UNITS AND MEASUREMENT – Guided Solutions
Work through every question with crystal‑clear steps, stress–strain visuals and exam‑oriented hints – all in a sleek deep‑space layout.
Q1 Q1. Fill in the blanks
Concept Brief
This question tests fundamental skills in unit conversion and dimensional understanding, which form the foundation of all physical calculations. In physics, quantities are expressed in the SI system of units, and it is often necessary to convert between different unit systems such as:
- cm → m
- mm → cm → m
- km h−1 → m s−1
- g cm−3 → kg m−3
Remember the key relations:
- \(1\ \mathrm{cm} = 10^{-2}\ \mathrm{m}\)
- \(1\ \mathrm{mm} = 10^{-3}\ \mathrm{m}\)
- \(1\ \mathrm{km\,h^{-1}} = \frac{1000}{3600}\ \mathrm{m\,s^{-1}}\)
- \(1\ \mathrm{g\,cm^{-3}} = 10^{3}\ \mathrm{kg\,m^{-3}}\)
Solution Map
This problem can be solved through the following logical steps:
- Convert given dimensions to the required units.
- Use appropriate geometric formulas.
- Apply SI conversion factors.
- Express the result in scientific notation where appropriate.
(a) The volume of a cube of side 1 cm is equal to ..... \( \mathrm{m^3} \)
Solution
Side of cube
\[ 1\ \text{cm} = 10^{-2}\ \text{m} \]Volume of cube
\[ V = a^3 \] \[ V = (10^{-2})^3 \] \[ V = 10^{-6}\ \text{m}^3 \]Answer: \(10^{-6}\ \mathrm{m^3}\)
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to .... \( \mathrm{(mm)^2} \)
Convert dimensions:
\[ r = 2.0\ \text{cm} = 20\ \text{mm} \] \[ h = 10.0\ \text{cm} = 100\ \text{mm} \]Total surface area of a cylinder:
\[ A = 2\pi r(h+r) \] \[ A = 2\pi \times 20 \times (100+20) \] \[ A = 4800\pi\ \text{mm}^2 \] \[ A \approx 1.5 \times 10^{4}\ \text{mm}^2 \]Answer: \(1.5\times10^4\ \mathrm{mm^2}\)
(c) A vehicle moving with a speed of \(18\ \mathrm{km\,h^{-1}}\) covers .... m in 1 s
Convert speed into SI units:
\[ 18\ \text{km h}^{-1} \] \[ = 18 \times \frac{1000}{3600} \] \[ = 5\ \text{m s}^{-1} \]Distance covered in \(1\ \text{s}\):
\[ d = vt \] \[ d = 5 \times 1 \] \[ d = 5\ \text{m} \]Answer: 5 m
(d) The relative density of lead is 11.3. Its density is .... \( \mathrm{g\,cm^{-3}} \) or .... \( \mathrm{kg\,m^{-3}} \)
Relative density is defined as:
\[ \text{Relative density} = \frac{\text{Density of substance}}{\text{Density of water}} \]Density of water
\[ = 1\ \text{g cm}^{-3} \]Therefore density of lead
\[ = 11.3 \times 1 \] \[ = 11.3\ \text{g cm}^{-3} \]Convert to SI units:
\[ 1\ \text{g cm}^{-3} = 10^3\ \text{kg m}^{-3} \] \[ 11.3\ \text{g cm}^{-3} \] \[ = 11.3 \times 10^3 \] \[ = 1.13\times10^4\ \text{kg m}^{-3} \]Answer:
\[ 11.3\ \text{g cm}^{-3} \] \[ 1.13\times10^4\ \text{kg m}^{-3} \]Why This Question Matters
- Builds mastery of unit conversion, one of the most common sources of mistakes in physics.
- Strengthens understanding of SI units and derived quantities.
- Tests familiarity with geometric formulas used in physics problems.
- Similar conversions frequently appear in JEE Main, NEET, BITSAT, and Olympiad exams.
Q2 Fill in the blanks by suitable conversion of units
Concept Brief
This question tests the ability to convert physical quantities between different unit systems such as SI units and CGS units. Such conversions are common in physics because many classical formulas were historically derived using the CGS system.
Key relations used in this problem:
- \(1\ \mathrm{kg} = 10^{3}\ \mathrm{g}\)
- \(1\ \mathrm{m} = 10^{2}\ \mathrm{cm}\)
- \(1\ \mathrm{m^{3}} = 10^{6}\ \mathrm{cm^{3}}\)
- \(1\ \mathrm{kg} = 10^{3}\ \mathrm{g}\)
- \(1\ \mathrm{ly} \approx 9.46\times10^{15}\ \mathrm{m}\)
- \(1\ \mathrm{h} = 3600\ \mathrm{s}\)
Solution Map
- Express the given quantity using fundamental units.
- Convert each unit using appropriate conversion factors.
- Combine powers of 10 systematically.
- Write the final answer in scientific notation.
(a) \(1\ \mathrm{kg\ m^2\ s^{-2}} = ....\ \mathrm{g\ cm^2\ s^{-2}}\)
Convert each unit separately.
\[ 1\ \text{kg} = 10^{3}\ \text{g} \] \[ 1\ \text{m} = 10^{2}\ \text{cm} \]Therefore
\[ 1\ \text{kg m}^2 \text{s}^{-2} \] \[ = (10^3\ \text{g})(10^2\ \text{cm})^2\text{s}^{-2} \] \[ = 10^3 \times 10^4\ \text{g cm}^2\text{s}^{-2} \] \[ = 10^7\ \text{g cm}^2\text{s}^{-2} \]Answer: \(10^{7}\ \mathrm{g\ cm^2\ s^{-2}}\)
(b) \(1\ \mathrm{m} = .....\ \mathrm{ly}\)
A light year is the distance travelled by light in one year.
\[ 1\ \text{ly} \approx 9.46\times10^{15}\ \text{m} \]Therefore
\[ 1\ \text{m} = \frac{1}{9.46\times10^{15}}\ \text{ly} \] \[ \approx 1.06\times10^{-16}\ \text{ly} \]Answer: \(1.06\times10^{-16}\ \mathrm{ly}\)
(c) \(3.0\ \mathrm{m\,s^{-2}} = ....\ \mathrm{km\,h^{-2}}\)
Convert metres to kilometres and seconds to hours.
\[ 3.0\ \text{m s}^{-2} \] \[ = 3.0 \times \frac{1\ \text{km}}{1000\ \text{m}} \times \left(\frac{3600\ \text{s}}{1\ \text{h}}\right)^2 \] \[ = 0.003 \times 3600^2 \] \[ = 0.003 \times 12\,960\,000 \] \[ \approx 3.89\times10^{4}\ \text{km h}^{-2} \]Answer: \(3.89\times10^{4}\ \mathrm{km\,h^{-2}}\)
(d) \[ G = 6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}} \] Convert into \( \mathrm{cm^3\,s^{-2}\,g^{-1}} \)
First express newton in base SI units.
\[ 1\ \text{N} = 1\ \text{kg m s}^{-2} \]Thus
\[ G = 6.67\times10^{-11} \frac{\text{m}^3}{\text{s}^2\text{kg}} \]Now convert SI units to CGS units.
\[ 1\ \text{m}^3 = 10^{6}\ \text{cm}^3 \] \[ 1\ \text{kg} = 10^{3}\ \text{g} \]Substituting
\[ G = 6.67\times10^{-11} \times \frac{10^{6}\ \text{cm}^3}{\text{s}^2 \times 10^{3}\ \text{g}} \] \[ G = 6.67\times10^{-8}\ \text{cm}^3\text{s}^{-2}\text{g}^{-1} \]Answer: \(6.67\times10^{-8}\ \mathrm{cm^3\ s^{-2}\ g^{-1}}\)
Why This Question Matters
- Develops mastery of SI ↔ CGS unit conversion.
- Strengthens understanding of dimensional consistency.
- Introduces the physical units of the gravitational constant \(G\).
- Similar conversion questions frequently appear in JEE Main, NEET, BITSAT, and Olympiad exams.
Q3 A calorie is a unit of heat (energy in transit) and it equals about \(4.2\ \mathrm{J}\). Suppose we employ a system of units in which
Concept Brief
This problem illustrates an important principle in physics: the numerical value of a physical quantity depends on the system of units used, even though the physical quantity itself remains unchanged.
Energy in SI units is expressed as:
\[ 1\ \text{J} = 1\ \text{kg m}^2 \text{s}^{-2} \]If we change the fundamental units of mass, length, and time, the numerical value of energy must change accordingly. This idea forms the basis of dimensional analysis, which is widely used to convert between unit systems.
Solution Map
- Write the dimensional form of energy.
- Express SI base units in terms of the new units.
- Substitute these relations into the joule definition.
- Identify the new unit of energy.
- Determine the numerical value of a calorie in the new system.
- unit of mass = \( \alpha\ \text{kg} \)
- unit of length = \( \beta\ \text{m} \)
- unit of time = \( \gamma\ \text{s} \)
Show that a calorie has magnitude \(4.2\,\alpha^{-1}\beta^{-2}\gamma^{2}\) in the new system.
Solution
Let the new fundamental units be:
\[ M' = \alpha\ \text{kg}, \quad L' = \beta\ \text{m}, \quad T' = \gamma\ \text{s} \]A calorie is related to the joule as
\[ 1\ \text{calorie} = 4.2\ \text{J} \]Since
\[ 1\ \text{J} = 1\ \text{kg m}^2 \text{s}^{-2} \]Step 1: Express SI units using new units
\[ 1\ \text{kg} = \frac{1}{\alpha} M' \] \[ 1\ \text{m} = \frac{1}{\beta} L' \] \[ 1\ \text{s} = \frac{1}{\gamma} T' \]Step 2: Substitute into joule definition
\[ 1\ \text{J} = 1\ \text{kg m}^2 \text{s}^{-2} \] \[ = \left(\frac{1}{\alpha} M'\right) \left(\frac{1}{\beta} L'\right)^2 \left(\frac{1}{\gamma} T'\right)^{-2} \]Step 3: Simplify
\[ \left(\frac{1}{\gamma}T'\right)^{-2} = \gamma^{2} {T'}^{-2} \] Therefore, \[ 1\ \text{J} = \frac{1}{\alpha}\cdot \frac{1}{\beta^{2}}\cdot \gamma^{2}\; M' {L'}^{2} {T'}^{-2} \] \[ = \alpha^{-1}\beta^{-2}\gamma^{2} \,(M' {L'}^{2} {T'}^{-2}) \]Step 4: Identify the new unit of energy
In the new system, the unit of energy is
\[ M' {L'}^{2} {T'}^{-2} \]Thus, the numerical value of \(1\ \text{J}\) becomes
\[ \alpha^{-1}\beta^{-2}\gamma^{2} \]Step 5: Convert calorie
\[ 1\ \text{calorie} = 4.2\ \text{J} \] \[ = 4.2\, \alpha^{-1}\beta^{-2}\gamma^{2} \,(M' {L'}^{2} {T'}^{-2}) \]Hence, in the new unit system the numerical value of a calorie is
\[ \boxed{4.2\,\alpha^{-1}\beta^{-2}\gamma^{2}} \]Why This Question Matters
- Introduces the concept that numerical values depend on chosen units.
- Develops strong understanding of dimensional analysis.
- This reasoning is frequently used in JEE Advanced, Olympiad, and theoretical physics problems.
- Forms the foundation for advanced topics like natural units and unit scaling in physics.
Q4 Explain the statement: “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison.”
Concept Brief
Terms such as large, small, fast, or massive are meaningful only when compared with a reference standard. In physics, every dimensional quantity must be interpreted relative to another quantity of the same dimension.
For example:
- A length may be large compared with atomic scales but small compared with astronomical distances.
- A speed may be large compared with everyday motion but small compared with relativistic speeds.
- A mass may be huge for laboratory objects but tiny on astronomical scales.
Therefore, scientific statements should include an explicit comparison or numerical scale.
Solution Map
- Identify the physical quantity involved (length, speed, mass, number, etc.).
- Choose an appropriate reference scale.
- Reframe the statement by expressing a quantitative comparison.
A dimensional quantity becomes meaningful only when compared with another quantity of the same type. For example, a length of \(10^{-10}\) m is extremely small compared with everyday objects, but it is typical for atomic dimensions.
Reframing the Statements
(a) Atoms are very small objects
Atoms have sizes of the order of
\[ 10^{-10}\ \text{m} \]which is extremely small compared with ordinary macroscopic lengths such as a millimetre (\(10^{-3}\) m) or a metre.
(b) A jet plane moves with great speed
A jet plane typically moves with a speed of about
\[ 200\text{–}300\ \text{m s}^{-1} \]which is much greater than the speed of ordinary road vehicles (\(\sim 30\ \text{m s}^{-1}\)).
(c) The mass of Jupiter is very large
The mass of Jupiter is approximately
\[ 1.9\times10^{27}\ \text{kg} \]which is about
\[ 318 \]times the mass of Earth.
(d) The air inside this room contains a large number of molecules
The air in a typical room contains of the order of
\[ 10^{25} \]molecules, which is enormously greater than the number of macroscopic objects we normally encounter in everyday life.
(e) A proton is much more massive than an electron
The mass of a proton is approximately
\[ 1.67\times10^{-27}\ \text{kg} \]which is about
\[ 1836 \]times the mass of an electron (\(9.11\times10^{-31}\ \text{kg}\)).
(f) The speed of sound is much smaller than the speed of light
The speed of sound in air is about
\[ 3.4\times10^{2}\ \text{m s}^{-1} \]whereas the speed of light in vacuum is
\[ 3.0\times10^{8}\ \text{m s}^{-1} \]Thus, the speed of sound is roughly
\[ 10^{6} \]times smaller than the speed of light.
Why This Question Matters
- Builds scientific precision in describing physical quantities.
- Develops understanding of orders of magnitude.
- Encourages quantitative thinking rather than vague descriptions.
- Similar reasoning appears frequently in JEE Main, NEET, Olympiad, and conceptual physics questions.
Q5 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 20 s to cover this distance?
Concept Brief
In some systems of units used in physics, certain fundamental constants are assigned the numerical value 1. Such systems simplify calculations and are widely used in advanced physics (for example, in relativity and particle physics).
In this problem, the unit of length is chosen such that the speed of light becomes
\[ c = 1 \]This means that the unit of length corresponds to the distance travelled by light in one second. Therefore, distance can be calculated directly from the time taken by light to travel that distance.
Solution Map
- Convert the given travel time of light into seconds.
- Use the relation \(d = ct\).
- Since \(c = 1\), the numerical value of distance equals the time taken.
Solution
Given:
- Speed of light in the new system: \(c' = 1\) (new length unit per second)
- Time taken by light from Sun to Earth: \(8\ \text{min}\ 20\ \text{s}\)
Step 1: Convert time to seconds
\[ t = 8 \times 60 + 20 \] \[ t = 480 + 20 \] \[ t = 500\ \text{s} \]Step 2: Apply distance relation
\[ d = ct \] Since \(c' = 1\), \[ d' = 1 \times 500 \] \[ d' = 500 \]Final Result
Thus, the distance between the Sun and the Earth in the new unit system is
\[ \boxed{500\ \text{new length units}} \]Why This Question Matters
- Introduces the idea of choosing convenient unit systems.
- Shows how physical constants can simplify calculations when normalized.
- Builds understanding of the relation \(d = ct\), important in astronomy and relativity.
- Similar conceptual problems appear in JEE Main, NEET, and Olympiad-level physics.
Q6 Which of the following is the most precise device for measuring length?
- (a) Vernier callipers with 20 divisions on the sliding scale
- (b) Screw gauge of pitch 1 mm and 100 divisions on the circular scale
- (c) Optical instrument measuring length within a wavelength of light
Concept Brief
The precision of a measuring instrument refers to the smallest change in the measured quantity that the instrument can detect. This smallest measurable value is called the least count.
An instrument with a smaller least count is considered more precise. Therefore, to determine the most precise instrument, we compare their least counts.
Solution Map
- Determine the least count of each instrument.
- Express all least counts in the same unit.
- Compare the values.
- The instrument with the smallest least count is the most precise.
Solution
(a) Vernier Callipers
For a typical vernier callipers with 20 divisions on the vernier scale,
\[ \text{Least count} \approx 0.01\ \text{cm} \](b) Screw Gauge
Given:
- Pitch = \(1\ \text{mm}\)
- Number of divisions = 100
Least count:
\[ \text{Least count} = \frac{\text{pitch}}{\text{number of divisions}} \] \[ = \frac{1}{100} \] \[ =0.01\ \text{mm} \] \[ =0.001\ \text{cm} \](c) Optical Instrument
An optical instrument capable of measuring length to within a wavelength of light has a least count approximately equal to the wavelength of visible light.
\[ \lambda \sim 10^{-5}\ \text{cm} \]Comparison
| Instrument | Least Count |
|---|---|
| Vernier callipers | \(10^{-2}\ \text{cm}\) |
| Screw gauge | \(10^{-3}\ \text{cm}\) |
| Optical instrument | \(10^{-5}\ \text{cm}\) |
Since \(10^{-5}\ \text{cm}\) is the smallest least count, the optical instrument is the most precise.
Final Answer
\[ \boxed{\text{(c) Optical instrument}} \]Why This Question Matters
- Clarifies the difference between precision and accuracy.
- Strengthens understanding of least count.
- Introduces the extremely small scale of wavelength-based measurements.
- Concepts of measuring instruments frequently appear in JEE Main, NEET, and laboratory-based questions.
Q7 A student measures the thickness of a human hair using a microscope of magnification 100. The average observed width in the microscope field is 3.5 mm. Estimate the actual thickness of the hair.
Concept Brief
Microscopes enlarge small objects so that their dimensions can be measured more easily. The amount by which an object appears larger is called the magnification.
Magnification is defined as
\[ M = \frac{\text{observed size}}{\text{actual size}} \]Therefore, if the magnified size of an object is known, its actual size can be obtained by dividing the observed size by the magnification factor.
Solution Map
- Write the magnification relation.
- Substitute the observed width and magnification.
- Calculate the actual thickness of the hair.
- Express the result in appropriate units.
Concept Brief
Microscopes enlarge small objects so that their dimensions can be measured more easily. The amount by which an object appears larger is called the magnification.
Magnification is defined as
\[ M = \frac{\text{observed size}}{\text{actual size}} \]Therefore, if the magnified size of an object is known, its actual size can be obtained by dividing the observed size by the magnification factor.
Solution Map
- Write the magnification relation.
- Substitute the observed width and magnification.
- Calculate the actual thickness of the hair.
- Express the result in appropriate units.
Q8 Answer the following:
(a) You are given a thread and a metre scale. How will you estimate the diameter of the thread?
(b) A screw gauge has pitch \(1.0\ \text{mm}\) and 200 divisions on the circular scale. Can accuracy be increased arbitrarily by increasing the number of divisions?
(c) Why does a set of 100 measurements give a more reliable estimate than 5 measurements?
Concept Brief
Many physical quantities are too small to measure directly with ordinary measuring instruments. In such cases, physicists use indirect measurement techniques and statistical averaging to obtain reliable estimates.
Two important ideas used in this problem are:
- Measurement amplification (e.g., wrapping a thread multiple times).
- Statistical averaging to reduce random errors.
Solution Map
- Use repeated measurement to estimate very small dimensions.
- Understand limitations of measuring instruments.
- Use statistical averaging to improve reliability of results.
(a) You are given a thread and a metre scale. How will you estimate the diameter of the thread?
Solution
Wrap the thread tightly around a cylindrical object (such as a pencil or ruler) so that the turns are closely packed without gaps.
Let:
- \(n\) = number of turns of the thread
- \(L\) = total length covered by the turns
Then the diameter of the thread is approximately
\[ d = \frac{L}{n} \]This method increases measurement accuracy because measuring the total length of many turns reduces the relative error.
(b) A screw gauge has pitch \(1.0\ \text{mm}\) and 200 divisions on the circular scale. Can accuracy be increased arbitrarily by increasing the number of divisions?
Solution
The least count of a screw gauge is
\[ \text{Least count} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} \] \[ = \frac{1}{200} = 0.005\ \text{mm} \]Increasing the number of divisions reduces the least count and improves precision. However, the accuracy cannot increase indefinitely because of practical limitations such as:
- finite resolving power of the human eye
- mechanical imperfections in the screw threads
- backlash errors
- zero errors
- thermal expansion and deformation
Thus, beyond a certain point, increasing divisions does not significantly improve the real accuracy of the instrument.
(c) Why does a set of 100 measurements give a more reliable estimate than 5 measurements?
Solution
Experimental measurements always contain small uncertainties known as random errors. These errors may be positive or negative.
When many observations are taken and averaged:
- positive and negative random errors tend to cancel each other
- the mean value approaches the true value
- the uncertainty in the average becomes smaller
Therefore, a set of 100 measurements provides a more reliable and statistically meaningful estimate than only 5 measurements.
Why This Question Matters
- Teaches practical techniques for measuring extremely small dimensions.
- Introduces the concepts of precision, least count, and experimental limitations.
- Builds understanding of statistical averaging in measurements.
- Such experimental reasoning questions frequently appear in JEE Main, NEET, and laboratory-based physics exams.
Q9 The photograph of a house occupies an area of \(1.75\ \text{cm}^2\) on a 35 mm slide. The slide is projected onto a screen where the area becomes \(1.55\ \text{m}^2\). Find the linear magnification.
Concept Brief
When an image is projected on a screen, its dimensions increase according to the linear magnification of the optical system.
If the linear magnification is \(m\), then
\[ \text{Area magnification} = m^2 \]This happens because area depends on the square of length. Therefore, if the area of the image and the area of the object are known, the linear magnification can be obtained from
\[ m = \sqrt{\frac{A_i}{A_o}} \]Solution Map
- Convert the object area to SI units.
- Calculate the areal magnification.
- Take the square root to obtain linear magnification.
Solution
Given:
- Area on slide (object): \(A_o = 1.75\ \text{cm}^2\)
- Area on screen (image): \(A_i = 1.55\ \text{m}^2\)
Step 1: Convert area to SI units
\[ 1\ \text{cm}^2 = 10^{-4}\ \text{m}^2 \] \[ A_o = 1.75 \times 10^{-4}\ \text{m}^2 \]Step 2: Calculate areal magnification
\[ m_A = \frac{A_i}{A_o} \] \[ m_A = \frac{1.55}{1.75 \times 10^{-4}} \] \[ m_A \approx 8.86 \times 10^{3} \]Step 3: Determine linear magnification
Since \[ m_A = m^2 \] \[ m = \sqrt{m_A} \] \[ m = \sqrt{8.86 \times 10^{3}} \] \[ m \approx 9.41 \times 10^{1} \] \[ m \approx 94 \]Final Result
The linear magnification of the projector–screen arrangement is
\[ \boxed{m \approx 94} \]Why This Question Matters
- Reinforces the relationship between linear magnification and area magnification.
- Builds familiarity with square scaling laws in geometry and optics.
- Highlights the importance of unit conversion before calculation.
- Similar scaling and magnification problems often appear in JEE Main, NEET, and optics-based questions.
Q10
State the number of significant figures in the following :
(a) \(\mathrm{0.007\ m^2}\)
(b) \(\mathrm{2.64 \times 10^{24}\ kg}\)
(c) \(\mathrm{0.2370\ g\ cm^{–3}}\)
(d) \(\mathrm{6.320\ J}\)
(e) \(\mathrm{6.032\ N\ m^{–2}}\)
(f) \(\mathrm{0.0006032\ m^2}\)
Concept Brief
Significant figures represent the number of meaningful digits in a measured quantity. They indicate the precision of the measurement.
The following rules are commonly used:
- All non-zero digits are significant.
- Zeros between non-zero digits are significant.
- Leading zeros (before the first non-zero digit) are not significant.
- Trailing zeros to the right of a decimal point are significant.
- In scientific notation, only the digits in the coefficient are significant.
Solution Map
- Ignore leading zeros.
- Count all non-zero digits.
- Include trailing decimal zeros where appropriate.
- For scientific notation, count digits in the coefficient only.
| Physical Quantity | Significant Figures | Explanation |
|---|---|---|
| \(0.007\ \mathrm{m^2}\) | 1 | Leading zeros are not significant; only the digit 7 counts. |
| \(2.64 \times 10^{24}\ \mathrm{kg}\) | 3 | In scientific notation, only digits in 2.64 are counted. |
| \(0.2370\ \mathrm{g\ cm^{-3}}\) | 4 | The trailing zero after the decimal point is significant. |
| \(6.320\ \mathrm{J}\) | 4 | The zero after the decimal indicates measurement precision. |
| \(6.032\ \mathrm{N\ m^{-2}}\) | 4 | The zero between 6 and 3 is significant. |
| \(0.0006032\ \mathrm{m^2}\) | 4 | Leading zeros are ignored; digits 6, 0, 3, 2 are significant. |
Why This Question Matters
- Teaches how measurement precision is represented numerically.
- Prevents common mistakes when reporting experimental results.
- Important for calculations involving rounding and error analysis.
- Questions on significant figures frequently appear in JEE Main, NEET, and laboratory-based physics exams.
Q11 The length, breadth and thickness of a rectangular sheet of metal are \(4.234\ \mathrm{m}\), \(1.005\ \mathrm{m}\), and \(2.01\ \mathrm{cm}\). Find the surface area and volume with correct significant figures.
The length, breadth and thickness of a rectangular sheet of metal are \(4.234\ \mathrm{m}\), \(1.005\ \mathrm{m}\), and \(2.01\ \mathrm{cm}\). Find the surface area and volume with correct significant figures.
Solution
Given:
- Length \(l = 4.234\ \text{m}\) (4 significant figures)
- Breadth \(b = 1.005\ \text{m}\) (4 significant figures)
- Thickness \(t = 2.01\ \text{cm}\)
Convert thickness to metres:
\[ t = 2.01\ \text{cm} = 2.01\times10^{-2}\ \text{m} = 0.0201\ \text{m} \] (3 significant figures)Step 1: Surface Area
Surface area of a rectangular block:
\[ SA = 2(lb + bt + tl) \] Substituting values: \[ SA = 2\Big[(4.234\times1.005) + (1.005\times0.0201) + (0.0201\times4.234)\Big] \] \[ =2[4.25517 + 0.0202005 + 0.0851034] \] \[ =2(4.3604739) \] \[ =8.7209478\ \text{m}^2 \]Since the least precise measurement (thickness) has 3 significant figures, the final surface area should be reported as
\[ \boxed{SA = 8.72\ \text{m}^2} \]Step 2: Volume
\[ V = lbt \] \[ V = (4.234)(1.005)(0.0201) \] \[ = 4.25517 \times 0.0201 \] \[ V = 0.085536\ \text{m}^3 \]Limiting significant figures = 3
\[ \boxed{V = 0.0855\ \text{m}^3} \]Why This Question Matters
- Reinforces rules of significant figures in calculations.
- Shows how measurement precision affects final results.
- Combines geometry with measurement concepts.
- Problems involving rounding and significant figures are common in JEE Main, NEET, and laboratory physics questions.
Q12 The mass of a box measured by a grocer’s balance is \(2.30\ \text{kg}\). Two gold pieces of masses \(20.15\ \text{g}\) and \(20.17\ \text{g}\) are added. Find:
- (a) the total mass of the box
- (b) the difference in the masses of the two pieces
Concept Brief
When measured quantities are added or subtracted, the final result must be rounded to the same decimal place as the least precise measurement.
This rule ensures that the reported result does not imply a precision greater than that of the original measurements.
Solution Map
- Convert all masses into the same unit.
- Add the masses to obtain the total mass.
- Apply the correct significant-figure rule for addition.
- Subtract the masses of the gold pieces and report the difference.
Solution
Given:
- Mass of box \(m_b = 2.30\ \text{kg}\)
- Mass of first gold piece \(m_1 = 20.15\ \text{g}\)
- Mass of second gold piece \(m_2 = 20.17\ \text{g}\)
Step 1: Convert grams to kilograms
\[ m_1 = 20.15\ \text{g} = 0.02015\ \text{kg} \] \[ m_2 = 20.17\ \text{g} = 0.02017\ \text{kg} \](a) Total Mass
\[ m_{\text{total}} = m_b + m_1 + m_2 \] \[ m_{\text{total}} = 2.30 + 0.02015 + 0.02017 \] \[ m_{\text{total}} = 2.34032\ \text{kg} \]The least precise measurement is \(2.30\ \text{kg}\), which is known only to two decimal places.
Therefore, the result must also be rounded to two decimal places.
\[ \boxed{m_{\text{total}} = 2.34\ \text{kg}} \](b) Difference in Masses
\[ \Delta m = m_2 - m_1 \] \[ \Delta m = 20.17 - 20.15 \] \[ \Delta m = 0.02\ \text{g} \]Both measurements are given to two decimal places, so the difference should also be reported to two decimal places.
\[ \boxed{\Delta m = 0.02\ \text{g}} \]Why This Question Matters
- Demonstrates correct handling of significant figures in addition and subtraction.
- Reinforces the importance of using consistent units in calculations.
- Helps students avoid reporting results with unrealistic precision.
- Significant-figure rounding problems are common in JEE Main, NEET, and laboratory physics questions.
Q13 A famous relation in physics relates the moving mass \(m\) of a particle to its rest mass \(m_0\) in terms of its speed \(v\) and the speed of light \(c\). A boy writes the relation as
\[ m = \frac{m_0}{\sqrt{1 - v^2}} \]Identify where the constant \(c\) should appear.
Concept Brief
In physics, every valid equation must satisfy the principle of dimensional consistency. This means that the dimensions of quantities on both sides of the equation must be the same.
Another important rule is that mathematical operations such as addition or subtraction can only be performed between dimensionless quantities.
In this question, dimensional analysis helps identify the correct placement of the constant \(c\), the speed of light.
Solution Map
- Check the dimensions of the expression inside the square root.
- Ensure the term subtracted from 1 is dimensionless.
- Introduce the speed of light \(c\) so that the ratio becomes dimensionless.
- Write the corrected physical relation.
Solution
The expression written by the boy is
\[ m = \frac{m_0}{\sqrt{1 - v^2}} \]However, the term inside the square root must be dimensionless.
The velocity \(v\) has dimensions
\[ [v] = LT^{-1} \] Therefore \[ [v^2] = L^2T^{-2} \]Since the number \(1\) is dimensionless, subtracting \(v^2\) from 1 is not valid unless \(v^2\) is also dimensionless.
Making the expression dimensionless
To make the expression dimensionless, we divide \(v^2\) by \(c^2\), where \(c\) is the speed of light and has the same dimensions as \(v\).
\[ \frac{v^2}{c^2} \]Now this ratio is dimensionless. Therefore, the correct expression becomes
\[ 1 - \frac{v^2}{c^2} \]Correct Relation
\[ \boxed{ m = \frac{m_0} {\sqrt{1 - \dfrac{v^2}{c^2}}} } \]This is the famous relativistic mass relation obtained from Einstein’s special theory of relativity.
Why This Question Matters
- Demonstrates how dimensional analysis can correct physical formulas.
- Shows the importance of dimensionless quantities in physics equations.
- Introduces the relativistic factor used in Einstein’s theory of special relativity.
- Dimensional reasoning problems are common in JEE Main, NEET, and Olympiad-level physics.
Q14 The unit of length convenient on the atomic scale is the angstrom. \(1\ \mathrm{Å} = 10^{-10}\ \mathrm{m}\). The size of a hydrogen atom is about \(0.5\ \mathrm{Å}\). Find the total atomic volume of one mole of hydrogen atoms.
Concept Brief
On atomic scales, lengths are commonly expressed in angstroms (Å).
\[ 1\ \mathrm{Å} = 10^{-10}\ \mathrm{m} \]Atoms are often approximated as spheres when estimating their volume. The volume of a sphere is
\[ V = \frac{4}{3}\pi r^3 \]To find the total volume occupied by atoms in one mole, we multiply the volume of a single atom by Avogadro’s number.
\[ N_A \approx 6.02\times10^{23} \]Solution Map
- Convert angstroms to metres.
- Calculate the radius of the hydrogen atom.
- Compute the volume of a single atom.
- Multiply by Avogadro’s number to obtain the total volume of one mole.
Solution
Given:
- Diameter of hydrogen atom \(d = 0.5\ \mathrm{Å}\)
- \(1\ \mathrm{Å} = 10^{-10}\ \mathrm{m}\)
Step 1: Convert diameter into metres
\[ d = 0.5 \times 10^{-10}\ \mathrm{m} \] Radius: \[ r = \frac{d}{2} \] \[ r = \frac{0.5\times10^{-10}}{2} \] \[ r = 2.5\times10^{-11}\ \mathrm{m} \]Step 2: Volume of one hydrogen atom
\[ V_{\text{atom}} = \frac{4}{3}\pi r^3 \] \[ V_{\text{atom}} = \frac{4}{3}\pi (2.5\times10^{-11})^3 \] \[ = \frac{4}{3}\pi (15.625\times10^{-33}) \] \[ \approx 6.55\times10^{-32}\ \mathrm{m^3} \]Step 3: Total volume of one mole
Avogadro number: \[ N_A = 6.02\times10^{23} \] \[ V_{\text{total}} = N_A V_{\text{atom}} \] \[ V_{\text{total}} = (6.02\times10^{23})(6.55\times10^{-32}) \] \[ = 39.4\times10^{-9} \] \[ V_{\text{total}} \approx 3.94\times10^{-8}\ \mathrm{m^3} \]Final Answer
\[ \boxed{V_{\text{total}} \approx 4\times10^{-8}\ \mathrm{m^3}} \]Why This Question Matters
- Builds intuition about atomic length scales.
- Introduces the concept of estimating atomic volume using simple models.
- Reinforces the importance of scientific notation.
- Combines ideas from physics and chemistry using Avogadro’s number.
- Similar estimation problems appear in JEE Main, NEET, and Olympiad-level physics.
Q15 One mole of an ideal gas at STP occupies \(22.4\ \text{L}\). The size of a hydrogen molecule is about \(1\ \text{Å}\). Find the ratio of molar volume to atomic volume of one mole of hydrogen. Why is this ratio so large?
One mole of an ideal gas at STP occupies \(22.4\ \text{L}\). The size of a hydrogen molecule is about \(1\ \text{Å}\). Find the ratio of molar volume to atomic volume of one mole of hydrogen. Why is this ratio so large?
Solution
Given:
- Molar volume \(V_m = 22.4\ \text{L} = 2.24\times10^{-2}\ \text{m}^3\)
- Diameter of hydrogen molecule \(d = 1\ \text{Å} = 10^{-10}\ \text{m}\)
Step 1: Radius of molecule
\[ r = \frac{d}{2} \] \[ r = \frac{10^{-10}}{2} \] \[ r = 5\times10^{-11}\ \text{m} \]Step 2: Volume of one molecule
\[ V_{\text{mol}} = \frac{4}{3}\pi r^3 \] \[ V_{\text{mol}} = \frac{4}{3}\pi (5\times10^{-11})^3 \] \[ = \frac{4}{3}\pi (125\times10^{-33}) \] \[ V_{\text{mol}} \approx 5.24\times10^{-31}\ \text{m}^3 \]Step 3: Total molecular volume of one mole
Avogadro number: \[ N_A = 6.02\times10^{23} \] \[ V_{\text{atomic}} = N_A V_{\text{mol}} \] \[ V_{\text{atomic}} = (6.02\times10^{23})(5.24\times10^{-31}) \] \[ V_{\text{atomic}} \approx 3.16\times10^{-7}\ \text{m}^3 \]Step 4: Ratio
\[ \frac{V_m}{V_{\text{atomic}}} = \frac{2.24\times10^{-2}}{3.16\times10^{-7}} \] \[ \approx 7.1\times10^{4} \] \[ \boxed{\frac{V_m}{V_{\text{atomic}}} \approx 10^{4}} \]Why is this ratio so large?
- Gas molecules are extremely small compared with the distances between them.
- Most of the volume of a gas consists of empty space.
- In kinetic theory, molecules are treated as nearly point particles with negligible volume.
- Therefore, the container volume is far larger than the combined molecular volume.
Why This Question Matters
- Develops intuition about the enormous empty space inside gases.
- Connects atomic-scale measurements with macroscopic gas properties.
- Uses estimation and scientific notation—important skills for JEE Main, NEET, and Olympiad physics.
- Supports the assumptions used in the kinetic theory of gases.
Q16 Why do nearby trees and houses appear to move rapidly in the opposite direction when viewed from a fast-moving train, while distant objects like hills or the Moon appear almost stationary?
Concept Brief
This observation occurs due to a visual effect known as motion parallax. Motion parallax refers to the apparent change in position of objects relative to each other when the observer moves.
The key idea is that the angular shift of an object in the observer’s field of view depends on its distance from the observer.
- Nearby objects show a large angular shift.
- Distant objects show a very small angular shift.
Solution Map
- Consider the motion of the observer (person in the train).
- Analyze how nearby objects change direction relative to the observer.
- Compare this with the angular shift of distant objects.
- Explain the resulting apparent motion.
Concept Brief
This observation occurs due to a visual effect known as motion parallax. Motion parallax refers to the apparent change in position of objects relative to each other when the observer moves.
The key idea is that the angular shift of an object in the observer’s field of view depends on its distance from the observer.
- Nearby objects show a large angular shift.
- Distant objects show a very small angular shift.
Solution Map
- Consider the motion of the observer (person in the train).
- Analyze how nearby objects change direction relative to the observer.
- Compare this with the angular shift of distant objects.
- Explain the resulting apparent motion.
Q17 Estimate the average mass density of the Sun using the data:
- Mass of Sun \(M = 2.0 \times 10^{30}\ \text{kg}\)
- Radius of Sun \(R = 7.0 \times 10^{8}\ \text{m}\)
Determine whether the density lies in the range of solids/liquids or gases.
Concept Brief
The average density of a celestial body can be estimated using the relation
\[ \rho = \frac{M}{V} \]where \(M\) is the mass and \(V\) is the volume. Since the Sun is approximately spherical, its volume is
\[ V = \frac{4}{3}\pi R^3 \]Although the Sun consists of extremely hot plasma, its enormous gravitational compression makes its average density comparable to that of liquids or solids.
Solution Map
- Calculate the volume of the Sun using the spherical volume formula.
- Substitute the mass and volume into the density formula.
- Compare the result with typical densities of solids, liquids, and gases.
Solution
Step 1: Volume of the Sun
\[ V = \frac{4}{3}\pi R^3 \] \[ R^3 = (7.0\times10^8)^3 \] \[ = 343\times10^{24} \] \[ = 3.43\times10^{26} \] \[ V = \frac{4}{3}\pi (3.43\times10^{26}) \] \[ V \approx 4.19\times3.43\times10^{26} \] \[ V \approx 1.44\times10^{27}\ \text{m}^3 \]Step 2: Average Density
\[ \rho = \frac{M}{V} \] \[ \rho = \frac{2.0\times10^{30}}{1.44\times10^{27}} \] \[ \rho \approx 1.39\times10^{3}\ \text{kg m}^{-3} \]Final Result
\[ \boxed{\rho \approx 1.4\times10^{3}\ \text{kg m}^{-3}} \]Density Comparison
Typical densities:- Gases at STP: \(\sim 1\ \text{kg m}^{-3}\)
- Liquids (e.g., water): \(\sim 10^{3}\ \text{kg m}^{-3}\)
- Solids: \(10^{3}–10^{4}\ \text{kg m}^{-3}\)
The calculated density of the Sun \(\sim 1.4\times10^{3}\ \text{kg m}^{-3}\) lies in the range of liquids or solids.
This may seem surprising because the Sun is made of extremely hot plasma, but its enormous gravitational compression keeps the matter highly dense.
Why This Question Matters
- Shows how simple physics formulas can estimate properties of stars.
- Builds intuition about the effect of gravitational compression.
- Connects microscopic physics (density) with astrophysical objects.
- Estimation problems like this frequently appear in JEE Main, NEET, and Olympiad-level physics.
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