Class 11 Physics Test Paper Chapter 1 | CBSE Board | Detailed Solutions & Marking Scheme1
VEDANT CLASSES – CLASS XI PHYSICS (THEORY)
Time Allowed: 1 Hour
Maximum Marks: 30
Chapter: Units and Measurement
GENERAL INSTRUCTIONS
All questions are compulsory.
The question paper consists of four sections:
Section A (1 mark each),
Section B (2 marks each),
Section C (3 marks each), and
Section D (5 marks each).There is no overall choice. However, an internal choice has been provided in the 5-mark question.
Use of calculators is not permitted. Log tables may be used if required.
SECTION A: OBJECTIVE & CONCEPTUAL QUESTIONS
(1 Mark Each)
Q1. State the number of significant figures in the measurement 0.000450 m.
(a) 6
(b) 4
(c) 3
(d) 5
Q2. Which of the following pairs of physical quantities has the same dimensional formula?
(a) Force and Power
(b) Stress and Pressure
(c) Momentum and Impulse
(d) Both (b) and (c)
Q3. The argument of an exponential function or a trigonometric function inside a physics equation must always be:
(a) Expressed in units of time [T]
(b) Dimensionless [M⁰L⁰T⁰]
(c) Equal to the dimensions of the left-hand side
(d) Variable depending on the system of units used
Q4. Assertion (A): A dimensionally correct equation may or may not be physically correct.
Reason (R): Dimensionless constants or numerical factors involved in a physical relationship cannot be determined using dimensional analysis.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
SECTION B: SHORT ANSWER QUESTIONS – TYPE I
(2 Marks Each)
Q5. Round off the following numbers to three significant figures:
(a) 25.354 s
(b) 0.04365 kg
(c) 3.475 m
(d) 6.285 J
Q6. Evaluate the following expression using the correct rules of significant figures:
Result = 8.432 × 2.1 / 0.00421
Q7. Each side of a uniform metallic cube is measured to be 4.214 m.
Calculate the total surface area of the cube to the appropriate number of significant figures.
SECTION C: SHORT ANSWER QUESTIONS – TYPE II
(3 Marks Each)
Q8. In a practical fluid dynamics laboratory setup, real gases show deviations from ideal gas behaviour. The governing equation is:
(P + a/V²)(V − b) = RT
where P is pressure, V is volume, R is the universal gas constant, and T is absolute temperature.
Using the principle of homogeneity, derive the dimensional formulas of the constants a and b.
Q9. The velocity v of a particle moving along a straight line varies with time according to:
v = At + B/(t + C) + D
Determine the dimensions of the constants A, B, C, and D.
Q10. The frequency of vibration (f) of a stretched string depends upon:
• Length of the string (l)
• Tension in the string (T)
• Mass per unit length (m)
Using the method of dimensions, derive the expression for frequency.
SECTION D: LONG ANSWER QUESTIONS
(5 Marks Each)
Q11. Answer both Part (A) and Part (B).
Part (A)
State the principle of homogeneity of dimensions.
Check the dimensional correctness of the equation:
K.E. = (1/2)mv² + mgh
where:
K.E. = kinetic energy
m = mass
v = velocity
g = acceleration due to gravity
h = height
(2.5 Marks)
Part (B)
Assume that the critical velocity (v_c) of a viscous liquid flowing through a cylindrical pipe depends on:
• Coefficient of viscosity (η)
• Density of fluid (ρ)
• Radius of pipe (r)
Using dimensional analysis, derive the expression for v_c.
Given:
[η] = [M¹L⁻¹T⁻¹]
(2.5 Marks)
OR
Q11. (Alternative)
Part (A)
Write any three limitations of dimensional analysis.
(3 Marks)
Part (B)
A student measures the mass of a metal block as 7.34 g and its volume as 1.2 cm³.
Calculate the density of the metal and express the answer with the correct number of significant figures.
(2 Marks)
MARKING SCHEME & SOLUTION KEY
SECTION A
Q1. (c) 3
Q2. (d) Both (b) and (c)
Q3. (b) Dimensionless [M⁰L⁰T⁰]
Q4. (a) Both A and R are true and R is the correct explanation.
SECTION B
Q5.
(a) 25.4 s
(b) 0.0436 kg
(c) 3.48 m
(d) 6.28 J
Q6.
8.432 × 2.1 = 17.7072
Result = 17.7072 / 0.00421 = 4205.748
The least number of significant figures is 2 (from 2.1).
Therefore:
Result = 4.2 × 10³
Q7.
Total surface area of a cube:
TSA = 6a²
= 6 × (4.214)²
= 6 × 17.757796
= 106.546776 m²
Since 4.214 has 4 significant figures:
TSA = 106.5 m²
SECTION C
Q8.
From:
(P + a/V²)(V − b) = RT
For subtraction:
[V] = [b]
Therefore:
[b] = [L³]
For addition:
[P] = [a/V²]
[a] = [P][V²]
[P] = [M¹L⁻¹T⁻²]
[V] = [L³]
[a] = [M¹L⁻¹T⁻²] × [L³]²
[a] = [M¹L⁵T⁻²]
Hence:
[a] = [M¹L⁵T⁻²]
[b] = [L³]
Q9.
Given:
v = At + B/(t + C) + D
[v] = [LT⁻¹]
For D:
[D] = [LT⁻¹]
For At:
[A][T] = [LT⁻¹]
[A] = [LT⁻²]
For (t + C):
[C] = [T]
For B/(t + C):
[B]/[T] = [LT⁻¹]
[B] = [L]
Therefore:
[A] = [LT⁻²]
[B] = [L]
[C] = [T]
[D] = [LT⁻¹]
Q10.
Assume:
f = klᵃTᵇmᶜ
Dimensions:
[f] = [T⁻¹]
[l] = [L]
[T] = [MLT⁻²]
[m] = [ML⁻¹]
Therefore:
[M⁰L⁰T⁻¹] = [L]ᵃ [MLT⁻²]ᵇ [ML⁻¹]ᶜ
[M⁰L⁰T⁻¹] = [Mᵇ⁺ᶜLᵃ⁺ᵇ⁻ᶜT⁻²ᵇ]
Comparing powers:
b + c = 0
a + b − c = 0
−2b = −1
b = 1/2
c = −1/2
a = −1
Hence:
f = kl⁻¹T¹ᐟ²m⁻¹ᐟ²
f = (k/l)√(T/m)
SECTION D
Q11. Part (A)
Principle of Homogeneity:
The dimensions of all terms in a physically valid equation must be the same.
For kinetic energy:
[K.E.] = [ML²T⁻²]
For (1/2)mv²:
[M][LT⁻¹]²
= [ML²T⁻²]
For mgh:
[M][LT⁻²][L]
= [ML²T⁻²]
Since all terms have identical dimensions, the equation is dimensionally correct.
Q11. Part (B)
Assume:
v_c = kηᵃρᵇrᶜ
Dimensions:
[v_c] = [LT⁻¹]
[η] = [ML⁻¹T⁻¹]
[ρ] = [ML⁻³]
[r] = [L]
Therefore:
[M⁰L¹T⁻¹] = [ML⁻¹T⁻¹]ᵃ [ML⁻³]ᵇ [L]ᶜ
[M⁰L¹T⁻¹] = [Mᵃ⁺ᵇL⁻ᵃ⁻³ᵇ⁺ᶜT⁻ᵃ]
Comparing powers:
a + b = 0
−a = −1
−a − 3b + c = 1
a = 1
b = −1
c = −1
Hence:
v_c = kη/(ρr)
OR
Q11. Alternative
Part (A)
Limitations of dimensional analysis:
It cannot determine numerical constants such as 2, π, etc.
It cannot derive equations containing trigonometric, exponential, or logarithmic functions.
It cannot establish equations involving addition or subtraction of terms.
Part (B)
Density = Mass/Volume
= 7.34 g / 1.2 cm³
= 6.1167 g/cm³
The least number of significant figures is 2.
Therefore:
Density = 6.1 g/cm³
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