116 cm

3. In the above problem, we have

(a) The mean width of a small rectangular plate which will be given by -

$\bar{w}$_avg = (sum of all data points) / (total number of data points)

$\bar{w}$_avg = [(5.036 cm) + (5.039 cm) + (5.041 cm) + (5.036 cm) + (5.036 cm)] / (5)

$\bar{w}$_avg = (25.188 cm) / (5)

$\bar{w}$_avg = 5.0376 cm

(b) The standard error in the width of a small rectangular plate which will be given by -

$\delta$w = _$\sqrt{}$(1 / N) (w_i - $\bar{w}$_avg)²

where, (w_i - $\bar{w}$_avg)² = [(5.036 cm) - (5.0376 cm)]² = 2.56 x 10^-6 cm²

[(5.039 cm) - (5.0376 cm)]² = 1.96 x 10^-6 cm²

[(5.041 cm) - (5.0376 cm)]² = 1.156 x 10^-5 cm²

[(5.036 cm) - (5.0376 cm)]² = 2.56 x 10^-6 cm²

[(5.036 cm) - (5.0376 cm)]² = 2.56 x 10^-6 cm²

then, we get

$\delta$w = _$\sqrt{}$(1 / 5) [(2.56 x 10^-6 cm²) + (1.96 x 10^-6 cm²) + (1.156 x 10^-5 cm²) + (2.56 x 10^-6 cm²) + (2.56 x 10^-6 cm²)]

$\delta$w = _$\sqrt{}$(1 / 5) (2.12 x 10^-5 cm²)

$\delta$w = _$\sqrt{}$4.24 x 10^-6 cm²

$\delta$w = 0.00206 cm

4. In the same problem, the average area of a small rectangular plate which will be given by -

A_avg = $\bar{L}$_avg x $\bar{w}$_avg

A_avg = (5.4232 cm) x (5.0376 cm)

A_avg = 27.3 cm²

5. The propagated error in the area of a small rectangular plate which will be given by -

$\delta$A = A_avg$\sqrt{}$($\delta$L / $\bar{L}$_avg)² + ($\delta$w / $\bar{w}$_avg)²

$\delta$A = (27.3 cm²) _$\sqrt{}$[(0.00116 cm) / (5.4232 cm)]² + [(0.00206 cm) / (5.0376 cm)]²

$\delta$A = (27.3 cm²) _$\sqrt{}$[(4.575 x 10^-8) + (1.672 x 10^-7)]

$\delta$A = (27.3 cm²) _$\sqrt{}$2.1295 x 10^-7

$\delta$A = [(27.3 cm²) (0.0004614)]

$\delta$A = 0.0126 cm²

Answer 2

2. Suppose that one measures the area of a small rectangular plate very accurately by making five measurements at different locations along its length and width with the following results in cm :

(a) The mean length of a small rectangular plate which will be given by -

$\bar{L}$_avg = (sum of all data points) / (total number of data points)

$\bar{L}$_avg = [(5.425 cm) + (5.422 cm) + (5.422 cm) + (5.423 cm) + (5.424 cm)] / (5)

$\bar{L}$_avg = (27.116 cm) / (5)

$\bar{L}$_avg = 5.4232 cm

(b) The standard error in the length which will be given by -

$\delta$L = _$\sqrt{}$(1 / N) (L_i - $\bar{L}$_avg)²

where, (L_i - $\bar{L}$_avg)² = [(5.425 cm) - (5.4232 cm)]² = 3.24 x 10^-6 cm²

[(5.422 cm) - (5.4232 cm)]² = 1.44 x 10^-6 cm²

[(5.422 cm) - (5.4232 cm)]² = 1.44 x 10^-6 cm²

[(5.423 cm) - (5.4232 cm)]² = 4 x 10^-8 cm²

[(5.424 cm) - (5.4232 cm)]² = 6.4 x 10^-7 cm²

then, we get

$\delta$L = _$\sqrt{}$(1 / 5) [(3.24 x 10^-6 cm²) + (1.44 x 10^-6 cm²) + (1.44 x 10^-6 cm²) + (4 x 10^-8 cm²) + (6.4 x 10^-7 cm²)]

$\delta$L = _$\sqrt{}$(1 / 5) (6.8 x 10^-6 cm²)

$\delta$L = _$\sqrt{}$1.36 x 10^-6 cm²

$\delta$L = 0.00116 cm

3. In the above problem, we have

(a) The mean width of a small rectangular plate which will be given by -

$\bar{w}$_avg = (sum of all data points) / (total number of data points)

$\bar{w}$_avg = [(5.036 cm) + (5.039 cm) + (5.041 cm) + (5.036 cm) + (5.036 cm)] / (5)

$\bar{w}$_avg = (25.188 cm) / (5)

$\bar{w}$_avg = 5.0376 cm

(b) The standard error in the width of a small rectangular plate which will be given by -

$\delta$w = _$\sqrt{}$(1 / N) (w_i - $\bar{w}$_avg)²

where, (w_i - $\bar{w}$_avg)² = [(5.036 cm) - (5.0376 cm)]² = 2.56 x 10^-6 cm²

[(5.039 cm) - (5.0376 cm)]² = 1.96 x 10^-6 cm²

[(5.041 cm) - (5.0376 cm)]² = 1.156 x 10^-5 cm²

[(5.036 cm) - (5.0376 cm)]² = 2.56 x 10^-6 cm²

[(5.036 cm) - (5.0376 cm)]² = 2.56 x 10^-6 cm²

then, we get

$\delta$w = _$\sqrt{}$(1 / 5) [(2.56 x 10^-6 cm²) + (1.96 x 10^-6 cm²) + (1.156 x 10^-5 cm²) + (2.56 x 10^-6 cm²) + (2.56 x 10^-6 cm²)]

$\delta$w = _$\sqrt{}$(1 / 5) (2.12 x 10^-5 cm²)

$\delta$w = _$\sqrt{}$4.24 x 10^-6 cm²

$\delta$w = 0.00206 cm

4. In the same problem, the average area of a small rectangular plate which will be given by -

A_avg = $\bar{L}$_avg x $\bar{w}$_avg

A_avg = (5.4232 cm) x (5.0376 cm)

A_avg = 27.3 cm²

5. The propagated error in the area of a small rectangular plate which will be given by -

$\delta$A = A_avg$\sqrt{}$($\delta$L / $\bar{L}$_avg)² + ($\delta$w / $\bar{w}$_avg)²

$\delta$A = (27.3 cm²) _$\sqrt{}$[(0.00116 cm) / (5.4232 cm)]² + [(0.00206 cm) / (5.0376 cm)]²

$\delta$A = (27.3 cm²) _$\sqrt{}$[(4.575 x 10^-8) + (1.672 x 10^-7)]

$\delta$A = (27.3 cm²) _$\sqrt{}$2.1295 x 10^-7

$\delta$A = [(27.3 cm²) (0.0004614)]

$\delta$A = 0.0126 cm²