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

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

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Riya is a Class 11 student performing a laboratory experiment to measure the density of an irregular stone. She uses a physical balance and finds the mass of the stone to be 5.74 g. To find the volume, she fills a measuring cylinder with water up to the 20.0 mL mark, immerses the stone, and the water level rises to 21.2 mL. She records the volume as 1.2 cm³. Her teacher asks her to report the density correctly, keeping significant figures in view. Riya calculates the density by dividing mass by volume and gets a long decimal. She is confused about how many digits to retain in her final answer. The teacher reminds her that the precision of the result cannot exceed the precision of the least precise measurement used in the calculation.

Question 1: How many significant figures are present in the mass measurement 5.74 g and the volume measurement 1.2 cm³ respectively?

The mass 5.74 g contains three significant figures: digits 5, 7, and 4 are all non-zero and therefore all significant.
The volume 1.2 cm³ contains two significant figures: digits 1 and 2 are both non-zero and significant; no zeros are present.
Since the two measurements have different numbers of significant figures (3 and 2), the less precise measurement (volume, 2 sig. figs.) will limit the precision of the derived density.

Question 2: What is the correct value of the density of the stone that Riya should report, and why?

Density = mass/volume = 5.74 g / 1.2 cm³ = 4.7833… g cm⁻³ by arithmetic division.
For division, the result must be rounded to the number of significant figures in the least precise input, which is 2 (from the volume 1.2 cm³).
Therefore, Riya should report the density as 4.8 g cm⁻³ (2 significant figures); reporting more digits would falsely imply greater precision than the volume measurement supports.

Question 3: If Riya had used a more precise measuring cylinder and recorded the volume as 1.20 cm³ instead of 1.2 cm³, how would this change the number of significant figures in the volume, and what would the correctly reported density be? Justify the difference.

Recording 1.20 cm³ (with a decimal point) makes the trailing zero significant; thus 1.20 cm³ has three significant figures (1, 2, and 0), compared to two in 1.2 cm³.
Density = 5.74 / 1.20 = 4.7833… g cm⁻³; now both numerator and denominator have three significant figures, so the result is rounded to three significant figures: 4.78 g cm⁻³.
This illustrates that a trailing zero after the decimal point carries real information about measurement precision; by improving the volume measurement from 2 to 3 sig. figs., the reported density gains one additional reliable digit, reflecting the higher precision of the instrument.

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Case Set 2

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During a physics lecture, a teacher writes three equations on the board and challenges students to identify which ones are dimensionally consistent. The equations are: (i) v = u + at², (ii) s = ut + ½at², and (iii) v² = u² + 2as. Here, v and u represent velocities, a is acceleration, s is displacement, and t is time. The teacher explains that the principle of homogeneity of dimensions requires every term in a correct equation to have identical dimensions. She reminds the class that passing the dimensional test does not guarantee physical correctness, but failing it proves the equation wrong. Students are asked to check each equation systematically using dimensional formulae.

Question 1: Write the dimensional formulae for velocity, acceleration, and displacement.

Velocity = displacement/time, so its dimensional formula is [M⁰ L T⁻¹].
Acceleration = velocity/time = length/time², so its dimensional formula is [M⁰ L T⁻²].
Displacement is a length, so its dimensional formula is [M⁰ L T⁰] = [L].

Question 2: Check the dimensional consistency of equation (ii): s = ut + ½at².

Dimension of s: [L]. Dimension of ut: [L T⁻¹][T] = [L]. Dimension of ½at²: [L T⁻²][T²] = [L]; the factor ½ is dimensionless.
All three terms — s, ut, and ½at² — have the same dimension [L], so the equation is dimensionally consistent.
This confirms equation (ii) cannot be ruled out on dimensional grounds; it represents the correct kinematic equation for displacement under uniform acceleration.

Question 3: Show that equation (i) v = u + at² is dimensionally inconsistent. What does this tell us about the equation, and what is the correct form?

Dimension of v: [L T⁻¹]. Dimension of u: [L T⁻¹]. Dimension of at²: [L T⁻²][T²] = [L], which differs from [L T⁻¹].
Since the term at² has dimension [L] while v and u have dimension [L T⁻¹], the terms do not share the same dimensions; the equation violates the principle of homogeneity and is therefore definitely wrong.
The correct equation is v = u + at, where the dimension of at = [L T⁻²][T] = [L T⁻¹], matching v and u; dimensional inconsistency is a definitive proof of error, whereas consistency is only a necessary, not sufficient, condition for correctness.

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Aditya is working in a school science fair project where he measures the dimensions of a rectangular metal sheet using different instruments. He uses a metre scale to measure the length as 4.234 m and the breadth as 1.005 m, and a vernier calliper to measure the thickness as 2.01 cm. He needs to report the surface area and volume of the sheet. His project guide explains that when multiplying several measurements, the result can be no more precise than the least precise measurement. Aditya is also asked to explain to younger students why reporting more digits than justified by the measurements is both superfluous and misleading.

Question 1: How many significant figures are present in each of the three measurements made by Aditya?

Length 4.234 m: all four digits (4, 2, 3, 4) are non-zero and significant — 4 significant figures.
Breadth 1.005 m: the zeros between 1 and 5 are between non-zero digits and are significant; all four digits (1, 0, 0, 5) count — 4 significant figures.
Thickness 2.01 cm: the zero between 2 and 1 is between two non-zero digits and is significant; all three digits (2, 0, 1) count — 3 significant figures.

Question 2: Calculate the area of the metal sheet to the correct number of significant figures.

Area = length × breadth = 4.234 × 1.005 = 4.255170 m².
Both length and breadth have 4 significant figures; for multiplication the result is limited to 4 significant figures.
Correctly reported area = 4.255 m² (4 significant figures); the extra digits 170 are dropped by rounding.

Question 3: Calculate the volume of the sheet to the correct significant figures and explain why the thickness measurement is the limiting factor for precision.

Thickness = 2.01 cm = 0.0201 m (3 significant figures); volume = area × thickness = 4.255 × 0.0201 = 0.0855255 m³ by arithmetic.
Among the three measurements, thickness has the fewest significant figures (3); for multiplication the result must be rounded to 3 significant figures: volume = 0.0855 m³.
The thickness controls the precision because it contributes the smallest number of reliable digits; improving the thickness measurement (e.g., to 4 sig. figs. using a micrometer) would allow the volume to be reported to 4 significant figures, demonstrating that the weakest measurement always limits the derived result.

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In a science museum, an exhibit compares the sizes of objects ranging from a hydrogen atom to the observable universe. The exhibit states that the diameter of a hydrogen atom is approximately 1.06 × 10⁻¹⁰ m, the diameter of the Earth is 1.28 × 10⁷ m, and the diameter of the observable universe is about 8.8 × 10²⁶ m. Visitors are challenged to compare these sizes using the concept of order of magnitude. A student, Meera, reads that the order of magnitude of a quantity is the power of 10 closest to its value when rounded to 1 or 10 in the coefficient. She decides to use this concept to answer the exhibit's questions.

Question 1: What is the order of magnitude of the diameter of the Earth (1.28 × 10⁷ m)?

The number is expressed as 1.28 × 10⁷ m; the coefficient is 1.28.
Since 1.28 ≤ 5, we round it to 1, giving 1 × 10⁷ m; therefore the order of magnitude is 7.
This means the Earth's diameter is approximately 10⁷ m, a useful single-number summary of the scale of the quantity.

Question 2: How many orders of magnitude larger is the Earth's diameter compared to the hydrogen atom's diameter?

Order of magnitude of hydrogen atom diameter (1.06 × 10⁻¹⁰ m): coefficient 1.06 ≤ 5, so order = −10.
Order of magnitude of Earth's diameter: 7 (as computed above).
Difference = 7 − (−10) = 17; the Earth's diameter is 17 orders of magnitude larger than the hydrogen atom's diameter.

Question 3: Find the order of magnitude of the observable universe's diameter (8.8 × 10²⁶ m) and determine how many orders of magnitude the universe is larger than the Earth. Comment on the significance of order-of-magnitude reasoning in physics.

Coefficient of 8.8 × 10²⁶ is 8.8; since 8.8 > 5, it rounds to 10, giving 10 × 10²⁶ = 10²⁷ m; order of magnitude = 27.
Difference from Earth (order 7): 27 − 7 = 20; the observable universe is 20 orders of magnitude larger than the Earth.
Order-of-magnitude reasoning is significant because it allows rapid estimation without detailed calculation, helps physicists judge the plausibility of results, identifies the relevant physical regime, and reveals the enormous range of scales — from 10⁻¹⁰ m (atomic) to 10²⁷ m (cosmic) — that physics must describe with a single unified framework.

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Case Set 5

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A physics teacher conducts a demonstration on the importance of scientific notation. She writes the length 4.700 m on the board and then converts it to centimetres, millimetres, and kilometres without using scientific notation: 470.0 cm, 4700 mm, and 0.004700 km. She asks students how many significant figures each representation has. Students are confused about 4700 mm, thinking it has only two significant figures because of the trailing zeros. The teacher then rewrites all four values in scientific notation and asks students to compare. She emphasises that a change of units must never change the number of significant figures, as that would misrepresent the precision of the original measurement.

Question 1: How many significant figures does 4.700 m have, and which digits are significant?

4.700 m has four significant figures.
The digits 4, 7, 0, and 0 are all significant; the trailing zeros after the decimal point are significant because they explicitly convey that the measurement is precise to the nearest 0.001 m.
If the trailing zeros were not significant, the value would simply be written as 4.7 m; writing 4.700 m is a deliberate choice to indicate higher precision.

Question 2: Why does writing 4700 mm (without a decimal point) create ambiguity about significant figures, and how does scientific notation resolve this?

4700 mm without a decimal point has trailing zeros in a whole number; by the rule for such numbers, trailing zeros are not significant, suggesting only 2 significant figures (4 and 7), which misrepresents the actual 4-figure precision.
Scientific notation removes the ambiguity: writing 4.700 × 10³ mm explicitly shows that all four digits — 4, 7, 0, 0 — in the base number are significant.
This is the key advantage of scientific notation for reporting measurements: every digit in the base number a of a × 10ᵇ is unambiguously significant, regardless of trailing zeros.

Question 3: Rewrite all four representations (4.700 m, 470.0 cm, 4700 mm, 0.004700 km) in scientific notation and verify that each has exactly four significant figures. What fundamental principle does this confirm?

4.700 m = 4.700 × 10⁰ m (4 sig. figs.); 470.0 cm = 4.700 × 10² cm (4 sig. figs.); 4700 mm = 4.700 × 10³ mm (4 sig. figs.); 0.004700 km = 4.700 × 10⁻³ km (4 sig. figs.).
In each case the base number is 4.700, which has four significant figures; the power of 10 is irrelevant to the count of significant figures.
This confirms the fundamental principle stated in the chapter: a choice of different units does not change the number of significant figures in a measurement, because significant figures reflect the precision of the instrument, not the scale of the unit chosen.

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Chapter 1: Units and Measurement | Case Study Questions

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Question 1: How many significant figures are present in the mass measurement 5.74 g and the volume measurement 1.2 cm³ respectively?

Question 2: What is the correct value of the density of the stone that Riya should report, and why?

Question 3: If Riya had used a more precise measuring cylinder and recorded the volume as 1.20 cm³ instead of 1.2 cm³, how would this change the number of significant figures in the volume, and what would the correctly reported density be? Justify the difference.

Passage

Question 1: Write the dimensional formulae for velocity, acceleration, and displacement.

Question 2: Check the dimensional consistency of equation (ii): s = ut + ½at².

Question 3: Show that equation (i) v = u + at² is dimensionally inconsistent. What does this tell us about the equation, and what is the correct form?

Passage

Question 1: How many significant figures are present in each of the three measurements made by Aditya?

Question 2: Calculate the area of the metal sheet to the correct number of significant figures.

Question 3: Calculate the volume of the sheet to the correct significant figures and explain why the thickness measurement is the limiting factor for precision.

Passage

Question 1: What is the order of magnitude of the diameter of the Earth (1.28 × 10⁷ m)?

Question 2: How many orders of magnitude larger is the Earth's diameter compared to the hydrogen atom's diameter?

Question 3: Find the order of magnitude of the observable universe's diameter (8.8 × 10²⁶ m) and determine how many orders of magnitude the universe is larger than the Earth. Comment on the significance of order-of-magnitude reasoning in physics.

Passage

Question 1: How many significant figures does 4.700 m have, and which digits are significant?

Question 2: Why does writing 4700 mm (without a decimal point) create ambiguity about significant figures, and how does scientific notation resolve this?

Question 3: Rewrite all four representations (4.700 m, 470.0 cm, 4700 mm, 0.004700 km) in scientific notation and verify that each has exactly four significant figures. What fundamental principle does this confirm?

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