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Unit 1: Units and Measurement | Physics | Class 11th
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Unit 1: Units and Measurement

1. Need for Measurement

Physics is the study of the fundamental laws of nature, and its primary goal is to explain the physical world around us. To achieve this, physics relies heavily on observation and experimentation. However, mere qualitative descriptions like "hot," "cold," "fast," or "slow" are subjective and insufficient for scientific inquiry. To establish and verify scientific laws, we need to make precise, objective, and reproducible measurements. This process of associating numbers with physical phenomena is called measurement.

Any quantity that can be measured is called a physical quantity. Measurement involves comparing a given physical quantity with a standard, internationally accepted reference of the same kind, known as a unit. The result of a measurement is always expressed as a numerical value followed by its corresponding unit.

Result of Measurement = Numerical Value (n) × Unit (u)

For example, if the length of a rod is measured as 5 meters, '5' is the numerical value and 'meter' is the unit. This quantitative approach is the bedrock of the scientific method, allowing for verification, comparison, and the formulation of mathematical laws.

2. Units of Measurement

A unit is a standard of measurement for a physical quantity. A well-defined system of units is crucial for global scientific communication and commerce. A good unit must possess the following characteristics:

2.1 Systems of Units

A complete set of units for all kinds of physical quantities (both fundamental and derived) is called a system of units. Historically, several systems have been in use:

The use of different systems created inconsistencies and difficulties in scientific collaboration. This led to the need for a single, globally accepted system.

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3. The International System of Units (SI)

To establish a universal standard, the General Conference on Weights and Measures (CGPM) developed and recommended the **International System of Units (Système Internationale d’Unités or SI)** in 1971. The SI is a rationalized and coherent system of units that is now the standard for scientific, technical, and commercial work worldwide.

The SI system is built upon seven fundamental (base) units and two supplementary units. All other units, known as derived units, can be expressed as combinations of these base units.

3.1 SI Base (Fundamental) Units and their Modern Definitions

These are the seven independent units that form the foundation of the SI system. Their definitions have evolved from physical artifacts to being based on fundamental constants of nature, ensuring their stability and universality.

1. Meter (m) - Unit of Length

The meter is defined by taking the fixed numerical value of the speed of light in vacuum, c, to be 299,792,458 when expressed in the unit m s-1, where the second is defined in terms of the caesium frequency.

2. Kilogram (kg) - Unit of Mass

The kilogram is defined by taking the fixed numerical value of the Planck constant, h, to be 6.62607015 × 10-34 when expressed in the unit J s, which is equal to kg m2 s-1.

3. Second (s) - Unit of Time

The second is defined by taking the fixed numerical value of the caesium frequency, ΔνCs, the unperturbed ground-state hyperfine transition frequency of the caesium-133 atom, to be 9,192,631,770 when expressed in the unit Hz, which is equal to s-1.

4. Ampere (A) - Unit of Electric Current

The ampere is defined by taking the fixed numerical value of the elementary charge, e, to be 1.602176634 × 10-19 when expressed in the unit C, which is equal to A s.

5. Kelvin (K) - Unit of Thermodynamic Temperature

The kelvin is defined by taking the fixed numerical value of the Boltzmann constant, kB, to be 1.380649 × 10-23 when expressed in the unit J K-1, which is equal to kg m2 s-2 K-1.

6. Mole (mol) - Unit of Amount of Substance

The mole is defined by taking the fixed numerical value of the Avogadro constant, NA, to be 6.02214076 × 1023 when expressed in the unit mol-1. One mole contains exactly 6.02214076 × 1023 elementary entities.

7. Candela (cd) - Unit of Luminous Intensity

The candela is defined by taking the fixed numerical value of the luminous efficacy of monochromatic radiation of frequency 540 × 1012 Hz, Kcd, to be 683 when expressed in the unit lm W-1, which is equal to cd sr W-1, or cd sr kg-1 m-2 s3.

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3.2 SI Supplementary Units

These units are for purely geometrical quantities and are dimensionless.

4. Derived Units

Derived units are the units of physical quantities that can be expressed in terms of the fundamental (base) units. They are formed by mathematically combining the base units. Below are some important derived quantities and their units.

Area (A)

Defining Formula: Area = Length × Breadth
Expression in Base Units: m × m = m2
Dimensional Formula: [L2]

Volume (V)

Defining Formula: Volume = Length × Breadth × Height
Expression in Base Units: m × m × m = m3
Dimensional Formula: [L3]

Density (ρ)

Defining Formula: Density = Mass / Volume
Expression in Base Units: kg / m3 = kg m-3
Dimensional Formula: [M L-3]

Velocity (v)

Defining Formula: Velocity = Displacement / Time
Expression in Base Units: m / s = m s-1
Dimensional Formula: [L T-1]

Acceleration (a)

Defining Formula: Acceleration = Change in velocity / Time
Expression in Base Units: (m s-1) / s = m s-2
Dimensional Formula: [L T-2]

Force (F)

Defining Formula: Force = Mass × Acceleration
Expression in Base Units: kg × (m s-2) = kg m s-2
Special SI Name: Newton (N)
Dimensional Formula: [M L T-2]

Work / Energy (W or E)

Defining Formula: Work = Force × Distance
Expression in Base Units: (kg m s-2) × m = kg m2 s-2
Special SI Name: Joule (J)
Dimensional Formula: [M L2 T-2]

Power (P)

Defining Formula: Power = Work / Time
Expression in Base Units: (kg m2 s-2) / s = kg m2 s-3
Special SI Name: Watt (W)
Dimensional Formula: [M L2 T-3]

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5. Significant Figures

Every measurement is subject to some degree of uncertainty due to the limitations of the measuring instrument and the skill of the observer. Significant figures are the digits in a measured value that are known with certainty, plus one additional digit that is estimated or uncertain. Reporting a measurement with the correct number of significant figures is crucial as it communicates the precision of the measurement.

Accuracy vs. Precision

Accuracy refers to how close a measured value is to the true or accepted value of the quantity. It indicates the correctness of the measurement.
Precision refers to how close multiple measurements of the same quantity are to each other. It indicates the resolution or limit of the measuring instrument and the consistency of the measurements.

A measurement can be precise but not accurate (e.g., a faulty scale consistently gives the same wrong reading), accurate but not precise (e.g., measurements fluctuate around the true value), or both accurate and precise.

5.1 Rules for Counting Significant Figures

  1. All non-zero digits are significant.
    Example: 285 cm has 3 significant figures. 1.345 has 4 significant figures.
  2. All zeros occurring between two non-zero digits are significant, no matter where the decimal point is.
    Example: 2005 kg has 4 significant figures. 5.03 m has 3 significant figures. 1.0008 has 5 significant figures.
  3. If the number is less than 1, the zero(s) on the right of the decimal point but to the left of the first non-zero digit are not significant. These are leading zeros and only indicate the position of the decimal point.
    Example: In 0.00345, there are 3 significant figures (3, 4, 5). In 0.0506, there are 3 significant figures (5, 0, 6).
  4. In a number with a decimal point, the trailing zeros (zeros at the end) are significant.
    Example: 12.300 has 5 significant figures. 0.004200 has 4 significant figures. These zeros indicate the level of precision of the measurement.
  5. In a number without a decimal point, the trailing zeros are not significant (ambiguous).
    Example: 4700 m has 2 significant figures. To avoid ambiguity, such numbers should be expressed in scientific notation.
    4.7 × 103 m has 2 significant figures.
    4.70 × 103 m has 3 significant figures.
    4.700 × 103 m has 4 significant figures.
  6. Exact numbers have an infinite number of significant figures. These include numbers obtained by counting (e.g., "20 students" in a class) or by definition (e.g., "1 inch = 2.54 cm exactly"). These numbers do not limit the precision of a calculation.
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5.2 Rules for Arithmetic Operations with Significant Figures

The result of a calculation involving measured values cannot be more precise than the least precise measurement used. This principle is reflected in the rules for arithmetic operations.

(a) In Multiplication or Division:

The final result should retain the same number of significant figures as the original number with the least number of significant figures.

Solved Example 1: Multiplication

Question: The length and breadth of a rectangle are measured as 5.7 cm and 3.42 cm respectively. Calculate the area of the rectangle to the correct number of significant figures.

Solution:
Given: length (l) = 5.7 cm (2 significant figures)
breadth (b) = 3.42 cm (3 significant figures)
Area (A) = l × b = 5.7 cm × 3.42 cm = 19.494 cm2.
The least number of significant figures in the input values is 2. Therefore, the result must be rounded to 2 significant figures.
Answer: Area = 19 cm2.

(b) In Addition or Subtraction:

The final result should retain the same number of decimal places as the original number with the fewest number of decimal places.

Solved Example 2: Addition

Question: The lengths of three rods are measured as 10.12 m, 5.3 m, and 0.235 m. Find their total length to the correct number of significant figures.

Solution:
Given: l1 = 10.12 m (2 decimal places)
l2 = 5.3 m (1 decimal place)
l3 = 0.235 m (3 decimal places)
Total length = 10.12 + 5.3 + 0.235 = 15.655 m.
The least number of decimal places in the input values is 1 (from 5.3 m). Therefore, the result must be rounded to 1 decimal place.
Answer: Total length = 15.7 m.

5.3 Rules for Rounding Off the Uncertain Digit

To report a result with the correct number of significant figures, rounding off is necessary. The standard rules are:

  1. If the digit to be dropped is less than 5, the preceding digit is left unchanged. (e.g., 2.74 is rounded to 2.7 for 2 sig figs)
  2. If the digit to be dropped is greater than 5, the preceding digit is increased by one. (e.g., 2.76 is rounded to 2.8 for 2 sig figs)
  3. If the digit to be dropped is 5 followed by non-zero digits, the preceding digit is increased by one. (e.g., 2.751 is rounded to 2.8 for 2 sig figs)
  4. If the digit to be dropped is 5 (or 5 followed by only zeros):
    • and the preceding digit is even, it is left unchanged. (e.g., 2.450 is rounded to 2.4 for 2 sig figs)
    • and the preceding digit is odd, it is increased by one. (e.g., 2.350 is rounded to 2.4 for 2 sig figs)
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6. Dimensions of Physical Quantities

The dimensions of a physical quantity are the powers to which the fundamental (base) units of Mass, Length, Time, etc., must be raised to represent the unit of that quantity. This provides a way to express the fundamental nature of a physical quantity without regard to its numerical value or the system of units used.

The dimensions of the seven base quantities are represented by symbols in square brackets: Mass [M], Length [L], Time [T], Electric Current [A], Thermodynamic Temperature [K], Amount of Substance [mol], and Luminous Intensity [cd].

A dimensional formula is an expression that shows how and which of the base quantities represent the dimensions of a physical quantity. For example, the dimensional formula for velocity is [M0 L1 T-1].

A dimensional equation is an equation obtained by equating a physical quantity with its dimensional formula. For example, [v] = [M0 L1 T-1].

6.1 Principle of Homogeneity of Dimensions

Principle Statement:

This fundamental principle states that the dimensions of all the terms on both sides of a physically correct equation must be identical. A direct consequence of this principle is that only physical quantities having the same dimensions can be added to, subtracted from, or equated with each other.

For an equation A = B + C, for it to be dimensionally valid, it must hold that [A] = [B] = [C].

This principle is the cornerstone of dimensional analysis and is used to check the correctness of equations and derive relationships between physical quantities.

7. Dimensional Analysis and its Applications

Dimensional analysis is a powerful analytical tool based on the Principle of Homogeneity of Dimensions. It has several critical applications in physics.

7.1 To Check the Dimensional Correctness of a Given Equation

This application serves as a first-pass check for the validity of any physical formula. If the dimensions on both sides of an equation do not match, the equation is definitively incorrect. However, a dimensionally correct equation is not necessarily correct, as dimensional analysis cannot verify dimensionless constants or the exact functional form.

Solved Example 3: Checking Consistency of Einstein's Mass-Energy Equivalence

Question: Check the dimensional correctness of the equation E = mc2, where E is energy, m is mass, and c is the speed of light.

Solution:
Dimension of Energy (E) = Dimension of Work = [M L2 T-2].

Now, for the right-hand side (RHS):
Dimension of mass (m) = [M].
Dimension of speed of light (c) = [L T-1].
Dimension of c2 = ([L T-1])2 = [L2 T-2].

Dimension of RHS = [m][c2] = [M] × [L2 T-2] = [M L2 T-2].

Since the dimensions of the LHS ([M L2 T-2]) are equal to the dimensions of the RHS ([M L2 T-2]), the equation is dimensionally correct.
Answer: The equation E = mc2 is dimensionally correct.

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7.2 To Convert a Physical Quantity from One System of Units to Another

This application relies on the principle that the magnitude of a physical quantity remains invariant regardless of the system of units used (n1U1 = n2U2). By expressing the units in terms of their fundamental dimensions, we can systematically find the conversion factor.

Solved Example 4: Unit Conversion of Joule to Erg

Question: Convert 1 Joule (SI unit of work) into ergs (CGS unit of work).

Solution:
Dimensional formula of Work/Energy = [M1 L2 T-2]. So, a=1, b=2, c=-2.

In SI system (System 1):
n1 = 1, M1 = 1 kg, L1 = 1 m, T1 = 1 s.
In CGS system (System 2):
n2 = ?, M2 = 1 g, L2 = 1 cm, T2 = 1 s.

Using the conversion formula: n2 = n1 (M1/M2)a (L1/L2)b (T1/T2)c
n2 = 1 × (1 kg / 1 g)1 × (1 m / 1 cm)2 × (1 s / 1 s)-2
n2 = 1 × (1000 g / 1 g)1 × (100 cm / 1 cm)2 × (1)
n2 = 1 × (1000) × (100)2 = 103 × 104 = 107.
Answer: 1 Joule = 107 ergs.

7.3 To Derive a Relationship Between Physical Quantities

If we have a hypothesis about the factors on which a physical quantity depends, dimensional analysis can help derive the formula that relates them (assuming a product-of-powers relationship).

Solved Example 5: Deriving the Formula for Centripetal Force

Question: Derive an expression for the centripetal force (F) acting on a particle of mass (m) moving with velocity (v) in a circle of radius (r).

Solution:
Assume the relationship is: F = k × ma vb rc, where k is a dimensionless constant.

Write the dimensions of all quantities:
[F] = [M L T-2]
[m] = [M]
[v] = [L T-1]
[r] = [L]

Substitute these into the assumed equation:
[M1 L1 T-2] = [M]a [L T-1]b [L]c
[M1 L1 T-2] = [Ma Lb+c T-b]

By the Principle of Homogeneity, equate the powers of M, L, and T on both sides:

  • For Mass (M): a = 1
  • For Length (L): b + c = 1 ---(i)
  • For Time (T): -2 = -b ---(ii)

From (ii), we get b = 2.
Substitute b in (i): 2 + c = 1c = -1.

Substitute a, b, and c back into the assumed relationship:
F = k × m1 v2 r-1
F = k (mv2 / r).
Experimentally, it is found that k = 1.
Answer: The derived relation is F = k(mv2/r).

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7.4 Limitations of Dimensional Analysis

While extremely useful, dimensional analysis is not a complete method and has certain limitations:

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8. Errors in Measurement

The measuring process is essentially a process of comparison. In spite of our best efforts, the measured value of a quantity is always somewhat different from its actual value, or true value. This difference in the true value and the measured value of a quantity is called error of measurement.

8.1 Absolute, Relative and Percentage Error

Let the measured values be a1, a2, ..., an. The arithmetic mean of these values is taken as the true value if it's not given: amean = (a1 + a2 + ... + an) / n.

Example 1: Absolute and Percentage Error

Question: The diameter of a wire is measured to be 1.02 mm, 1.03 mm, and 1.01 mm. Find the percentage error.

Solution:
Mean diameter d = (1.02 + 1.03 + 1.01) / 3 = 1.02 mm.
Absolute errors: Δd1 = 0, Δd2 = 0.01 mm, Δd3 = 0.01 mm.
Mean absolute error Δdmean = (0 + 0.01 + 0.01) / 3 = 0.0067 mm.
Percentage error = (0.0067 / 1.02) × 100 = 0.66%.

Example 2: Combination of Errors

Question: Two resistors R1 = 100 ± 3 Ω and R2 = 200 ± 4 Ω are connected in series. Find the equivalent resistance.

Solution:
Equivalent resistance in series Rs = R1 + R2.
Rs = 100 + 200 = 300 Ω.
Error ΔRs = ΔR1 + ΔR2 = 3 + 4 = 7 Ω.
So, Rs = 300 ± 7 Ω.

Example 3: Error in Product

Question: The mass and volume of a body are M = (22.00 ± 0.05) g and V = (4.0 ± 0.1) cm3. Find the maximum percentage error in density.

Solution:
Density ρ = M / V.
Percentage error in ρ = Percentage error in M + Percentage error in V
(Δρ / ρ) × 100 = (0.05 / 22.00) × 100 + (0.1 / 4.0) × 100 = 0.23% + 2.5% = 2.73%.

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Important Questions and Answers

Conceptual Questions

Q1: What is a unit? State the essential characteristics of a standard unit.

A: A unit is a definite magnitude of a physical quantity, defined and adopted by convention or by law, that is used as a standard for measurement of the same kind of quantity.
The essential characteristics of a standard unit are:
1. Invariability: It should not change with time or under different physical conditions.
2. Availability: It should be easily available for comparison.
3. Reproducibility: It should be easily reproducible.
4. International Acceptance: It should be universally accepted for global consistency.

Q2: Distinguish between accuracy and precision of a measurement.

A: Accuracy refers to the closeness of a measured value to the true value of the physical quantity. Precision refers to the resolution or the limit to which the quantity is measured, indicating the consistency of repeated measurements.

Q3: Can a quantity have dimensions but still be a constant? Give an example.

A: Yes. Such quantities are called dimensional constants.
Example: The Universal Gravitational Constant (G) has dimensions [M-1 L3 T-2] but its value is constant throughout the universe (6.67 × 10-11 N m2 kg-2).

Q4: Why must the arguments of trigonometric and logarithmic functions be dimensionless?

A: These functions have series expansions. For example, sin(x) = x - x3/3! + x5/5! - .... According to the Principle of Homogeneity, we can only add or subtract quantities with the same dimensions. If 'x' had dimensions, then x, x3, x5 etc., would all have different dimensions, and we could not add or subtract them. Therefore, 'x' must be dimensionless.

Q5: Why are SI units considered a coherent system of units?

A: The SI system is considered coherent because all derived units can be obtained by simple multiplication or division of the base units without the need for any numerical conversion factors. For example, the unit of force (Newton) is directly derived as kg × m / s2, with a proportionality constant of 1.

Q6: State the Principle of Homogeneity of Dimensions.

A: The Principle of Homogeneity of Dimensions states that the dimensions of all the terms in a physically correct equation must be the same. This means we can only add, subtract, or equate quantities that have the same physical dimensions.

Q7: Is a dimensionally correct equation necessarily a physically correct equation?

A: No. A dimensionally correct equation is not necessarily a physically correct equation. For example, the equation for kinetic energy K = 3mv2 is dimensionally correct ([M L2 T-2] on both sides), but it is physically incorrect because the dimensionless constant should be 1/2, not 3. Dimensional correctness is a necessary but not a sufficient condition for an equation to be physically correct.

Q8: What is the primary limitation of using dimensional analysis to derive formulas?

A: The primary limitation is its inability to determine the value of dimensionless constants of proportionality (like k, 2π, etc.). These constants must be determined through experiments or more advanced theories.

Q9: Why was the kilogram redefined in terms of Planck's constant instead of the physical prototype?

A: The physical prototype (Le Grand K) was found to be losing mass over time, which made it an unstable standard. Redefining the kilogram in terms of a fundamental constant of nature, Planck's constant (h), ensures that its value is universal, stable, and can be reproduced with high accuracy in any well-equipped laboratory, making it a more reliable standard.

Q10: What is the number of significant figures in 3.0800 and 0.000532?

A:
✔ In 3.0800, there are 5 significant figures (zeros between non-zero digits are significant, and trailing zeros after a decimal are significant).
✔ In 0.000532, there are 3 significant figures (leading zeros are not significant).

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Numerical Problems

Q11: The length, breadth, and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Calculate the area and volume of the sheet to correct significant figures.

A:
Given: Length (l) = 4.234 m (4 sig figs), Breadth (b) = 1.005 m (4 sig figs), Thickness (t) = 2.01 cm = 0.0201 m (3 sig figs).

Area (A) = 2(lb + bt + tl) (Assuming total surface area)
lb = 4.234 × 1.005 = 4.25517 m2
bt = 1.005 × 0.0201 = 0.0202005 m2
tl = 0.0201 × 4.234 = 0.0851034 m2
Now, rounding each product to least sig figs (3): 4.26, 0.0202, 0.0851.
Sum = 4.26 + 0.0202 + 0.0851 = 4.3653 m2. Rounding to 2 decimal places: 4.37 m2.
Total Area = 2 × 4.37 = 8.74 m2.

Volume (V) = l × b × t
V = 4.234 × 1.005 × 0.0201 = 0.0855289 m3.
The least number of significant figures is 3 (from thickness). So, rounding the volume to 3 significant figures: 0.0855 m3.
Answer: Total Area ≈ 8.74 m2, Volume ≈ 0.0855 m3.

Q12: 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?

A:
Given: P = a3b2 / (c1/2 d1).
The relative error in P is given by:
ΔP/P = 3(Δa/a) + 2(Δb/b) + (1/2)(Δc/c) + 1(Δd/d)
The percentage error in P is:
(ΔP/P)×100 = 3(Δa/a×100) + 2(Δb/b×100) + (1/2)(Δc/c×100) + 1(Δd/d×100)
= 3(1%) + 2(3%) + (1/2)(4%) + 1(2%)
= 3% + 6% + 2% + 2% = 13%.
Answer: The percentage error in the quantity P is 13%.

Q13: Check whether the equation λ = h/mv is dimensionally correct, where λ is wavelength, h is Planck's constant, m is mass, and v is velocity.

A:
Dimension of Wavelength (λ) [LHS] = [L].

For RHS:
Dimension of Planck's constant (h) = [Energy]/[Frequency] = [M L2 T-2]/[T-1] = [M L2 T-1].
Dimension of mass (m) = [M].
Dimension of velocity (v) = [L T-1].

Dimension of RHS = [h]/([m][v]) = [M L2 T-1] / ([M] × [L T-1])
= [M1-1 L2-1 T-1-(-1)] = [M0 L1 T0] = [L].
Since LHS [L] = RHS [L], the equation is dimensionally correct.
Answer: The equation is dimensionally correct.

Q14: The frequency (ν) of vibration of a stretched string depends on its length (l), its mass per unit length (m), and the tension (T) in the string. Obtain the expression for frequency by the method of dimensions.

A:
Assume: ν = k × la mb Tc, where k is a dimensionless constant.

Dimensions of quantities:
[ν] = [T-1]
[l] = [L]
[m] (mass per length) = [M]/[L] = [M L-1]
[T] (tension, a force) = [M L T-2]

Substitute into the equation:
[M0 L0 T-1] = [L]a [M L-1]b [M L T-2]c
[M0 L0 T-1] = [Mb+c La-b+c T-2c]

Equating powers:
For M: b + c = 0b = -c
For L: a - b + c = 0
For T: -1 = -2cc = 1/2

Substitute c into first equation: b = -1/2.
Substitute b and c into second equation: a - (-1/2) + (1/2) = 0a + 1 = 0a = -1.

So, ν = k × l-1 m-1/2 T1/2 = (k/l)√(T/m).
Answer: The expression for frequency is ν = (k/l)√(T/m).

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Q15: A cube has a side of length 1.2 × 10-2 m. Calculate its volume to correct significant figures.

A:
Given: Side (l) = 1.2 × 10-2 m (2 significant figures).
Volume (V) = l3 = (1.2 × 10-2 m)3
V = (1.2)3 × (10-2)3 m3 = 1.728 × 10-6 m3.
Since the input value has 2 significant figures, the result must be rounded to 2 significant figures.
V = 1.7 × 10-6 m3.
Answer: The volume of the cube is 1.7 × 10-6 m3.

Q16: The value of G in the CGS system is 6.67 × 10-8 dyne cm2 g-2. Calculate its value in the SI system.

A:
Dimensional formula for G = [M-1 L3 T-2]. So, a=-1, b=3, c=-2.

In CGS system (System 1):
n1 = 6.67 × 10-8, M1 = 1 g, L1 = 1 cm, T1 = 1 s.
In SI system (System 2):
n2 = ?, M2 = 1 kg, L2 = 1 m, T2 = 1 s.

Using n2 = n1 (M1/M2)a (L1/L2)b (T1/T2)c:
n2 = 6.67 × 10-8 × (1 g / 1 kg)-1 × (1 cm / 1 m)3 × (1 s / 1 s)-2
n2 = 6.67 × 10-8 × (1 g / 1000 g)-1 × (1 cm / 100 cm)3 × (1)
n2 = 6.67 × 10-8 × (10-3)-1 × (10-2)3
n2 = 6.67 × 10-8 × 103 × 10-6
n2 = 6.67 × 10-8+3-6 = 6.67 × 10-11.
Answer: The value of G in the SI system is 6.67 × 10-11 N m2 kg-2.

Q17: Subtract 0.23 J from 7.3 J and express the result with correct number of significant figures.

A:
Given: 7.3 J (1 decimal place) and 0.23 J (2 decimal places).
Subtraction: 7.3 - 0.23 = 7.07 J.
According to the rule for subtraction, the result should have the same number of decimal places as the number with the fewest decimal places. Here, 7.3 J has 1 decimal place.
So, we must round 7.07 J to 1 decimal place.
Answer: 7.1 J.

Q18: The radius of a sphere is measured to be (5.3 ± 0.1) cm. Calculate the percentage error in the calculation of its volume.

A:
Given: Radius (r) = 5.3 cm, Absolute error in radius (Δr) = 0.1 cm.
Volume of sphere (V) = (4/3)πr3.
The relative error in volume is given by: ΔV/V = 3(Δr/r) (since 4/3 and π are constants).
The percentage error in volume is: (ΔV/V)×100 = 3(Δr/r)×100.
First, find percentage error in radius: (Δr/r)×100 = (0.1 / 5.3) × 100 ≈ 1.886%.
Now, percentage error in volume = 3 × 1.886% ≈ 5.658%.
Rounding to one decimal place: 5.7%.
Answer: The percentage error in the volume is approximately 5.7%.

Q19: A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s, and 92 s. If the minimum division in the measuring clock is 1 s, what is the reported mean time? What is the relative error?

A:
Measured times: 90 s, 91 s, 95 s, 92 s.
Mean time (tmean) = (90 + 91 + 95 + 92) / 4 = 368 / 4 = 92 s.

Absolute errors (Δti = |tmean - ti|):
Δt1 = |92 - 90| = 2 s
Δt2 = |92 - 91| = 1 s
Δt3 = |92 - 95| = 3 s
Δt4 = |92 - 92| = 0 s

Mean absolute error (Δtmean) = (2 + 1 + 3 + 0) / 4 = 6 / 4 = 1.5 s.
Rounding to one significant figure (as per convention for errors): 2 s.
The reported mean time is 92 ± 2 s.

Relative error = Δtmean / tmean = 2 / 92 ≈ 0.0217.
Answer: The reported mean time is 92 ± 2 s. The relative error is approximately 0.02.

Q20: The resistance R = V/I where V = (100 ± 5) V and I = (10 ± 0.2) A. Find the percentage error in R.

A:
Given: V = 100 V, ΔV = 5 V, I = 10 A, ΔI = 0.2 A.
The percentage error in V = (ΔV/V) × 100 = (5/100) × 100 = 5%.
The percentage error in I = (ΔI/I) × 100 = (0.2/10) × 100 = 2%.
Since R = V/I, the maximum percentage error in R is the sum of the percentage errors in V and I.
Percentage error in R = (Percentage error in V) + (Percentage error in I)
= 5% + 2% = 7%.
Answer: The percentage error in R is 7%.

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Unit 1: Units and Measurement | Physics | Class 11th
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NEET Based Objective Questions

Q21: The dimensional formula of magnetic flux is:

A: [M L2 T-2 A-1]
Solution: Magnetic flux Φ = B × A. B = F / (I × L) = [M L T-2] / [A L] = [M T-2 A-1]. Thus, Φ = [M T-2 A-1] × [L2] = [M L2 T-2 A-1].

Q22: A physical quantity x is given by x = 2k3l2 / (m√n). The percentage errors in measurements of k, l, m and n are 1%, 2%, 3% and 4% respectively. The value of x is uncertain by:

A: 12%
Solution: % error in x = 3(% error in k) + 2(% error in l) + 1(% error in m) + (1/2)(% error in n)
= 3(1%) + 2(2%) + 3% + (1/2)(4%) = 3% + 4% + 3% + 2% = 12%.

Q23: Planck's constant (h), speed of light in vacuum (c) and Newton's gravitational constant (G) are three fundamental constants. Which of the following combinations of these has the dimension of length?

A: √(hG / c3)
Solution: Let L ∝ ha cb Gc. Writing dimensions: [L1] = [M L2 T-1]a [L T-1]b [M-1 L3 T-2]c.
Comparing powers: M: a - c = 0 → a = c. L: 2a + b + 3c = 1. T: -a - b - 2c = 0.
Solving these gives a = 1/2, b = -3/2, c = 1/2. Thus L = √(hG / c3).

Q24: A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is:

A: 0.5 mm
Solution: Least count = Pitch / Number of divisions on circular scale.
0.01 mm = Pitch / 50 → Pitch = 0.01 × 50 = 0.5 mm.

Q25: In an experiment, four quantities a, b, c and d are measured with percentage errors 1%, 2%, 3% and 4% respectively. Quantity P is calculated as P = (a3 b2) / (cd). % error in P is:

A: 14%
Solution: (ΔP/P) × 100 = 3(1%) + 2(2%) + 3% + 4% = 3 + 4 + 3 + 4 = 14%.

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Unit 1: Units and Measurement | Physics | Class 11th
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Q26: The dimensions of (μ0 ε0)-1/2 are:

A: [L T-1]
Solution: c = 1 / √(μ0 ε0) where c is the speed of light. Therefore, the dimensions are that of velocity, i.e., [L T-1].

Q27: Which of the following is the most precise device for measuring length?

A: An optical instrument that can measure length to within a wavelength of light.
Solution: Precision depends on least count. A vernier caliper has LC of 0.1 mm, screw gauge has 0.01 mm. Wavelength of light is around 5000 Å or 5 × 10-4 mm, which gives the smallest least count and highest precision.

Q28: If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be:

A: [E V-2 T-2]
Solution: Surface tension S = Force / Length = Energy / Area. Area = L2 = (V T)2. So, S = E / (V2 T2) = [E V-2 T-2].

Q29: A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of -0.004 cm, the correct diameter of the ball is:

A: 0.529 cm
Solution: Measured value = Main scale reading + (Circular scale division × Least count) = 0.5 cm + (25 × 0.001 cm) = 0.525 cm.
Correct diameter = Measured value - Zero error = 0.525 cm - (-0.004 cm) = 0.529 cm.

Q30: The unit of thermal conductivity is:

A: W m-1 K-1
Solution: The rate of heat flow is dQ/dt = K A (dT/dx).
K = (dQ/dt) / (A (dT/dx)). Unit of dQ/dt is Watt (W). Unit of Area is m2. Unit of temperature gradient is K/m.
Thus, Unit of K = W / (m2 · (K/m)) = W m-1 K-1.