This chapter introduces the fundamental concepts of measurement and units.
Units:
*Fundamental UnitsThese are independent units for fundamental physical quantities like length, mass, time, temperature, electric current, luminous intensity, and amount of substance.
* **Derived Units:** These units are derived from the fundamental units and are used to measure derived quantities like velocity, acceleration, force, etc.
System of Units:
SI System: The International System of Units is the most widely used system of units.
*Measurement of Length:
Various instruments like Vernier calipers, screw gauges, and micrometers are used to measure length with high accuracy.
*Measurement of Mass:
* A common balance is used to measure mass.
*Measurement of Time:
*Time is measured with the help of clocks and stopwatches.
Significant Figures:**
The precision of a measurement showned by Significant figures
Rules for significant figures are used to determine the precision of calculations involving measured quantities.
Errors in Measurement
*Systematic Errors: These errors occur due to faulty calibration of instruments or experimental procedures.
*Random Errors: These errors arise from unpredictable fluctuations in experimental conditions.
Dimensional Analysis:
* Dimensional analysis is a technique used to check the correctness of physical equations.
It involves checking the consistency of dimensions on both sides of an equation.
1. Fill in the blanks
(i) The volume of a cube of side 1 cm is equal to…………m3.
Ans : for a cube with a side of 1 cm, the volume is 1 cm x 1 cm x 1 cm = 1 cm³.
(ii) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ……..(mm)2.
Ans :The surface area of a solid cylinder A =2πr² + 2πrh
2 x 22/7 x 2 x 10 (10 x 10 + 2 x 10) mm2 =Surface area ≈ 15079.64 mm² (rounded to two decimal places)
(iii) A vehicle moving with a speed of 18 km/h covers ………. m in 1 s.
Ans : We know 1 hour (h) is equal to 3600 seconds (s).
We know 1 kilometre (km) is equal to 1000 metres (m)
Speed (m/s) = 18 km/h ×(1000 m/km) × (1 h / 3600 s)
Calculate the speed: Speed (m/s) ≈ 5 m/s (rounded to one decimal place)
(iv) The relative density of lead is 11.3. Its density is …….. g cm-3 or ………. kg m-3
Ans : The density of lead is 11.3 g cm⁻³ or 1.13 x 10⁴ kg m⁻³.
2. 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 = …. (cm)3 s-2 g-1.
Ans :(a) Convert kg to g: 1 kg = 1000 g
Convert m² to cm²: 1 m = 100 cm, so 1 m² = (100 cm)² = 10,000 cm²
Apply the conversions: 1 kg m² s⁻² = (1000 g) ×(10,000 cm²) ×(1 s⁻²) = 10,000,000 g cm² s⁻²
(b) Light year (ly) is a unit of distance, so direct conversion between meters and light years is needed.
1 light year is roughly 9.461 x 10¹² metres.
(c)1 km = 1000 m
1 hour (h) = 3600 seconds (s)
Therefore, multiply by (1000 m/km) and (1 h / 3600 s) to convert: 3.0 m/s² × (1000 m/km) ×(1 h / 3600 s) ≈ 38,880 km/h² (rounded to significant digits)
(d)G= 6.67×10-11 Nm2/kg2
1N=1kgm/s then G=6.67×10-11kg-1 m3s-2
Or G=6.67×10-11 m3/kgs2=6.67×10-11×(102)3/(103 )2
G=6.67×10-8 cm-3s-2g-1
3. A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kgm2 s-2. Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals j8 m, the. unit of time is ys. Show that a calorie has a magnitude 4.2 α-1 β-2 γ2 in terms of the new units.
Ans :1 calorie ≈ 4.2 Joules (J)
- 1 Joule (J) = 1 kg m² s⁻² (in standard units)
1 kg (mass) = α kg (new unit)
1 m (length) = β m (new unit)
1 s (time) = γ s (new unit)
- Substitute the new unit conversions for mass, length, and time in the standard Joule unit: 1 J = 1 (α kg) * (β m)² * (γ s⁻²)²
- 1 calorie =4.2
We need to find the equivalent of 1 calorie in terms of the new units.
Steps:
- Convert Joule (J) to new units:
- Substitute the new unit conversions for mass, length, and time in the standard Joule unit: 1 J = 1 (α kg) * (β m)² * (γ s⁻²)² (square the time conversion because there are two seconds in the unit)
- Convert calorie to new units:
- Multiply the Joule conversion by the calorie-Joule equivalence: Calorie (new units) = 4.2 J * (α kg * (β m)² * (γ s⁻²)²)
- Simplify: Calorie (new units) ≈ 4.2 * α * (β⁸) * γ² (combine constant factors and exponents with the same base)
Therefore, the magnitude of a calorie in terms of the new units is 4.2 α⁻¹ β⁻² γ².
4. 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.
Ans : To call a dimensional quantity “large” or “small” is meaningless without specifying a standard for comparison because the concept of size is relative. For example, a person might consider a house to be large, but compared to a skyscraper, it would seem small. Similarly, a small insect might seem large compared to a microscopic organism. Therefore, it’s essential to have a reference point to accurately assess the magnitude of a dimensional quantity.
Here are the reframed statements considering the above principle:
(a) Atoms are very small compared to macroscopic objects (e.g., a grain of sand).
(b) A jet plane moves with great speed compared to a car or a bicycle.
(c) The mass of Jupiter is very large compared to the mass of Earth.
(d) The air inside this room contains a large number of molecules compared to the number of stars in the Milky Way galaxy.
(e) A proton is much more massive than an electron compared to their respective masses.
(f) The speed of sound is much smaller than the speed of light compared to their respective values.
By specifying a standard for comparison in each statement, we provide a more accurate and meaningful understanding of the relative magnitude of the dimensional quantities involved.
5. 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 ?
Ans : The velocity of light in empty space is about 3 * 10^8 meters per second.
The time taken by light to cover the distance between the Sun and the Earth is 8 min 20 s.
Therefore, the distance between the Sun and the Earth is 3 * 10^8 * (8 * 60 + 20) = 1.5 * 10^11 m.
In the new unit of length, the speed of light is unity.
Therefore, the distance between the Sun and the Earth in the new unit is 1.5 * 10^11 / 3 * 10^8 = 500.
So, the distance from the Sun to Earth is 500 in the new way we’re measuring things.
6. 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 ?
Ans : (c)
Explanation:
The precision of a measuring instrument depends on its least count. The smaller the least count, the more precise the instrument.
Vernier callipers:
Least count =
1 main scale division
——————————————————-
number of vernier scale divisions
= 1 mm / 20
= 0.05 mm
Screw gauge:
Least count =
Pitch
———————————————————
number of divisions on circular scale
= 1 mm / 100
= 0.01 mm
Optical instrument:
Least count = Wavelength of light ≈ 600 nm
= 6 × 10⁻⁷ m
As we can see, the optical instrument has the smallest least count, making it the most precise of the three.
7. 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 ?
Ans : The microscope makes the hair look 100 times bigger.
The microscope revealed an average breadth of 3.5 mm for the hair.
Solution:
The actual thickness of the hair can be calculated by dividing the observed width, as magnified by the microscope, by the magnification power.
Actual thickness = Observed width
——————————-
Magnification
Actual thickness = 3.5 mm
——————-
100
Actual thickness =
0.035 mm
So, the hair is about 0.035 mm thick.
8 .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 a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?
(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ?
Ans : (a) Estimating the diameter of a thread using a thread and a meter scale:
Method:
Wind the thread: Wrap the thread tightly around a cylindrical object (like a pencil) many times, ensuring the coils are adjacent to each other.
Measure the length: Measure the total length of the wound thread using the meter scale.
Calculate the diameter: Divide the total length of the thread by the number of turns. The result is an estimate of the thread’s diameter.
Formula:
Diameter = (Total length of thread) / (Number of turns)
(b) Increasing the accuracy of a screw gauge:
Yes, increasing the number of divisions on the circular scale of a screw gauge can improve its accuracy. This is because the smaller the divisions, the more precise the readings can be. However, there are limitations:
Human error: Even with more divisions, human limitations in reading the scale accurately can still affect precision.
Physical limitations: There is a limit to how small the divisions can be before they become difficult to distinguish.
(c) Why 100 measurements are more reliable than 5:
Taking a larger number of measurements (100) generally leads to a more reliable estimate than a smaller number (5). This is due to the law of large numbers. Random errors, which can occur in individual measurements, tend to cancel each other out when a large number of measurements are averaged. This reduces the overall uncertainty in the estimate.
Therefore, by taking more measurements, we can improve the statistical reliability of the average value obtained.
9. The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2 . What is the linear magnification of the projector-screen arrangement.
Ans : We’re given the areas of the house on the slide and the screen.
We need to find the linear magnification.
Solution
Area magnification is the ratio of the area of the image to the area of the object.
To find the linear magnification, take the square root of the area magnification.
Given:
Area of object (house on slide) = 1.75 cm² = 1.75 * 10⁻⁴ m²
Area of image (house on screen) = 1.55 m²
Calculations:
Area magnification:
Area magnification = Area of image / Area of object
Area magnification = 1.55 m² / 1.75 * 10⁻⁴ m²
Area magnification ≈ 8857.14
Linear magnification:
Linear magnification = √(Area magnification)
Linear magnification = √8857.14
Linear magnification ≈ 94.11
Therefore, the linear magnification of the projector-screen arrangement is approximately 94.11.
10. State the number of significant figures in the following :
(a) 0.007 m2
(b) 2.64 × 1024 kg
(c) 0.2370 g cm–3
(d) 6.320 J
(e) 6.032 N m–2
(f) 0.0006032 m2
Ans : (a) The zeros at the start of 0.007 m² don’t count, so it only has one significant figure.
(b)2.64 × 10²⁴ kg is expressed with 3 significant figures.
(c)0.2370 g cm⁻³ has four significant figures because the zero at the end counts.
(d)The value 6.320 Joules has four significant digits due to the presence of the trailing zero.
(e) The value 6.032 N m⁻² possesses four significant figures, comprising all non-zero digits and the zero positioned between non-zero digits.
(f) The expression 0.0006032 m² has four significant digits.
11. 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.
Ans : Converting Units and Calculating Area and Volume
Step 1: Convert units:
Ensure all units are consistent. Let’s convert centimeters to meters:
Thickness = 2.01 cm = 0.0201 m
Step 2: Calculate area:
Area = Length * Breadth
Area = 4.234 m * 1.005 m
Area ≈ 4.254 m²
Step 3: Calculate volume:
Volume = Length * Breadth * Thickness
Volume = 4.234 m * 1.005 m * 0.0201 m
Volume ≈ 0.0857 m³
Considering Significant Figures
The least number of significant figures in any of the given measurements is 3 (in the thickness).
Hence, the calculated values should be rounded to a precision of three significant digits..
Area: 4.25 m²
Volume: 0.0857 m³
12. The mass of a box measured by a grocer’s balance is 2.30 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 pieces to correct significant figures ?
Ans :
(a) Total Mass of the Box
Step 1: Convert units:
Convert grams to kilograms for consistency:
20.15 g = 0.02015 kg
20.17 g = 0.02017 kg
Step 2: Calculate total mass:
Total mass = Mass of box + Mass of gold piece 1 + Mass of gold piece 2
Total mass = 2.30 kg + 0.02015 kg + 0.02017 kg
Total mass ≈ 2.34032 kg
Considering significant figures:
The mass of the box has the fewest significant figures, which is 3.
Thus, the total mass should be expressed with three significant figures.
Total mass ≈ 2.34 kg
(b) Difference in the Masses of the Pieces
Step 1: Calculate the difference:
Difference = Mass of gold piece 2 – Mass of gold piece 1
Difference = 0.02017 kg – 0.02015 kg
Difference = 0.00002 kg
Considering significant figures:
Both masses have 4 significant figures, so the difference should also have 4 significant figures.
Final answer:
Difference = 0.00002 kg
13. A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo 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 due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes
Guess where to put the missing c.
Ans : The correct equation relating moving mass (m) to rest mass (m₀), speed (v), and the speed of light (c) is:
m = m₀ / √(1 – v²/c²)
The boy’s mistake was likely in placing the constant c. It should be in the denominator under the square root, not in the numerator.
14. 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 ?
Ans : Understanding the Problem:
Given: Size of a hydrogen atom = 0.5 Å = 0.5 × 10⁻¹⁰ m
To find: Total atomic volume of a mole of hydrogen atoms in m³
1. Volume of a single hydrogen atom:
Assuming a spherical shape, the volume of a single atom is:
V_atom = (4/3)πr³
Given the diameter is 0.5 × 10⁻¹⁰ m, the radius (r) is 0.25 × 10⁻¹⁰ m.
So, V_atom = (4/3)π(0.25 × 10⁻¹⁰)³ ≈ 6.54 × 10⁻³¹ m³
2. Total volume of a mole of hydrogen atoms:
One mole contains Avogadro’s number (6.022 × 10²³) of atoms.
Total volume = V_atom × Avogadro’s number
Total volume ≈ 6.54 × 10⁻³¹ m³ × 6.022 × 10²³ ≈ 3.94 × 10⁻⁷ m³
Therefore, the total atomic volume of a mole of hydrogen atoms is approximately 3.94 × 10⁻⁷ m³.
15. One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large ?
Ans : Calculating the Ratio
Step 1: Convert units:
Convert liters to cubic meters: 22.4 L =
22.4 * 10⁻³ m³
Convert angstroms to meters: 1 Å = 10⁻¹⁰ m
Step 2: Calculate the atomic volume of a single hydrogen molecule:
Assuming the hydrogen molecule is spherical, we can use the formula for the volume of a sphere:
Volume =
(4/3) * π * r³
Since the diameter is 1 Å = 10⁻¹⁰ m, the radius (r) is half of that: 0.5 * 10⁻¹⁰ m
Volume ≈ 5.24 * 10⁻³¹ m³
Step 3: Calculate the total atomic volume of a mole of hydrogen molecules:
Total atomic volume = Volume of a single molecule * Number of molecules in a mole
Total atomic volume ≈ 5.24 * 10⁻³¹ m³ * 6.022 * 10²³
Total atomic volume ≈ 3.15 * 10⁻⁷ m³
Step 4: Calculate the ratio of molar volume to atomic volume:
Ratio = Molar volume
———————-
Atomic volume
Ratio =
(22.4 * 10⁻³) m³ / (3.15 * 10⁻⁷ m³)
Ratio ≈ 7.08 * 10⁴
Therefore, the ratio of molar volume to the atomic volume of a mole of hydrogen is approximately 7.08 * 10⁴.
Why is this ratio so large?
The reason for this large ratio is the significant amount of empty space between the hydrogen molecules in a gas. The molecules are constantly moving and colliding, but they occupy only a small fraction of the total volume. This is why the molar volume (which includes the empty space) is much larger than the actual atomic volume of the molecules.
16. Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).
Ans : The observation you’ve described is a result of the concept of relative motion.
When you’re on a moving train, you and everything inside the train are moving at the same speed as the train. This is your reference frame.
Nearby objects (trees, houses) are moving at a much slower speed or are stationary relative to the ground. When you look at them from the moving train, their relative motion appears much faster and in the opposite direction to the train’s motion. This is because their speed relative to you (on the train) is the sum of their speed relative to the ground and the train’s speed in the opposite direction.
Distant objects (hilltops, the Moon, stars) are so far away that their relative motion compared to the train is negligible. Even though they are moving at incredibly high speeds (like the Earth’s rotation or the Moon’s orbit), their distance makes this motion appear insignificant from your perspective on the train. Therefore, they seem stationary relative to you.
In essence, the apparent motion of an object depends on the reference frame from which it’s observed.
17. The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 ×1030 kg, radius of the Sun = 7.0 × 108 m.
Ans : The mass of the Sun is 2.0 ×1030 kg and the radius of the Sun is 7.0 × 108 m.
The average density of the Sun is mass/volume.
The volume of the Sun is (4/3) * π * r^3, where r is the radius of the Sun.
Therefore, the average density of the Sun is
(2.0 ×1030 kg) / ((4/3) * π * (7.0 × 108 m)^3) = 1400 kg/m^3.
The density of solids and liquids is typically in the range of 1000 kg/m^3 to 10000 kg/m^3. The density of gases is typically in the range of 0.1 kg/m^3 to 10 kg/m^3.
Therefore, the average density of the Sun is in the range of densities of solids and liquids. This is because the Sun is mostly composed of hydrogen and helium, which are elements that have a relatively low atomic mass. Even though the Sun is very hot, the pressure at its core is very high, which causes the hydrogen and helium atoms to be packed together very tightly. This results in a density that is similar to the density of solids and liquids.
The following table summarizes the data:
Property Value
Mass of the Sun 2.0 ×1030 kg
Radius of the Sun 7.0 × 108 m
Average density of the Sun 1400 kg/m^3
Density of solids and liquids 1000 kg/m^3 to 10000 kg/m^3
Density of gases 0.1 kg/m^3 to 10 kg/m^3
As can be seen from the table, the average density of the Sun is in the range of densities of solids and liquids. This is because the Sun is mostly composed of hydrogen and helium, which are elements that have a relatively low atomic mass. Even though the Sun is very hot, the pressure at its core is very high, which causes the hydrogen and helium atoms to be packed together very tightly. This results in a density that is similar to the density of solids and liquids.
