Measurement Assignment (With Solutions)

UMOJA WA WAZAZI TANZANIA

WARI SECONDARY SCHOOL

PHYSICS MEASUREMENT TEST 2

FORM FIVE

Time: 2:00hrs November, 2022.

INSTRUCTIONS

1. This paper consists of four (4).
2. Answer all questions.

01.

(a) Define the following and give one example for each.

(i) Dimension.
(ii) Dimensional analysis.
(iii) Dimensional variable.
(iv) Dimensional constant.
(v) Dimensionless variable.
(vi) Dimensionless constant.

(b)

(i) State (1) significance and (3) limitations of dimensional analysis

(ii) From the equation below find the dimension for a and b.

\[(P + \frac{a}{v^2}) (v - b)\]

Where
P: Pressure
V: Volume

(iii) The velocity v of the particle depends on time t according to the relation

\[v = at + \frac{b}{t+c}\]

What are the dimension of a, b and c.

02.

(a)

(i) Name two physical quantities have same dimension as that of work and vector)

(ii) Name two physical (scalar and vector) quantities that have some dimension.

(iii) State the principle of homogeneity.

(iv) Can dimension analysis tell you the physical relation is completely right? State the reason(s).

(b) By using the method of dimension, derive the expression for energy (E) of the body executing S.H.M, assuming that energy depends on Mass (M), frequency (f) and amplitude (A) of the body.

03.

(a) Convert the following system of unity by using dimensional analysis.

(i) 5gms^{-2} to newton (N)

(ii) 19 \times 10^{10} N/m^2 to gcm^{-1}s^{-2}

(b)

(i) List down two causes of systematic errors.

(ii) Differentiate between precise experiments from accurate experiment.

(iii) Differentiate Dimension and unit.

(iv) Stone of weight (25.0 \pm 0.1)N in air, weight of stone in water is (10 \pm 0.02)N finding the percentage error in measuring specific gravity.

04.

(a)

(i) The rate of flow of the fluid (\frac{v}{L}) in pipe, depend on length (l) of the pipe, pressure difference (Δp), coefficient of viscosity (η) and radius r of the pipe, use the dimension analysis to develop relation between flow rate and other terms.

(ii) The length, width and breadth of the wooden piece was recorded as

\[l = (15.12 \pm 0.01) cm, b = (10.15 \pm 0.01) cm\]

and

\[n = (5.29 \pm 0.01) cm.\]

Find the percentage error in volume of the block.

(iii) Physical quantity is given by P = ab^2c^3d^4. The percentage errors in measurement of a, b, c, and d are all 0.5%. calculate the percentage error in P.

Measurement Assignment 2

Submission: Saturday 17th June 2023 at 0900AM

Question 1: Error Analysis

(a) Definitions

Error: The difference between the measured value and the true value of a quantity.

Mistake: An incorrect action or calculation resulting from human oversight.

(b) Error Comparison

Characteristic	Random Error	Systematic Error
Nature	Unpredictable fluctuations	Consistent deviation in one direction
Cause	Unknown/uncontrolled variables	Instrument calibration or experimental setup
Reduction	Take multiple measurements	Calibrate instruments
Example	Slight variations in stopwatch reaction time	Zero error in vernier calipers

Question 2: True Value Determination

Measurements: 21.02, 20.99, 20.92, 21.11, 20.69

Solution:

Mean value (x̄) = (21.02 + 20.99 + 20.92 + 21.11 + 20.69)/5 = 20.946

Absolute deviations:

• |21.02 - 20.946| = 0.074
• |20.99 - 20.946| = 0.044
• |20.92 - 20.946| = 0.026
• |21.11 - 20.946| = 0.164
• |20.69 - 20.946| = 0.256

Mean absolute error = (0.074 + 0.044 + 0.026 + 0.164 + 0.256)/5 = 0.1128

Final result: (20.946 ± 0.1128) units

Question 3: Percentage Error Propagation

Given: P = (a³b²)/(c^¼d)
% errors: a=1%, b=3%, c=4%, d=2%

Solution:

ΔP/P = 3(Δa/a) + 2(Δb/b) + ¼(Δc/c) + Δd/d
= 3(1%) + 2(3%) + ¼(4%) + 2%
= 3% + 6% + 1% + 2% = 12%

Percentage error in P: 12%

Question 4: Error Calculation

J = (I²R)/(W + m) × t/θ
Given:
I = 2.5 ± 0.05, R = 11.36 ± 0.01
W = 21 ± 1, m = 155 ± 1
θ = 28 ± 0.5, t = 298 ± 0.5

Solution:

Numerical value:

J = (2.5² × 11.36)/(21 + 155) × 298/28 ≈ 4.2943

Error calculation:

ΔJ/J = 2(ΔI/I) + ΔR/R + (ΔW + Δm)/(W + m) + Δt/t + Δθ/θ
= 2(0.05/2.5) + 0.01/11.36 + (1+1)/176 + 0.5/298 + 0.5/28
≈ 0.04 + 0.0009 + 0.0114 + 0.0017 + 0.0179 ≈ 0.0718 (7.18%)

Final result: J = 4.2943 ± 0.308

Question 9: Resistor Combinations

Given: R₁ = (2 ± 0.5)Ω, R₂ = (4 ± 0.5)Ω

(i) Series Arrangement

R_series = R₁ + R₂ = 2 + 4 = 6Ω
ΔR_series = ΔR₁ + ΔR₂ = 0.5 + 0.5 = 1Ω

Result: (6 ± 1)Ω

Percentage error: (1/6)×100% ≈ 16.67%

(ii) Parallel Arrangement

1/R_parallel = 1/R₁ + 1/R₂ ⇒ R = (2×4)/(2+4) ≈ 1.333Ω

ΔR/R = ΔR₁/R₁ + ΔR₂/R₂ + (ΔR₁+ΔR₂)/(R₁+R₂)
= 0.5/2 + 0.5/4 + (0.5+0.5)/6 ≈ 0.25 + 0.125 + 0.1667 ≈ 0.5417 (54.17%)

Absolute error: 0.5417 × 1.333 ≈ 0.722Ω

Final result: (1.333 ± 0.722)Ω

Marking Guide

Question	Marks Allocation
1(a) Definitions	3 marks (error) + 2 marks (mistake)
1(b) Comparison	3 marks (random) + 3 marks (systematic) + 2 marks examples
2 True value calculation	5 marks (mean) + 5 marks (error)
3 Percentage error	10 marks (full derivation)
4 Numerical value + error	5 marks (value) + 5 marks (error)
9 Resistor combinations	5 marks each (series & parallel)