Chapter 1: Units and Measurement Application Questions | physics Class 11 CBSE

ByteNotes

Chapter 1: Units and Measurement Application Questions | physics Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A student measures the side of a cube to be 7.203 m. Calculate the total surface area and volume of the cube, expressing each result to the appropriate number of significant figures.

The measured length has 4 significant figures, so both derived results must be rounded to 4 significant figures.
Total surface area = 6 × (7.203)² = 6 × 51.883209 = 311.299254 m², which rounds to 311.3 m².
Volume = (7.203)³ = 373.714754 m³, which rounds to 373.7 m³.
Reporting more digits (e.g., 311.299254 m²) would be misleading, implying precision beyond what the single measurement of 7.203 m can support.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A substance has a mass of 5.74 g and occupies a volume of 1.2 cm³. Express its density to the correct number of significant figures and justify your answer.

Mass = 5.74 g has 3 significant figures; volume = 1.2 cm³ has 2 significant figures; for division, the result is limited to the fewer, i.e., 2 significant figures.
Density = 5.74/1.2 = 4.7833… g cm⁻³; rounded to 2 significant figures, density = 4.8 g cm⁻³.
Reporting 4.78 g cm⁻³ (three sig. figs.) would be incorrect because the volume measurement 1.2 cm³ cannot support that level of precision.
This example illustrates that the precision of the least precise measurement governs the precision of the derived quantity.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

The length and breadth of a thin rectangular sheet are measured as 16.2 cm and 10.1 cm respectively using a metre scale. Find the area and express the uncertainty in the result.

Both measurements have 3 significant figures; the absolute uncertainties are l = 16.2 ± 0.1 cm (±0.6%) and b = 10.1 ± 0.1 cm (±1%).
Area = 16.2 × 10.1 = 163.62 cm²; the percentage error in the product = 0.6% + 1% = 1.6%, giving absolute error = 1.6% of 163.62 ≈ 2.6 cm².
The final result is expressed as 164 ± 3 cm², where 3 cm² is the uncertainty rounded appropriately.
Expressing the area as 163.62 cm² without an uncertainty statement falsely implies a precision far beyond what the original measurements support.

ByteNotes

Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 4

Applied Concept

State the number of significant figures in each of the following: (a) 0.007 m² (b) 2.64 × 10²⁴ kg (c) 0.2370 g cm⁻³ (d) 6.320 J (e) 6.032 N m⁻² (f) 0.0006032 m².

(a) 0.007 m²: leading zeros are not significant; only the digit 7 is significant — 1 significant figure.
(b) 2.64 × 10²⁴ kg: all three digits 2, 6, 4 in the base number are significant — 3 significant figures; the power of 10 is irrelevant.
(c) 0.2370 g cm⁻³: leading zero is not significant; the trailing zero after the decimal is significant; digits are 2, 3, 7, 0 — 4 significant figures.
(d) 6.320 J: the trailing zero after the decimal is significant — 4 significant figures. (e) 6.032 N m⁻²: the internal zero is significant — 4 significant figures. (f) 0.0006032 m²: leading zeros are not significant; digits are 6, 0, 3, 2 — 4 significant figures.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A grocer’s balance measures the mass of a box as 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added. Find (a) the total mass and (b) the difference in masses, each to correct significant figures.

Converting: 20.15 g = 0.02015 kg and 20.17 g = 0.02017 kg.
(a) Total mass = 2.30 + 0.02015 + 0.02017 = 2.34032 kg. For addition, round to the fewest decimal places: 2.30 has 2 decimal places, so total = 2.34 kg.
(b) Difference in masses = 20.17 − 20.15 = 0.02 g. Both values have 2 decimal places in grams, so the result is 0.02 g (1 significant figure).
Subtraction of nearly equal numbers reduces significant figures drastically, from 4 sig. figs. in each input to only 1 in the difference.

ByteNotes

Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 6

Applied Concept

A physics student writes the relativistic mass formula as m = m₀ / (1 − v²)^(1/2). Use dimensional analysis to identify where the missing constant c must be inserted.

For the formula to be dimensionally consistent, the term inside the square root must be dimensionless, since subtracting a dimensioned quantity from 1 (a pure number) is not allowed.
The quantity v² has dimensions [L² T⁻²]; to make it dimensionless it must be divided by a quantity with the same dimensions.
The speed of light c has dimensions [L T⁻±], so c² has dimensions [L² T⁻²]; dividing v² by c² makes the ratio v²/c² dimensionless.
The corrected formula is m = m₀ / (1 − v²/c²)^(1/2), which is dimensionally consistent and is the correct relativistic mass relation from special relativity.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

The photograph of a house occupies 1.75 cm² on a 35 mm slide. When projected onto a screen, the house image occupies 1.55 m². Calculate the linear magnification of the projector-screen arrangement.

Area magnification = screen area / slide area = 1.55 m² / 1.75 cm². Converting: 1.75 cm² = 1.75 × 10⁻⁴ m².
Area magnification = 1.55 / (1.75 × 10⁻⁴) = 8857 (approximately).
Linear magnification = √(area magnification) = √8857 ≈ 94.1.
Reporting this as 94 (three significant figures, limited by 1.75 cm² which has 3 sig. figs.) gives the linear magnification of the projector-screen arrangement as approximately 94.

ByteNotes

Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 8

Applied Concept

A length is reported as 4.700 m. Rewrite this in cm, mm, and km using scientific notation and identify the number of significant figures in each representation.

4.700 m = 4.700 × 10² cm; the base number 4.700 has four significant figures, all of which are retained.
4.700 m = 4.700 × 10³ mm; again four significant figures; the power of 10 does not affect the significant-figure count.
4.700 m = 4.700 × 10⁻³ km; still four significant figures.
Scientific notation eliminates the confusion that would arise from writing 4700 mm (which ambiguously appears to have only 2 sig. figs.) by making all four digits explicitly significant in the base number.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

With reference to the kinematic equation, verify that the equation v² = u² + 2as is dimensionally consistent.

Dimension of v²: [v²] = [L T⁻¹]² = [L² T⁻²]. Dimension of u²: [u²] = [L T⁻¹]² = [L² T⁻²].
Dimension of 2as: [a][s] = [L T⁻²][L] = [L² T⁻²]; the factor 2 is dimensionless and does not change the dimension.
All three terms v², u², and 2as have dimensions [L² T⁻²], so the principle of homogeneity is satisfied.
The equation is dimensionally consistent; however, this test only confirms dimensional correctness, not physical or mathematical exactness — the correct derivation must come from kinematics.

ByteNotes

Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 10

Applied Concept

A student is asked to add 436.32 g, 227.2 g, and 0.301 g. She reports the answer as 663.821 g. Is her answer correct in terms of significant figures? What should the correct answer be?

For addition, the result must be rounded to the same number of decimal places as the term with the fewest decimal places.
436.32 g has 2 decimal places, 227.2 g has 1 decimal place, and 0.301 g has 3 decimal places; the least is 1 decimal place (from 227.2 g).
Arithmetic sum = 663.821 g, which must be rounded to 1 decimal place, giving 663.8 g.
The student’s answer of 663.821 g is incorrect because it implies a precision to 3 decimal places, far exceeding the precision of the least precise measurement (227.2 g).

ByteNotes

Premium Access

Unlock all Application Questions

You are viewing 10 free application questions from this chapter. Upgrade to Premium to unlock the remaining 12 questions.

Upgrade to Premium12 questions locked

Chapter sections

Chapter 1: Units and Measurement | Application Questions

A student measures the side of a cube to be 7.203 m. Calculate the total surface area and volume of the cube, expressing each result to the appropriate number of significant figures.

A substance has a mass of 5.74 g and occupies a volume of 1.2 cm³. Express its density to the correct number of significant figures and justify your answer.

The length and breadth of a thin rectangular sheet are measured as 16.2 cm and 10.1 cm respectively using a metre scale. Find the area and express the uncertainty in the result.

State the number of significant figures in each of the following: (a) 0.007 m² (b) 2.64 × 10²⁴ kg (c) 0.2370 g cm⁻³ (d) 6.320 J (e) 6.032 N m⁻² (f) 0.0006032 m².

A grocer’s balance measures the mass of a box as 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added. Find (a) the total mass and (b) the difference in masses, each to correct significant figures.

A physics student writes the relativistic mass formula as m = m₀ / (1 − v²)^(1/2). Use dimensional analysis to identify where the missing constant c must be inserted.

The photograph of a house occupies 1.75 cm² on a 35 mm slide. When projected onto a screen, the house image occupies 1.55 m². Calculate the linear magnification of the projector-screen arrangement.

A length is reported as 4.700 m. Rewrite this in cm, mm, and km using scientific notation and identify the number of significant figures in each representation.

With reference to the kinematic equation, verify that the equation v² = u² + 2as is dimensionally consistent.

A student is asked to add 436.32 g, 227.2 g, and 0.301 g. She reports the answer as 663.821 g. Is her answer correct in terms of significant figures? What should the correct answer be?

Unlock all Application Questions