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Chapter 2: Units and Measurements

Class 11 Physics | 26 Questions

1fill blanks🎯 HIGH⭐ Important

Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to _____ m³
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to _____ (mm)²
(c) A vehicle moving with a speed of 18 km h⁻¹ covers _____ m in 1 s
(d) The relative density of lead is 11.3. Its density in SI units is _____ g cm⁻³ or _____ kg m⁻³.

✅ Answer

(a) 10⁻⁶ m³
(b) 1.508 × 10⁴ (mm)²
(c) 5 m
(d) 11.3 g cm⁻³ or 1.13 × 10⁴ kg m⁻³

NEET Relevance

Basic unit conversions and understanding of physical quantities are fundamental and frequently tested in NEET, often as part of larger problems or direct MCQs.

Key Concepts

Unit conversionVolumeSurface AreaSpeedDensityRelative Density

This question has appeared in previous NEET exams.

2numericalMEDIUM⭐ Important

The unit of length convenient on the atomic scale is an angstrom and is denoted by Å: 1 Å = 10⁻¹⁰ m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms?

✅ Answer

The total atomic volume of a mole of hydrogen atoms is approximately 3.14 × 10⁻⁷ m³.

Solution Steps

  1. Step 1: Convert atomic radius to meters

    Given the size of a hydrogen atom (radius) r = 0.5 Å.
    We know that 1 Å = 10⁻¹⁰ m.
    So, r = 0.5 × 10⁻¹⁰ m = 5 × 10⁻¹¹ m.

  2. Step 2: Calculate the volume of one hydrogen atom

    Assuming a hydrogen atom is spherical, its volume Vatom is given by the formula V = (4/3)πr³.
    Vatom = (4/3) × 3.14 × (5 × 10⁻¹¹ m)³
    Vatom = (4/3) × 3.14 × 125 × 10⁻³³ m³
    Vatom = (4 × 3.14 × 125 / 3) × 10⁻³³ m³
    Vatom = (1570 / 3) × 10⁻³³ m³
    Vatom ≈ 523.33 × 10⁻³³ m³ = 5.2333 × 10⁻³¹ m³.

  3. Step 3: Calculate the total volume of a mole of hydrogen atoms

    A mole of hydrogen atoms contains Avogadro's number (NA) of atoms.
    Avogadro's number NA = 6.022 × 10²³ mol⁻¹.
    Total volume Vtotal = NA × Vatom
    Vtotal = 6.022 × 10²³ × 5.2333 × 10⁻³¹ m³
    Vtotal ≈ (6.022 × 5.2333) × 10^(23 - 31) m³
    Vtotal ≈ 31.51 × 10⁻⁸ m³
    Vtotal ≈ 3.151 × 10⁻⁷ m³.
    Rounding to two significant figures (as 0.5 Å has one significant figure, but 1 Å = 10⁻¹⁰ m implies more precision, and typically physical constants are given with more precision, let's keep 3 significant figures based on 0.5 Å and π value):
    Vtotal ≈ 3.14 × 10⁻⁷ m³.

Final Answer: Verify units and significant figures.

NEET Relevance

While direct calculation of this complexity might be rare in NEET, the underlying concepts of unit conversion, Avogadro's number, and basic geometric volume calculations are highly relevant. Questions involving these principles in simpler forms are common.

Key Concepts

Atomic radiusVolume of a sphereAvogadro's numberUnit conversion

This question has appeared in previous NEET exams.

3numerical🎯 HIGH⭐ Important

One astronomical unit (AU) is the average distance of the Earth from the Sun, approximately 1.496 × 10¹¹ m. The speed of light is about 3 × 10⁸ m/s. Express the speed of light in AU per minute.

✅ Answer

The speed of light in AU per minute is approximately 0.12 AU/minute.

Solution Steps

  1. Step 1: Given values

    Speed of light (c) = 3 × 10⁸ m/s
    1 Astronomical Unit (AU) = 1.496 × 10¹¹ m

  2. Step 2: Convert meters to AU

    To convert meters to AU, we use the conversion factor:
    1 m = 1 / (1.496 × 10¹¹) AU
    So, c = (3 × 10⁸ m/s) × (1 AU / (1.496 × 10¹¹ m)) c = (3 × 10⁸ / 1.496 × 10¹¹) AU/s c = (3 / 1.496) × 10^(8 - 11) AU/s c ≈ 2.0053 × 10⁻³ AU/s

  3. Step 3: Convert seconds to minutes

    To convert AU/s to AU/minute, we multiply by the number of seconds in a minute:
    1 minute = 60 seconds
    So, cAU/min = cAU/s × 60 s/minute cAU/min = (2.0053 × 10⁻³ AU/s) × (60 s/minute) cAU/min = (2.0053 × 60) × 10⁻³ AU/minute cAU/min = 120.318 × 10⁻³ AU/minute cAU/min ≈ 0.120318 AU/minute

  4. Step 4: Final Answer

    Rounding to two significant figures (based on the given speed of light 3 × 10⁸ m/s):
    Speed of light ≈ 0.12 AU/minute.

Final Answer: Verify units and significant figures.

NEET Relevance

Unit conversion is a fundamental skill in physics and is very frequently tested in NEET. Questions involving conversion between different units of length, time, or other physical quantities are common MCQs.

Key Concepts

Unit conversionSpeedAstronomical Unit

This question has appeared in previous NEET exams.

4long answerLOW

Explain this statement clearly: “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:
(a) atoms are very small objects
(b) a jet plane moves with great speed
(c) the mass of Jupiter is very large
(d) the air inside this room contains a large number of molecules
(e) a proton is much more massive than an electron
(f) the speed of sound is much smaller than the speed of light

✅ Answer

The statement “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison” highlights a fundamental principle of measurement. Measurement is essentially a process of comparison. When we measure a physical quantity, we compare it with a standard, internationally accepted reference value called a unit. The terms 'large' or 'small' are relative. For example, a distance of 1 kilometer is large compared to the length of a pencil but extremely small compared to the distance between the Earth and the Sun. Without a reference for comparison, these terms have no physical meaning.

Reframing the statements:

(a) atoms are very small objects

  • Reasoning: The statement is ambiguous. 'Small' compared to what?
  • Reframed Statement: The size of an atom (approx. 10⁻¹⁰ m) is much smaller than the size of the tip of a sharp pin (approx. 10⁻⁵ m).

(b) a jet plane moves with great speed

  • Reasoning: 'Great speed' is relative. It's fast compared to a bicycle but slow compared to the speed of light.
  • Reframed Statement: The speed of a jet plane (approx. 900 km/h) is much greater than the speed of a superfast train (approx. 300 km/h).

(c) the mass of Jupiter is very large

  • Reasoning: 'Very large' needs a reference for comparison.
  • Reframed Statement: The mass of Jupiter (approx. 1.9 × 10²⁷ kg) is very large compared to the mass of the Earth (approx. 6 × 10²⁴ kg).

(d) the air inside this room contains a large number of molecules

  • Reasoning: The term 'large number' is not specific.
  • Reframed Statement: The number of molecules in the air inside this room is much larger than the number of molecules in one mole of air (Avogadro's number, 6.023 × 10²³).

(e) a proton is much more massive than an electron

  • Reasoning: This statement is already well-defined because it provides a direct comparison between two specific quantities (mass of a proton and mass of an electron).
  • Reframed Statement: No reframing is necessary. The statement is complete as it compares the mass of a proton to the mass of an electron.

(f) the speed of sound is much smaller than the speed of light

  • Reasoning: Similar to the previous statement, this is a direct and clear comparison between two speeds.
  • Reframed Statement: No reframing is necessary. The statement correctly compares the speed of sound with the speed of light.

NEET Relevance

This is a fundamental conceptual question. While direct questions of this type are not asked in NEET, the underlying principle of comparing quantities and understanding orders of magnitude is essential for problem-solving and estimation.

Key Concepts

Principle of MeasurementUnits and StandardsRelative ComparisonOrder of Magnitude
5numericalMEDIUM⭐ Important

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

✅ Answer

Let the new unit of length be denoted by 'Lnew'.
According to the problem, the speed of light in vacuum (c) is 1 in this new system.
So, c = 1 Lnew / s.

The time taken for light to travel from the Sun to the Earth is given as: t = 8 min 20 s

First, we convert this time into a single unit (seconds): t = (8 × 60) s + 20 s t = 480 s + 20 s t = 500 s

Now, we can use the fundamental relationship between distance, speed, and time:
Distance = Speed × Time

Substituting the values in the new unit system:
Distance (Sun to Earth) = c × t
Distance (Sun to Earth) = (1 Lnew / s) × 500 s
Distance (Sun to Earth) = 500 Lnew

Therefore, the distance between the Sun and the Earth in terms of the new unit of length is 500 units.

Solution Steps

  1. Step 1: Identify Given Information

    The problem states that a new unit of length is defined such that the speed of light, c, is 1. We can write this as c = 1 new unit of length / second. The time taken for light to travel from the Sun to Earth is t = 8 min 20 s.

  2. Step 2: Convert Time to Seconds

    The time is given in minutes and seconds. To use it in calculations, we convert the entire duration into seconds. t = 8 minutes + 20 seconds t = (8 × 60) seconds + 20 seconds t = 480 s + 20 s = 500 s.

  3. Step 3: Apply the Distance Formula

    The relationship between distance, speed, and time is: Distance = Speed × Time.

  4. Step 4: Calculate the Distance in the New Unit

    Using the values from the new system:
    Speed (c) = 1 new unit/s
    Time (t) = 500 s
    Distance = (1 new unit/s) × 500 s = 500 new units.
    The distance between the Sun and the Earth is 500 new units of length.

Final Answer: Verify units and significant figures.

NEET Relevance

Questions involving hypothetical or new unit systems are a classic problem type in NEET. They test the student's understanding of dimensional analysis and the ability to manipulate units, which is a crucial skill.

Key Concepts

Unit SystemsSpeed of LightDistance-Speed-Time RelationUnit Conversion

This question has appeared in previous NEET exams.

6short answer🎯 HIGH⭐ Important

Which of the following is the most precise device for measuring length:
(a) a vernier callipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light ?

✅ Answer

The precision of a measuring instrument is determined by its least count. The smaller the least count, the more precise the instrument. Let's calculate the least count for each device.

(a) Vernier Callipers:
The least count (LC) of a vernier callipers is given by:
LC = (Value of one smallest division on the main scale) / (Total number of divisions on the vernier scale)
Assuming the smallest division on the main scale is 1 mm (standard):
LC = 1 mm / 20 = 0.05 mm

(b) Screw Gauge:
The least count (LC) of a screw gauge is given by:
LC = Pitch / (Number of divisions on the circular scale)
Given, Pitch = 1 mm and Number of divisions = 100.
LC = 1 mm / 100 = 0.01 mm

(c) Optical Instrument:
The instrument can measure length to within a wavelength of light. The wavelength of visible light (λ) is of the order of 10⁻⁷ m.
Let's take an average wavelength, λ ≈ 5000 Å = 5000 × 10⁻¹⁰ m = 5 × 10⁻⁷ m.
Converting this to millimeters:
LC ≈ 5 × 10⁻⁷ m × (1000 mm / 1 m) = 5 × 10⁻⁴ mm = 0.0005 mm

Comparison:

  • Least Count of Vernier Callipers = 0.05 mm
  • Least Count of Screw Gauge = 0.01 mm
  • Least Count of Optical Instrument ≈ 0.0005 mm

Since 0.0005 mm < 0.01 mm < 0.05 mm, the optical instrument has the smallest least count.

Conclusion:
The optical instrument is the most precise device for measuring length among the three options.

Solution Steps

  1. Step 1: Define Precision and Least Count

    The precision of a measuring instrument is its ability to measure the smallest change in the quantity. It is determined by its least count (LC). A smaller least count implies higher precision.

  2. Step 2: Calculate Least Count of Vernier Callipers

    For a vernier callipers, LC = (Smallest main scale division) / (Number of vernier scale divisions). Assuming the main scale is in millimeters (1 MSD = 1 mm), LC = 1 mm / 20 = 0.05 mm.

  3. Step 3: Calculate Least Count of Screw Gauge

    For a screw gauge, LC = Pitch / (Number of circular scale divisions). Given Pitch = 1 mm and number of divisions = 100, LC = 1 mm / 100 = 0.01 mm.

  4. Step 4: Estimate Least Count of Optical Instrument

    The precision is given as the wavelength of light (λ). The range of visible light is approximately 4000 Å to 7000 Å. Let's take λ ≈ 5000 Å = 5 × 10⁻⁷ m. Converting to mm: LC ≈ 5 × 10⁻⁷ m = 0.0005 mm.

  5. Step 5: Compare the Least Counts

    Comparing the calculated least counts:
    LC (Vernier) = 0.05 mm
    LC (Screw Gauge) = 0.01 mm
    LC (Optical Instrument) ≈ 0.0005 mm
    The smallest value is that of the optical instrument.

  6. Step 6: Conclusion

    Since the optical instrument has the smallest least count, it is the most precise device among the given options.

NEET Relevance

Questions on calculating the least count of Vernier Callipers and Screw Gauge, and identifying errors (zero error), are very frequently asked in NEET. Understanding the concept of precision is crucial for this topic.

Key Concepts

PrecisionLeast CountVernier CallipersScrew GaugeMeasurement Error

This question has appeared in previous NEET exams.

7numericalLOW

A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair?

✅ Answer

The problem provides the magnification of the microscope and the observed (magnified) thickness of the hair. We need to find the actual thickness.

Given:
Magnification of the microscope, M = 100
Observed thickness (average width in the field of view), dobserved = 3.5 mm
The number of observations (20) is used to calculate the reliable average of 3.5 mm and is not directly used in the final calculation.

The formula for linear magnification is:
Magnification (M) = (Observed size) / (Actual size)

We need to find the actual size, which is the actual thickness of the hair (dactual).
Rearranging the formula:
Actual size (dactual) = (Observed size) / Magnification

Substituting the given values: dactual = dobserved / M dactual = 3.5 mm / 100 dactual = 0.035 mm

Thus, the estimated thickness of the human hair is 0.035 mm.

Solution Steps

  1. Step 1: List the Given Information

    Identify the known quantities from the problem statement:
    - Magnification (M) = 100
    - Observed average thickness (dobserved) = 3.5 mm
    (Note: The 20 observations are for statistical reliability of the average and not a variable in the final calculation).

  2. Step 2: Recall the Magnification Formula

    The formula for linear magnification relates the observed size to the actual size:
    M = (Observed size) / (Actual size)

  3. Step 3: Rearrange the Formula

    To find the actual thickness of the hair, we rearrange the formula:
    Actual size = (Observed size) / M

  4. Step 4: Substitute Values and Calculate

    Substitute the given values into the rearranged formula:
    Actual thickness = 3.5 mm / 100
    Actual thickness = 0.035 mm

  5. Step 5: State the Final Answer

    The estimated thickness of the hair is 0.035 mm.

Final Answer: Verify units and significant figures.

NEET Relevance

While the concept of magnification is part of the NEET syllabus under 'Optics', this specific type of simple calculation is more of a basic application problem. It's less likely to appear as a standalone question but the principle is fundamental.

Key Concepts

MagnificationMicroscopeMeasurementObserved vs Actual Size
8numerical🎯 HIGH⭐ Important

A physical quantity P is related to four observables a, b, c and d as follows: P = a3 b2 / (sqrt(c) d). The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage error in the quantity P?

✅ Answer

The percentage error in the quantity P is 13%.

Solution Steps

  1. Step 1: Recall the formula for error propagation

    For a physical quantity P = (ax by) / (cz dw), the maximum fractional error is given by:
    (ΔP/P) = x(Δa/a) + y(Δb/b) + z(Δc/c) + w(Δd/d)
    And the maximum percentage error is:
    (ΔP/P) x 100% = x(Δa/a x 100%) + y(Δb/b x 100%) + z(Δc/c x 100%) + w(Δd/d x 100%)

  2. Step 2: Identify the given expression and powers

    The given expression is P = a3 b2 / (sqrt(c) d) = a3 b2 c^(-1/2) d^(-1).
    Comparing this with the general form, we have: x = 3 (for a) y = 2 (for b) z = 1/2 (for c, since sqrt(c) = c^(1/2)) w = 1 (for d)

  3. Step 3: List the given percentage errors

    Percentage error in a (Δa/a x 100%) = 1%
    Percentage error in b (Δb/b x 100%) = 3%
    Percentage error in c (Δc/c x 100%) = 4%
    Percentage error in d (Δd/d x 100%) = 2%

  4. Step 4: Calculate the total percentage error in P

    Substitute the values into the percentage error formula:
    (ΔP/P) x 100% = 3(Δa/a x 100%) + 2(Δb/b x 100%) + (1/2)(Δc/c x 100%) + 1(Δd/d x 100%)
    (ΔP/P) x 100% = 3(1%) + 2(3%) + (1/2)(4%) + 1(2%)
    (ΔP/P) x 100% = 3% + 6% + 2% + 2%
    (ΔP/P) x 100% = 13%

Final Answer: Verify units and significant figures.

NEET Relevance

Error analysis, especially percentage error calculation for quantities involving powers, is a very common topic in NEET. Questions often involve simple formulas and direct application of the error propagation rules.

Key Concepts

Error propagationPercentage errorPowers in error calculation

This question has appeared in previous NEET exams.

9short answer🎯 HIGH⭐ Important

A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:
(a) y = a sin (2πt/T)
(b) y = a sin vt
(c) y = (a/T) sin (t/a)
(d) y = a sqrt(2) (sin (2πt/T) + cos (2πt/T))
(a = maximum displacement of the particle, v = speed, T = time period of motion).
Rule out the wrong formulas on dimensional grounds.

✅ Answer

To rule out wrong formulas on dimensional grounds, we apply the principle of dimensional homogeneity, which states that the dimensions of terms on both sides of an equation must be the same. Also, arguments of trigonometric functions must be dimensionless.

Given dimensions:
- Displacement, y: [L]
- Maximum displacement, a: [L]
- Speed, v: [LT⁻¹]
- Time, t: [T]
- Time period, T: [T]

Let's analyze each formula:

(a) y = a sin (2πt/T)

  • LHS (y): [L]
  • RHS (a sin (2πt/T)):
  • Dimension of a: [L]
  • Dimension of (2πt/T): 2π is a dimensionless constant. Dimension of t is [T]. Dimension of T is [T]. So, (2πt/T) has dimension [T]/[T] = [M⁰L⁰T⁰] (dimensionless).
  • Since sin (dimensionless quantity) is dimensionless, the dimension of RHS is [L] x [M⁰L⁰T⁰] = [L].
  • Conclusion: LHS dimensions = RHS dimensions ([L] = [L]). The argument of sin is dimensionless. This formula is dimensionally correct.

(b) y = a sin vt

  • LHS (y): [L]
  • RHS (a sin vt):
  • Dimension of a: [L]
  • Dimension of vt: Dimension of v is [LT⁻¹]. Dimension of t is [T]. So, vt has dimension [LT⁻¹][T] = [L].
  • The argument of the sine function (vt) has dimension [L], which is not dimensionless. Trigonometric functions must have dimensionless arguments.
  • Conclusion: This formula is dimensionally incorrect because the argument of the sine function is not dimensionless.

(c) y = (a/T) sin (t/a)

  • LHS (y): [L]
  • RHS ((a/T) sin (t/a)):
  • Dimension of (a/T): [L]/[T] = [LT⁻¹]
  • Dimension of (t/a): Dimension of t is [T]. Dimension of a is [L]. So, (t/a) has dimension [T]/[L] = [L⁻¹T].
  • The argument of the sine function (t/a) has dimension [L⁻¹T], which is not dimensionless.
  • The dimension of RHS would be [LT⁻¹] x [M⁰L⁰T⁰] (if sin argument was dimensionless) = [LT⁻¹]. This does not match the LHS dimension [L].
  • Conclusion: This formula is dimensionally incorrect because the argument of the sine function is not dimensionless, and the dimensions of LHS and RHS do not match.

(d) y = a sqrt(2) (sin (2πt/T) + cos (2πt/T))

  • LHS (y): [L]
  • RHS (a sqrt(2) (sin (2πt/T) + cos (2πt/T))):
  • Dimension of a: [L]
  • sqrt(2) is a dimensionless constant.
  • Dimension of (2πt/T): As in (a), this is dimensionless ([T]/[T] = [M⁰L⁰T⁰]).
  • Since sin (dimensionless) and cos (dimensionless) are dimensionless, the sum (sin (2πt/T) + cos (2πt/T)) is also dimensionless.
  • The dimension of RHS is [L] x [M⁰L⁰T⁰] x [M⁰L⁰T⁰] = [L].
  • Conclusion: LHS dimensions = RHS dimensions ([L] = [L]). The arguments of the trigonometric functions are dimensionless. This formula is dimensionally correct.

Formulas ruled out on dimensional grounds are (b) and (c).

    NEET Relevance

    Dimensional analysis is a fundamental concept frequently tested in NEET. Questions often involve checking dimensional consistency of formulas or deriving dimensions of physical quantities. It's a quick way to eliminate incorrect options in MCQs.

    Key Concepts

    Dimensional analysisPrinciple of homogeneity of dimensionsDimensions of trigonometric arguments

    This question has appeared in previous NEET exams.

    10numericalMEDIUM

    A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance? (Given: speed of light c = 3 x 108 m/s)

    ✅ Answer

    The distance between the Sun and the Earth in the new unit is 500 new units of length.

    Solution Steps

    1. Step 1: Understand the new unit definition

      The speed of light in vacuum (c) is chosen to be unity in the new system of units. This means: c = 1 newunit/s
      We know that in SI units, c = 3 x 108 m/s.
      Therefore, 1 newunit/s = 3 x 108 m/s.
      This implies: 1 newunit = 3 x 108 m.

    2. Step 2: Convert the given time into seconds

      Time taken by light to cover the distance = 8 min 20 s.
      Convert minutes to seconds: 8 min = 8 x 60 s = 480 s.
      Total time (t) = 480 s + 20 s = 500 s.

    3. Step 3: Calculate the distance in standard SI units (meters)

      Distance (D) = Speed of light (c) x Time (t)
      D = (3 x 108 m/s) x (500 s)
      D = 1500 x 108 m
      D = 1.5 x 1011 m

    4. Step 4: Convert the distance to the new unit of length

      From Step 1, we know that 1 newunit = 3 x 108 m.
      To convert distance from meters to the new unit, we divide by the value of 1 newunit in meters:
      Distance in newunits = D / (3 x 108 m/newunit)
      Distance in newunits = (1.5 x 1011 m) / (3 x 108 m/newunit)
      Distance in newunits = (1.5 / 3) x (1011 / 108) newunits
      Distance in newunits = 0.5 x 103 newunits
      Distance in newunits = 500 newunits.

    Final Answer: Verify units and significant figures.

    NEET Relevance

    While direct questions on defining new units based on fundamental constants are less common, the underlying principle of unit conversion and understanding the relationship between different unit systems is highly relevant for NEET. This problem tests conceptual understanding of units.

    Key Concepts

    Unit conversionSpeed of lightDefinition of new units
    11short answer🎯 HIGH⭐ Important

    Which of the following has the highest number of significant figures?
    (a) 0.007 m2
    (b) 2.64 x 1024 kg
    (c) 0.2370 g cm-3
    (d) 6.320 J
    (e) 6.032 N m-2

    ✅ Answer

    To determine the number of significant figures for each option, we apply the standard rules:
    1. All non-zero digits are significant.
    2. Zeros between two non-zero digits are significant.
    3. Leading zeros (zeros before non-zero digits) are not significant.
    4. Trailing zeros (zeros at the end of a number) are significant if the number contains a decimal point.
    5. In scientific notation (A x 10B), all digits in A are significant.

    Let's analyze each option:

    (a) 0.007 m2

    • The leading zeros (0.00) are not significant.
    • Only the digit '7' is significant.
    • Number of significant figures = 1.

    (b) 2.64 x 1024 kg

    • In scientific notation, all digits in the mantissa (2.64) are significant.
    • The digits '2', '6', and '4' are all non-zero.
    • Number of significant figures = 3.

    (c) 0.2370 g cm-3

    • The leading zero (0.) is not significant.
    • The digits '2', '3', '7' are non-zero and thus significant.
    • The trailing zero '0' after the decimal point is significant.
    • Number of significant figures = 4.

    (d) 6.320 J

    • The digits '6', '3', '2' are non-zero and thus significant.
    • The trailing zero '0' after the decimal point is significant.
    • Number of significant figures = 4.

    (e) 6.032 N m-2

    • The digits '6', '3', '2' are non-zero and thus significant.
    • The zero '0' between '6' and '3' is a trapped zero and thus significant.
    • Number of significant figures = 4.

    Comparing the number of significant figures:
    (a) 1
    (b) 3
    (c) 4
    (d) 4
    (e) 4

    Options (c), (d), and (e) all have 4 significant figures, which is the highest among the given choices.

    Therefore, 0.2370 g cm-3, 6.320 J, and 6.032 N m-2 all have the highest number of significant figures (4).

    NEET Relevance

    Significant figures and rounding off are crucial for numerical calculations in physics and chemistry, and frequently appear in NEET. Questions often involve identifying significant figures in given numbers or applying rules in calculations.

    Key Concepts

    Significant figuresRules for significant figures

    This question has appeared in previous NEET exams.

    12numerical🎯 HIGH⭐ Important

    Fill in the blanks by suitable conversion of units:
    (a) 1 kg m2 s-2 = ..... g cm2 s-2
    (b) 1 m = ..... ly
    (c) 3.0 m s-2 = ..... km h-2
    (d) G = 6.67 x 10-11 N m2 kg-2 = ..... cm3 s-2 g-1

    ✅ Answer

    The filled blanks are:
    (a) 1 kg m2 s-2 = 107 g cm2 s-2
    (b) 1 m = 1.057 x 10-16 ly
    (c) 3.0 m s-2 = 3.888 x 104 km h-2
    (d) G = 6.67 x 10-11 N m2 kg-2 = 6.67 x 10-8 cm3 s-2 g-1

    Solution Steps

    1. Step 1: Part (a): 1 kg m^2 s^-2 to g cm^2 s^-2

      We need to convert kilograms (kg) to grams (g) and meters (m) to centimeters (cm).
      1 kg = 103 g
      1 m = 102 cm

      1 kg m2 s-2 = (1 kg) x (1 m)2 x s-2
      = (103 g) x (102 cm)2 x s-2
      = (103 g) x (104 cm2) x s-2
      = 10^(3+4) g cm2 s-2
      = 107 g cm2 s-2

      So, 1 kg m2 s-2 = 107 g cm2 s-2.

    2. Step 2: Part (b): 1 m to ly (light-year)

      A light-year (ly) is the distance light travels in one year.
      Speed of light (c) = 3 x 108 m/s
      1 year = 365.25 days = 365.25 x 24 hours = 365.25 x 24 x 60 minutes = 365.25 x 24 x 60 x 60 seconds
      1 year = 3.15576 x 107 s

      1 ly = c x 1 year = (3 x 108 m/s) x (3.15576 x 107 s)
      1 ly = 9.46728 x 1015 m

      Now, we need to find how many light-years are in 1 meter:
      1 m = 1 / (9.46728 x 1015) ly
      1 m ≈ 0.10562 x 10-15 ly
      1 m ≈ 1.057 x 10-16 ly

      So, 1 m = 1.057 x 10-16 ly.

    3. Step 3: Part (c): 3.0 m s^-2 to km h^-2

      We need to convert meters (m) to kilometers (km) and seconds (s) to hours (h).
      1 km = 103 m => 1 m = 10-3 km
      1 hour = 3600 s => 1 s = 1/3600 h

      3.0 m s-2 = 3.0 x (1 m) x (1 s)-2
      = 3.0 x (10-3 km) x (1/3600 h)-2
      = 3.0 x (10-3 km) x (3600)2 h-2
      = 3.0 x 10-3 x (1.296 x 107) km h-2
      = 3.888 x 10^(7-3) km h-2
      = 3.888 x 104 km h-2

      So, 3.0 m s-2 = 3.888 x 104 km h-2.

    4. Step 4: Part (d): G = 6.67 x 10^-11 N m^2 kg^-2 to cm^3 s^-2 g^-1

      First, express Newton (N) in fundamental SI units: 1 N = 1 kg m s-2.
      So, G = 6.67 x 10-11 (kg m s-2) m2 kg-2
      G = 6.67 x 10-11 kg^(1-2) m^(1+2) s-2
      G = 6.67 x 10-11 kg-1 m3 s-2

      Now, convert kg to g and m to cm:
      1 kg = 103 g => 1 kg-1 = (103 g)-1 = 10-3 g-1
      1 m = 102 cm => 1 m3 = (102 cm)3 = 106 cm3

      G = 6.67 x 10-11 x (10-3 g-1) x (106 cm3) x s-2
      G = 6.67 x 10-11 x 10^(-3+6) cm3 s-2 g-1
      G = 6.67 x 10-11 x 103 cm3 s-2 g-1
      G = 6.67 x 10^(-11+3) cm3 s-2 g-1
      G = 6.67 x 10-8 cm3 s-2 g-1

      So, G = 6.67 x 10-11 N m2 kg-2 = 6.67 x 10-8 cm3 s-2 g-1.

    Final Answer: Verify units and significant figures.

    NEET Relevance

    Unit conversion is a fundamental skill required for almost all numerical problems in NEET physics and chemistry. Questions often involve converting between SI and CGS units, or other derived units. Mastering these conversions is essential.

    Key Concepts

    Unit conversionSI unitsCGS unitsLight-yearGravitational constant

    This question has appeared in previous NEET exams.

    13numericalMEDIUM

    A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance? (Speed of light = 3 × 10⁸ m s⁻¹)

    ✅ Answer

    The distance between the Sun and the Earth in the new unit is 500 new units.

    Solution Steps

    1. Step 1: Convert time to seconds

      Given time taken by light to travel from Sun to Earth = 8 min 20 s.
      Convert minutes to seconds: 8 min = 8 × 60 s = 480 s.
      Total time (t) = 480 s + 20 s = 500 s.

    2. Step 2: Calculate distance in meters

      The speed of light in vacuum (c) = 3 × 10⁸ m s⁻¹.
      Distance (D) = Speed × Time
      D = c × t = (3 × 10⁸ m s⁻¹) × (500 s)
      D = 1500 × 10⁸ m = 1.5 × 10¹¹ m.

    3. Step 3: Define the new unit of length

      In the new system, the speed of light in vacuum is unity. Let the new unit of length be 'Lnew'.
      So, c = 1 Lnew / s.
      Since c = 3 × 10⁸ m s⁻¹, we have:
      1 Lnew / s = 3 × 10⁸ m / s
      Therefore, 1 Lnew = 3 × 10⁸ m.

    4. Step 4: Convert the distance to the new unit

      We found the distance D = 1.5 × 10¹¹ m.
      To convert this to the new unit, divide by the value of 1 Lnew in meters:
      Dnew = D / (1 Lnew in meters)
      Dnew = (1.5 × 10¹¹ m) / (3 × 10⁸ m / Lnew)
      Dnew = (1.5 × 10¹¹ / 3 × 10⁸) Lnew
      Dnew = (0.5 × 10³) Lnew
      Dnew = 500 Lnew.
      So, the distance between the Sun and the Earth is 500 new units.

    Final Answer: Verify units and significant figures.

    NEET Relevance

    Questions involving unit conversion and the definition of new units based on fundamental constants (like speed of light) are common in NEET, often testing conceptual understanding of dimensions and units.

    Key Concepts

    Unit conversionSpeed, distance, time relationshipDefining new units
    14short answer🎯 HIGH⭐ Important

    Which of the following has the highest number of significant figures?
    (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

    ✅ Answer

    Let's determine the number of significant figures for each option:

    (a) 0.007 m²: Leading zeros (zeros before non-zero digits) are never significant. So, only '7' is significant. Number of significant figures = 1.

    (b) 2.64 × 10²⁴ kg: The power of 10 does not affect the number of significant figures. We count the significant figures in 2.64. All non-zero digits are significant. Number of significant figures = 3.

    (c) 0.2370 g cm⁻³: Leading zeros are not significant. Trailing zeros (zeros at the end of a number) are significant if the number contains a decimal point. So, '2', '3', '7', and '0' are significant. Number of significant figures = 4.

    (d) 6.320 J: All non-zero digits are significant. Trailing zeros are significant if the number contains a decimal point. So, '6', '3', '2', and '0' are significant. Number of significant figures = 4.

    (e) 6.032 N m⁻²: All non-zero digits are significant. Zeros between two non-zero digits are significant. So, '6', '0', '3', and '2' are significant. Number of significant figures = 4.

    Comparing the number of significant figures:
    (a) 1
    (b) 3
    (c) 4
    (d) 4
    (e) 4

    Options (c), (d), and (e) all have the highest number of significant figures, which is 4.

    NEET Relevance

    Rules for significant figures are a fundamental concept and frequently tested in NEET, often as direct MCQs or as part of numerical problems where the final answer needs to be reported with correct significant figures.

    Key Concepts

    Significant figures rules

    This question has appeared in previous NEET exams.

    15numerical🎯 HIGH⭐ Important

    The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

    ✅ Answer

    The area of the sheet is 4.255 m² and the volume of the sheet is 0.0855 m³.

    Solution Steps

    1. Step 1: List given measurements and their significant figures

      Length (l) = 4.234 m (4 significant figures)
      Breadth (b) = 1.005 m (4 significant figures)
      Thickness (t) = 2.01 cm (3 significant figures)

    2. Step 2: Convert all measurements to a consistent unit (meters)

      l = 4.234 m b = 1.005 m t = 2.01 cm = 2.01 / 100 m = 0.0201 m (3 significant figures)

    3. Step 3: Calculate the area of the sheet

      The question asks for 'the area of the sheet'. Assuming it refers to the area of one of the largest faces (length × breadth), or total surface area. Given the context of volume calculation, it's most likely asking for the area of the top/bottom face (l x b) or perhaps total surface area. Let's calculate the area of the largest face (length x breadth) first, as it's a common intermediate step.
      Area (A) = l × b
      A = 4.234 m × 1.005 m
      A = 4.25527 m²

      Rule for multiplication/division: The result should be rounded to the same number of significant figures as the measurement with the fewest significant figures. Both length and breadth have 4 significant figures. So the result should also have 4 significant figures.
      A = 4.255 m² (rounded to 4 significant figures).

    4. Step 4: Calculate the volume of the sheet

      Volume (V) = l × b × t
      V = 4.234 m × 1.005 m × 0.0201 m
      V = 0.085537674 m³

      Now, apply the significant figures rule for multiplication. The number of significant figures in l is 4, in b is 4, and in t is 3. The measurement with the fewest significant figures is thickness (t) with 3 significant figures. Therefore, the volume should be rounded to 3 significant figures.
      V = 0.0855 m³ (rounded to 3 significant figures).

    Final Answer: Verify units and significant figures.

    NEET Relevance

    This type of question is very important for NEET. It tests the application of significant figure rules in calculations involving multiplication and division, which is a common source of errors if not handled correctly. Expect MCQs where you need to choose the option with the correct number of significant figures.

    Key Concepts

    Significant figures in multiplication and divisionUnit conversionArea and Volume calculation

    This question has appeared in previous NEET exams.

    16numerical🎯 HIGH⭐ Important

    The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the gold pieces to the correct significant figures?

    ✅ Answer

    (a) The total mass of the box is 2.340 kg.
    (b) The difference in the masses of the gold pieces is 0.02 g.

    Solution Steps

    1. Step 1: List given masses and their significant figures/decimal places

      Mass of the box (Mbox) = 2.300 kg (3 decimal places)
      Mass of first gold piece (m1) = 20.15 g (2 decimal places)
      Mass of second gold piece (m2) = 20.17 g (2 decimal places)

    2. Step 2: Convert all masses to a consistent unit (kg for total mass)

      Mbox = 2.300 kg m1 = 20.15 g = 20.15 / 1000 kg = 0.02015 kg m2 = 20.17 g = 20.17 / 1000 kg = 0.02017 kg

    3. Step 3: Calculate (a) Total mass of the box

      Total mass (Mtotal) = Mbox + m1 + m2
      Mtotal = 2.300 kg + 0.02015 kg + 0.02017 kg
      Mtotal = 2.34032 kg

      Rule for addition/subtraction: The result should be rounded to the same number of decimal places as the measurement with the fewest decimal places.
      - Mbox: 2.300 kg (3 decimal places)
      - m1: 0.02015 kg (5 decimal places)
      - m2: 0.02017 kg (5 decimal places)

      The fewest decimal places is 3 (from Mbox). So, the total mass should be rounded to 3 decimal places.
      Mtotal = 2.340 kg.

    4. Step 4: Calculate (b) Difference in the masses of the gold pieces

      Difference in mass (Δm) = m2 - m1 (or m1 - m2, magnitude is what matters)
      Δm = 20.17 g - 20.15 g
      Δm = 0.02 g

      Rule for addition/subtraction: The result should be rounded to the same number of decimal places as the measurement with the fewest decimal places.
      - m1: 20.15 g (2 decimal places)
      - m2: 20.17 g (2 decimal places)

      Both masses have 2 decimal places. So, the difference should also be reported to 2 decimal places.
      Δm = 0.02 g.

    Final Answer: Verify units and significant figures.

    NEET Relevance

    This question is highly relevant for NEET as it directly tests the rules for significant figures in addition and subtraction, which are distinct from multiplication/division rules. These rules are frequently applied in experimental data analysis and can be a source of errors if not understood well.

    Key Concepts

    Significant figures in addition and subtractionUnit conversion

    This question has appeared in previous NEET exams.

    17numerical🎯 HIGH⭐ Important

    A physical quantity P is related to four observables a, b, c and d as follows:
    P = a³b² / (√c d)
    The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage error in the quantity P?

    ✅ Answer

    The percentage error in the quantity P is 13%.

    Solution Steps

    1. Step 1: Write down the given relation and percentage errors

      The physical quantity P is given by the relation: P = a³b² / (√c d)
      This can be written as: P = a³b²c⁻¹ᐟ²d⁻¹

      Given percentage errors:
      Percentage error in a (Δa/a × 100%) = 1%
      Percentage error in b (Δb/b × 100%) = 3%
      Percentage error in c (Δc/c × 100%) = 4%
      Percentage error in d (Δd/d × 100%) = 2%

    2. Step 2: Apply the formula for error propagation in products and powers

      For a quantity Q = xp yq zr, the maximum fractional error is given by:
      ΔQ/Q = |p|(Δx/x) + |q|(Δy/y) + |r|(Δz/z)

      For percentage error, multiply by 100%:
      (ΔQ/Q × 100%) = |p|(Δx/x × 100%) + |q|(Δy/y × 100%) + |r|(Δz/z × 100%)

      In our case, P = a³b²c⁻¹ᐟ²d⁻¹.
      The powers are: pa = 3, pb = 2, pc = -1/2, pd = -1.
      When calculating maximum possible error, we always add the absolute values of the fractional errors, so the negative signs of the powers become positive.

    3. Step 3: Calculate the percentage error in P

      Percentage error in P (ΔP/P × 100%) = |3|(Δa/a × 100%) + |2|(Δb/b × 100%) + |-1/2|(Δc/c × 100%) + |-1|(Δd/d × 100%)

      Substitute the given percentage errors:
      ΔP/P × 100% = 3 × (1%) + 2 × (3%) + (1/2) × (4%) + 1 × (2%)
      ΔP/P × 100% = 3% + 6% + 2% + 2%
      ΔP/P × 100% = 13%

    Final Answer: Verify units and significant figures.

    NEET Relevance

    This is an extremely important question for NEET. Error analysis, especially percentage error propagation in complex formulas, is a very common topic. Expect direct MCQs on calculating percentage error, or questions where you need to apply this concept to experimental data.

    Key Concepts

    Error propagationPercentage errorPowers in error calculation

    This question has appeared in previous NEET exams.

    18numerical🎯 HIGH⭐ Important

    A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance? (Speed of light c = 3 × 10⁸ m s⁻¹)

    ✅ Answer

    The distance between the Sun and the Earth in the new unit is 500 new units.

    Solution Steps

    1. Step 1: Understand the new unit definition

      The problem states that a new unit of length is chosen such that the speed of light in vacuum is unity. This means c = 1 newunit/s. We are given the standard speed of light c = 3 × 10⁸ m/s. Therefore, 1 newunit/s = 3 × 10⁸ m/s, which implies 1 newunit = 3 × 10⁸ m.

    2. Step 2: Convert time to seconds

      Light takes 8 min and 20 s to cover the distance between the Sun and the Earth.
      Time (t) = 8 minutes + 20 seconds
      Time (t) = (8 × 60) seconds + 20 seconds
      Time (t) = 480 seconds + 20 seconds
      Time (t) = 500 seconds.

    3. Step 3: Calculate distance in standard units (meters)

      Using the formula Distance = Speed × Time:
      Distance (d) = c × t
      Distance (d) = (3 × 10⁸ m/s) × (500 s)
      Distance (d) = 1500 × 10⁸ m
      Distance (d) = 1.5 × 10¹¹ m.

    4. Step 4: Convert distance to the new unit

      From Step 1, we know that 1 newunit = 3 × 10⁸ m.
      To convert the distance from meters to the new unit, we divide the distance in meters by the value of 1 newunit in meters:
      Distance in new units = (Distance in meters) / (Value of 1 newunit in meters)
      Distance in new units = (1.5 × 10¹¹ m) / (3 × 10⁸ m/newunit)
      Distance in new units = (1.5 / 3) × (10¹¹ / 10⁸) newunits
      Distance in new units = 0.5 × 10³ newunits
      Distance in new units = 500 newunits.

    Final Answer: Verify units and significant figures.

    NEET Relevance

    This question tests the fundamental understanding of units, conversion, and the definition of physical constants. Similar conceptual questions or calculations involving unit conversions are common in NEET, especially in the context of dimensional analysis or basic physics problems.

    Key Concepts

    Units and dimensionsSpeed of lightUnit conversionBasic kinematics

    This question has appeared in previous NEET exams.

    19numerical🎯 HIGH⭐ Important

    Which of the following has the highest number of significant figures?
    (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

    ✅ Answer

    The option with the highest number of significant figures is (c) 0.2370 g cm⁻³ and (e) 6.032 N m⁻², both having 4 significant figures.

    Solution Steps

    1. Step 1: Recall rules for significant figures
      1. All non-zero digits are significant.
      2. Zeros between two non-zero digits are significant.
      3. Leading zeros (zeros before non-zero digits) are not significant.
      4. Trailing zeros (zeros at the end of a number) are significant if the number contains a decimal point.
      5. In scientific notation (A × 10ⁿ), all digits in A are significant.
    2. Step 2: Analyze option (a) 0.007 m²

      The leading zeros (0.00) are not significant. Only the digit '7' is significant.
      Number of significant figures = 1.

    3. Step 3: Analyze option (b) 2.64 × 10²⁴ kg

      In scientific notation, all digits in the mantissa (2.64) are significant.
      Number of significant figures = 3 (2, 6, 4).

    4. Step 4: Analyze option (c) 0.2370 g cm⁻³

      The leading zero (0.) is not significant. The digits '2', '3', '7' are significant. The trailing zero '0' after the decimal point is significant.
      Number of significant figures = 4 (2, 3, 7, 0).

    5. Step 5: Analyze option (d) 6.320 J

      The digits '6', '3', '2' are significant. The trailing zero '0' after the decimal point is significant.
      Number of significant figures = 4 (6, 3, 2, 0).

    6. Step 6: Analyze option (e) 6.032 N m⁻²

      The digits '6', '3', '2' are significant. The zero '0' between non-zero digits '6' and '3' is significant.
      Number of significant figures = 4 (6, 0, 3, 2).

    7. Step 7: Compare and conclude

      Comparing the number of significant figures for each option:
      (a) 0.007 m²: 1 significant figure
      (b) 2.64 × 10²⁴ kg: 3 significant figures
      (c) 0.2370 g cm⁻³: 4 significant figures
      (d) 6.320 J: 4 significant figures
      (e) 6.032 N m⁻²: 4 significant figures
      Options (c), (d), and (e) all have the highest number of significant figures, which is 4.

    Final Answer: Verify units and significant figures.

    NEET Relevance

    Questions on identifying the number of significant figures are very common in NEET, often appearing as direct MCQs or as a part of calculations where the final answer needs to be reported with the correct significant figures. This is a fundamental concept in error analysis.

    Key Concepts

    Significant figuresRules for counting significant figures

    This question has appeared in previous NEET exams.

    20numerical🎯 HIGH⭐ Important

    The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to appropriate significant figures.

    ✅ Answer

    The area of the sheet is 4.255 m² and the volume of the sheet is 0.0855 m³ (both to appropriate significant figures).

    Solution Steps

    1. Step 1: List given measurements and their significant figures

      Length (l) = 4.234 m (4 significant figures)
      Breadth (b) = 1.005 m (4 significant figures)
      Thickness (t) = 2.01 cm (3 significant figures)

    2. Step 2: Convert all measurements to a consistent unit (meters)

      Length (l) = 4.234 m
      Breadth (b) = 1.005 m
      Thickness (t) = 2.01 cm = 2.01 × 10⁻² m = 0.0201 m

    3. Step 3: Calculate the area of the sheet

      Area (A) = Length × Breadth
      A = 4.234 m × 1.005 m
      A = 4.25517 m²

      Rule for multiplication/division: The result should be rounded to the same number of significant figures as the measurement with the fewest significant figures. Both length and breadth have 4 significant figures.
      Therefore, the area should be rounded to 4 significant figures.
      A = 4.255 m².

    4. Step 4: Calculate the volume of the sheet

      Volume (V) = Length × Breadth × Thickness
      V = 4.234 m × 1.005 m × 0.0201 m
      V = 4.25517 m² × 0.0201 m
      V = 0.085528917 m³

      Rule for multiplication/division: The result should be rounded to the same number of significant figures as the measurement with the fewest significant figures. The thickness (0.0201 m) has 3 significant figures (the leading zeros are not significant). Length and breadth have 4 significant figures.
      Therefore, the volume should be rounded to 3 significant figures.
      V = 0.0855 m³.

    Final Answer: Verify units and significant figures.

    NEET Relevance

    This question is highly relevant for NEET as it combines unit conversion with the application of significant figure rules in calculations (multiplication). Such problems are frequently asked to test precision and accuracy in measurements.

    Key Concepts

    Significant figuresRules for significant figures in multiplication/divisionUnit conversionArea and volume calculation

    This question has appeared in previous NEET exams.

    21numerical🎯 HIGH⭐ Important

    The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the gold pieces to the correct significant figures?

    ✅ Answer

    (a) The total mass of the box is 2.340 kg.
    (b) The difference in the masses of the gold pieces is 0.02 g.

    Solution Steps

    1. Step 1: List given masses and their precision

      Mass of the box (Mbox) = 2.300 kg (3 decimal places)
      Mass of first gold piece (m1) = 20.15 g (2 decimal places)
      Mass of second gold piece (m2) = 20.17 g (2 decimal places)

    2. Step 2: Convert all masses to a consistent unit (e.g., kg or g)

      It's generally easier to perform addition/subtraction with a consistent unit. Let's convert everything to kilograms for part (a) and grams for part (b).

      For part (a) (Total mass):
      Mbox = 2.300 kg m1 = 20.15 g = 20.15 / 1000 kg = 0.02015 kg m2 = 20.17 g = 20.17 / 1000 kg = 0.02017 kg

    3. Step 3: Calculate the total mass of the box (Part a)

      Total mass (Mtotal) = Mbox + m1 + m2
      Mtotal = 2.300 kg + 0.02015 kg + 0.02017 kg
      Mtotal = 2.34032 kg

      Rule for addition/subtraction: The result should be rounded to the same number of decimal places as the measurement with the fewest decimal places.
      - Mbox: 2.300 kg (3 decimal places)
      - m1: 0.02015 kg (5 decimal places)
      - m2: 0.02017 kg (5 decimal places)

      The least number of decimal places is 3 (from Mbox).
      Therefore, the total mass should be rounded to 3 decimal places.
      Mtotal = 2.340 kg.

    4. Step 4: Calculate the difference in masses of the gold pieces (Part b)

      Difference in masses (Δm) = |m2 - m1|
      Δm = |20.17 g - 20.15 g|
      Δm = 0.02 g

      Rule for addition/subtraction: The result should be rounded to the same number of decimal places as the measurement with the fewest decimal places.
      - m1: 20.15 g (2 decimal places)
      - m2: 20.17 g (2 decimal places)

      Both masses have 2 decimal places. The result (0.02 g) also has 2 decimal places, so no rounding is needed.
      Δm = 0.02 g.

    Final Answer: Verify units and significant figures.

    NEET Relevance

    This question is highly relevant for NEET as it specifically tests the application of significant figure rules for both addition and subtraction, which are distinct from multiplication/division rules. Understanding these rules is crucial for reporting experimental results accurately.

    Key Concepts

    Significant figuresRules for significant figures in addition/subtractionUnit conversion

    This question has appeared in previous NEET exams.

    22numerical🎯 HIGH⭐ Important

    A physical quantity P is related to four observables a, b, c and d as follows: P = a³b²/(√c d). The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage error in the quantity P?

    ✅ Answer

    The percentage error in the quantity P is 13%.

    Solution Steps

    1. Step 1: Recall the formula for propagation of errors

      For a quantity X = Ap Bq / (Cr Ds), the maximum fractional error is given by:
      ΔX/X = |p(ΔA/A)| + |q(ΔB/B)| + |r(ΔC/C)| + |s(ΔD/D)|

      And the maximum percentage error is:
      % Error in X = |p(% Error in A)| + |q(% Error in B)| + |r(% Error in C)| + |s(% Error in D)|

    2. Step 2: Identify the given relation and powers

      The given relation is P = a³b²/(√c d).
      This can be written as P = a³ b² c⁻¹ᐟ² d⁻¹.

      Comparing with X = Ap Bq Cr Ds:
      - For 'a': power p = 3
      - For 'b': power q = 2
      - For 'c': power r = -1/2 (since √c = c¹ᐟ² and it's in the denominator, so c⁻¹ᐟ²)
      - For 'd': power s = -1 (since d = d¹ and it's in the denominator, so d⁻¹)

    3. Step 3: List the given percentage errors

      Percentage error in a (Δa/a × 100%) = 1%
      Percentage error in b (Δb/b × 100%) = 3%
      Percentage error in c (Δc/c × 100%) = 4%
      Percentage error in d (Δd/d × 100%) = 2%

    4. Step 4: Apply the formula for percentage error in P

      Percentage error in P = |3 × (% Error in a)| + |2 × (% Error in b)| + |(-1/2) × (% Error in c)| + |(-1) × (% Error in d)|

      Note: When calculating maximum possible error, the absolute values of the powers are used, as errors always add up.
      Percentage error in P = (3 × 1%) + (2 × 3%) + (1/2 × 4%) + (1 × 2%)

    5. Step 5: Calculate the total percentage error

      Percentage error in P = (3%) + (6%) + (2%) + (2%)
      Percentage error in P = 13%.

    Final Answer: Verify units and significant figures.

    NEET Relevance

    This is a critically important question for NEET. Error analysis, especially the propagation of errors in multiplication, division, and powers, is a very frequent topic in the exam. Students must be proficient in applying this formula quickly and accurately.

    Key Concepts

    Error analysisPropagation of errorsPercentage errorPowers in error calculation

    This question has appeared in previous NEET exams.

    23numerical🎯 HIGH⭐ Important

    A physical quantity P is related to four observables a, b, c and d as follows:
    P = a3 b2 / (sqrt(c) d)
    The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P?

    ✅ Answer

    The percentage error in the quantity P is 13%.

    Solution Steps

    1. Step 1: Recall the formula for propagation of errors

      If a quantity P is given by P = (ax by) / (cz dw), then the maximum fractional error in P is given by:
      ΔP/P = x(Δa/a) + y(Δb/b) + z(Δc/c) + w(Δd/d)
      And the maximum percentage error is given by:
      (ΔP/P) × 100% = [x(Δa/a) + y(Δb/b) + z(Δc/c) + w(Δd/d)] × 100%

    2. Step 2: Identify the given relation and powers

      The given relation is P = a3 b2 / (sqrt(c) d) = a3 b2 c^(-1/2) d^(-1).
      Comparing this with the general form, we have: x = 3 (for a) y = 2 (for b) z = 1/2 (for c, as sqrt(c) = c^(1/2)) w = 1 (for d)

    3. Step 3: List the given percentage errors

      Percentage error in a (Δa/a × 100%) = 1%
      Percentage error in b (Δb/b × 100%) = 3%
      Percentage error in c (Δc/c × 100%) = 4%
      Percentage error in d (Δd/d × 100%) = 2%

    4. Step 4: Apply the error propagation formula for percentage error

      Percentage error in P = [3 × (Δa/a) + 2 × (Δb/b) + (1/2) × (Δc/c) + 1 × (Δd/d)] × 100%
      Note: Even if a term is in the denominator or has a negative power, its contribution to the total percentage error is always added, as errors are always additive.

    5. Step 5: Substitute the given values

      Percentage error in P = 3 × (1%) + 2 × (3%) + (1/2) × (4%) + 1 × (2%)

    6. Step 6: Calculate the individual contributions

      Contribution from a = 3 × 1% = 3%
      Contribution from b = 2 × 3% = 6%
      Contribution from c = (1/2) × 4% = 2%
      Contribution from d = 1 × 2% = 2%

    7. Step 7: Sum the contributions to find the total percentage error

      Total Percentage error in P = 3% + 6% + 2% + 2% = 13%

    Final Answer: Verify units and significant figures.

    NEET Relevance

    Error analysis and propagation of errors are very frequently tested topics in NEET. Questions involving percentage errors in quantities with powers are common MCQs.

    Key Concepts

    Error analysisPropagation of errorsPercentage errorPowers in error calculation

    This question has appeared in previous NEET exams.

    24long answer🎯 HIGH⭐ Important

    A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:
    (a) y = a sin (2πt/T)
    (b) y = a sin vt
    (c) y = (a/T) sin (t/a)
    (d) y = a sqrt(2) (sin (2πt/T) + cos (2πt/T))
    (a = maximum displacement of the particle, v = speed of the particle, T = time period of motion).
    Rule out the wrong formulas on dimensional grounds.

    ✅ Answer

    We will use the principle of homogeneity of dimensions, which states that the dimensions of all terms on both sides of a physical equation must be the same. Also, arguments of trigonometric functions must be dimensionless.

    Given dimensions:
    - Displacement (y): [L]
    - Maximum displacement (a): [L]
    - Speed (v): [LT⁻¹]
    - Time (t): [T]
    - Time period (T): [T]

    Let's analyze each formula:

    (a) y = a sin (2πt/T)
    * LHS: Dimension of y = [L]
    * RHS:
    * Dimension of a = [L]
    * Dimension of argument (2πt/T): 2π is a dimensionless constant.
    Dimension of (t/T) = [T]/[T] = [M⁰L⁰T⁰] (dimensionless).
    * Since the argument is dimensionless, sin(dimensionless) is dimensionless.
    * Dimension of RHS = [L] × [M⁰L⁰T⁰] = [L]
    * Conclusion: LHS dimensions = RHS dimensions ([L] = [L]). The formula is dimensionally correct.

    (b) y = a sin vt
    * LHS: Dimension of y = [L]
    * RHS:
    * Dimension of a = [L]
    * Dimension of argument (vt): Dimension of v = [LT⁻¹], Dimension of t = [T].
    Dimension of (vt) = [LT⁻¹] × [T] = [L].
    * Since the argument (vt) has dimension [L] (length), it is not dimensionless. A trigonometric function's argument must be dimensionless.
    * Conclusion: The argument of the sine function is not dimensionless. Therefore, this formula is dimensionally incorrect.

    (c) y = (a/T) sin (t/a)
    * LHS: Dimension of y = [L]
    * RHS:
    * Dimension of (a/T): Dimension of a = [L], Dimension of T = [T].
    Dimension of (a/T) = [L]/[T] = [LT⁻¹].
    * Dimension of argument (t/a): Dimension of t = [T], Dimension of a = [L].
    Dimension of (t/a) = [T]/[L] = [L⁻¹T].
    * Since the argument (t/a) has dimension [L⁻¹T], it is not dimensionless.
    * Conclusion: The argument of the sine function is not dimensionless. Also, the overall dimension of RHS ([LT⁻¹]) does not match the dimension of LHS ([L]). Therefore, this formula is dimensionally incorrect.

    (d) y = a sqrt(2) (sin (2πt/T) + cos (2πt/T))

    • LHS: Dimension of y = [L]
    • RHS:
    • Dimension of a = [L]
    • sqrt(2) is a dimensionless constant.
    • Dimension of argument (2πt/T): As shown in (a), Dimension of (2πt/T) = [M⁰L⁰T⁰] (dimensionless).
    • Since the arguments of both sin and cos are dimensionless, sin(2πt/T) and cos(2πt/T) are both dimensionless.
    • The sum of two dimensionless quantities (sin(...) + cos(...)) is also dimensionless.
    • Dimension of RHS = [L] × [M⁰L⁰T⁰] × ([M⁰L⁰T⁰] + [M⁰L⁰T⁰]) = [L] × [M⁰L⁰T⁰] = [L]
    • Conclusion: LHS dimensions = RHS dimensions ([L] = [L]). The formula is dimensionally correct.

    Formulas ruled out on dimensional grounds are (b) and (c).

      Solution Steps

      1. Step 1: State the Principle of Homogeneity of Dimensions

        The principle of homogeneity of dimensions states that a physical equation is dimensionally correct if the dimensions of all the terms on both sides of the equation are the same. Additionally, the arguments of trigonometric, logarithmic, and exponential functions must be dimensionless.

      2. Step 2: List the dimensions of given quantities

        y (displacement) = [L] a (maximum displacement) = [L] v (speed) = [LT⁻¹] t (time) = [T]
        T (time period) = [T]

      3. Step 3: Analyze formula (a): y = a sin (2πt/T)

        LHS: Dim(y) = [L]
        RHS: Dim(a) = [L]
        Argument of sin: Dim(2πt/T) = Dim(t)/Dim(T) = [T]/[T] = [M⁰L⁰T⁰] (dimensionless). Since the argument is dimensionless, sin(2πt/T) is dimensionless.
        Dim(RHS) = Dim(a) × Dim(sin(2πt/T)) = [L] × [M⁰L⁰T⁰] = [L].
        Since Dim(LHS) = Dim(RHS) = [L], formula (a) is dimensionally correct.

      4. Step 4: Analyze formula (b): y = a sin vt

        LHS: Dim(y) = [L]
        RHS: Dim(a) = [L]
        Argument of sin: Dim(vt) = Dim(v) × Dim(t) = [LT⁻¹] × [T] = [L].
        Since the argument (vt) has dimension [L] (length), it is not dimensionless. This violates the rule for trigonometric functions.
        Therefore, formula (b) is dimensionally incorrect.

      5. Step 5: Analyze formula (c): y = (a/T) sin (t/a)

        LHS: Dim(y) = [L]
        RHS: Dim(a/T) = Dim(a)/Dim(T) = [L]/[T] = [LT⁻¹].
        Argument of sin: Dim(t/a) = Dim(t)/Dim(a) = [T]/[L] = [L⁻¹T].
        Since the argument (t/a) has dimension [L⁻¹T], it is not dimensionless. This violates the rule for trigonometric functions.
        Also, the overall dimension of RHS is [LT⁻¹], which does not match Dim(LHS) = [L].
        Therefore, formula (c) is dimensionally incorrect.

      6. Step 6: Analyze formula (d): y = a sqrt(2) (sin (2πt/T) + cos (2πt/T))

        LHS: Dim(y) = [L]
        RHS: Dim(a) = [L]. sqrt(2) is a dimensionless constant.
        Argument of sin and cos: Dim(2πt/T) = Dim(t)/Dim(T) = [T]/[T] = [M⁰L⁰T⁰] (dimensionless). Thus, sin(2πt/T) and cos(2πt/T) are dimensionless.
        The sum of two dimensionless quantities (sin(...) + cos(...)) is also dimensionless.
        Dim(RHS) = Dim(a) × Dim(sqrt(2)) × Dim(sum of dimensionless terms) = [L] × [M⁰L⁰T⁰] × [M⁰L⁰T⁰] = [L].
        Since Dim(LHS) = Dim(RHS) = [L], formula (d) is dimensionally correct.

      7. Step 7: Conclude the wrong formulas

        Based on dimensional analysis, formulas (b) and (c) are dimensionally incorrect.

      NEET Relevance

      Dimensional analysis is a fundamental concept and a very common topic for MCQs in NEET. Questions involving checking dimensional correctness of formulas or deriving relations using dimensions appear frequently.

      Key Concepts

      Dimensional analysisPrinciple of homogeneity of dimensionsDimensions of physical quantitiesDimensionless quantitiesTrigonometric function arguments

      This question has appeared in previous NEET exams.

      25short answerMEDIUM

      A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m₀ of a particle in terms of its speed v and the speed of light c. (This relation first arose as a consequence of special relativity by Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes: m = m₀ / sqrt(1 - v²)
      Guess where to put the missing c.

      ✅ Answer

      The principle of homogeneity of dimensions states that all terms in a physical equation must have the same dimensions. In the given relation, the term `v²` appears inside the square root, subtracted from a dimensionless constant `1`. For the subtraction `1 - v²` to be dimensionally consistent, `v²` must also be dimensionless.

      However, `v` is speed, so its dimension is [LT⁻¹], and `v²` has dimension [L²T⁻²]. This is not dimensionless.

      To make `v²` dimensionless, it must be divided by another quantity with the same dimensions, i.e., [L²T⁻²]. The speed of light `c` has the dimension of speed, [LT⁻¹], so `c²` has the dimension [L²T⁻²].

      Therefore, to make the term `v²` dimensionless, it must be divided by `c²`. The correct relation should be:

      m = m₀ / sqrt(1 - v²/c²)

      This ensures that the term `v²/c²` is dimensionless ([L²T⁻²]/[L²T⁻²] = [M⁰L⁰T⁰]), allowing it to be subtracted from the dimensionless constant `1`.

      NEET Relevance

      While the specific formula is from special relativity, the core concept tested is dimensional analysis, which is highly relevant for NEET. Questions on dimensional consistency are common.

      Key Concepts

      Dimensional analysisPrinciple of homogeneity of dimensionsDimensionless quantitiesSpecial relativity (mass-energy relation - qualitative)
      26numericalMEDIUM⭐ Important

      The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10⁻¹⁰ m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms?

      ✅ Answer

      The total atomic volume of a mole of hydrogen atoms is approximately 3.14 × 10⁻⁷ m³.

      Solution Steps

      1. Step 1: Convert the size of a hydrogen atom to meters

        Given size of hydrogen atom = 0.5 Å
        We know 1 Å = 10⁻¹⁰ m
        Radius (r) of a hydrogen atom = 0.5 × 10⁻¹⁰ m = 5 × 10⁻¹¹ m

      2. Step 2: Calculate the volume of a single hydrogen atom

        Assuming a hydrogen atom is spherical, its volume (Vatom) is given by the formula for the volume of a sphere:
        Vatom = (4/3)πr³
        Vatom = (4/3) × 3.14159 × (5 × 10⁻¹¹ m)³
        Vatom = (4/3) × 3.14159 × (125 × 10⁻³³ m³)
        Vatom = (500/3) × 3.14159 × 10⁻³³ m³
        Vatom ≈ 523.598 × 10⁻³³ m³ ≈ 5.236 × 10⁻³¹ m³

      3. Step 3: Determine the number of hydrogen atoms in a mole

        A mole of any substance contains Avogadro's number (NA) of particles.
        Avogadro's number (NA) = 6.022 × 10²³ atoms/mole

      4. Step 4: Calculate the total atomic volume for a mole of hydrogen atoms

        Total atomic volume (Vtotal) = Vatom × NA
        Vtotal = (5.236 × 10⁻³¹ m³) × (6.022 × 10²³)
        Vtotal = (5.236 × 6.022) × 10^(-31 + 23) m³
        Vtotal = 31.536 × 10⁻⁸ m³
        Vtotal ≈ 3.1536 × 10⁻⁷ m³
        Rounding to two significant figures (as 0.5 Å has one significant figure, but 3.14 is often used for pi), we can say:
        Vtotal ≈ 3.14 × 10⁻⁷ m³

      Final Answer: Verify units and significant figures.

      NEET Relevance

      This question combines unit conversion, basic geometry, and the mole concept, all of which are relevant for NEET. Such calculations often appear in physics and chemistry sections.

      Key Concepts

      Unit conversion (Angstrom to meter)Volume of a sphereAvogadro's numberMole conceptAtomic scale calculations

      This question has appeared in previous NEET exams.

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