MCQs
1.1 The instrument that is most suitable for measuring the thickness of a few sheets of cardboard is:
(a) metre rule
(b) measuring tape
(c) Vernier Callipers
(d) micrometer screw gauge
Correct Answer: (d) micrometer screw gauge
Reason: A micrometer screw gauge is specifically designed for precise measurements of small thicknesses, such as sheets of cardboard.
1.2 One femtometre is equal to:
(a) \(10^{-9}\) m
(b) \(10^{-15}\) m
(c) \(10^{9}\) m
(d) \(10^{15}\) m
Correct Answer: (b) \(10^{-15}\) m
Reason: A femtometre is a unit of length equivalent to \(10^{-15}\) meters, used for extremely small distances.
1.3 A light year is a unit of:
(a) light
(b) time
(c) distance
(d) speed
Correct Answer: (c) distance
Reason: A light year measures the distance light travels in one year, making it a unit of distance.
1.4 Which one is a non-physical quantity?
(a) distance
(b) density
(c) colour
(d) temperature
Correct Answer: (c) colour
Reason: Colour is a perception and not a measurable physical quantity like distance, density, or temperature.
1.5 When using a measuring cylinder, one precaution to take is to:
(a) check for the zero error
(b) look at the meniscus from below the level of the water surface
(c) take several readings by looking from more than one direction
(d) position the eye in line with the bottom of the meniscus
Correct Answer: (d) position the eye in line with the bottom of the meniscus
Reason: Aligning the eye with the bottom of the meniscus ensures accurate readings by avoiding parallax errors.
1.6 Volume of water consumed by you per day is estimated in:
(a) millilitre
(b) litre
(c) kilogram
(d) cubic metre
Correct Answer: (b) litre
Reason: Daily water consumption is commonly measured in litres for practical usage.
1.7 A displacement can be used to measure:
(a) mass of a liquid
(b) mass of a solid
(c) volume of a liquid
(d) volume of a solid
Correct Answer: (d) volume of a solid
Reason: Displacement is a method to measure the volume of a solid by observing the water it displaces.
1.8 Two rods with lengths 12.321 cm and 10.3 cm are placed side by side, the difference in their lengths is:
(a) 2.02 cm
(b) 2.0 cm
(c) 2 cm
(d) 2.021 cm
Correct Answer: (a) 2.02 cm
Reason: The exact difference is 2.021 cm, rounded to 2.02 cm based on significant figures.
1.9 Four students measure the diameter of a cylinder with Vernier Callipers. Which of the following readings is correct?
(a) 3.4 cm
(b) 3.475 cm
(c) 3.47 cm
(d) 3.5 cm
Correct Answer: (b) 3.475 cm
Reason: Vernier callipers provide precise measurements up to three decimal places, making 3.475 cm the correct choice.
1.10 Which of the following measures are likely to represent the thickness of a sheet of this book?
(a) \(6 \times 10^{-25}\) m
(b) \(1 \times 10^{-4}\) m
(c) \(1.2 \times 10^{-15}\) m
(d) \(4 \times 10^{-2}\) m
Correct Answer: (b) \(1 \times 10^{-4}\) m
Reason: The thickness of a sheet of paper is typically measured in micrometers or millimeters, which corresponds to \(10^{-4}\) meters.
1.11 In a Vernier Callipers ten smallest divisions of the Vernier scale are equal to nine smallest divisions of the main scale. If the smallest division of the main scale is half a millimeter, the Vernier constant is equal to:
(a) 0.5 mm
(b) 0.1 mm
(c) 0.05 mm
(d) 0.001 mm
Correct Answer: (c) 0.05 mm
Reason: The Vernier constant is calculated as the difference between one smallest division of the main scale and one smallest division of the Vernier scale. Here, the constant is \(0.5 \, \text{mm} \div 10 = 0.05 \, \text{mm}\).
Short Questions
1.1 Can a non-physical quantity be measured? If yes, then how?
A non-physical quantity like happiness or beauty cannot be measured directly, but it can be estimated using subjective or indirect scales, like surveys or ratings.
1.2 What is measurement? Name its two parts.
Measurement is the process of determining the size, length, or amount of something using a standard unit. Its two parts are the numerical value and the unit (e.g., 5 meters).
1.3 Why do we need a standard unit for measurements?
We need a standard unit for measurements to ensure accuracy, consistency, and easy communication of results across the world.
1.4 Write the name of 3 base quantities and 3 derived quantities.
Base quantities: Length, mass, time
Derived quantities: Speed, force, volume
1.5 Which SI unit will you use to express the height of your desk?
The height of a desk is expressed in meters (m).
1.6 Write the name and symbols of all SI base units.
Length – Meter (m)
Mass – Kilogram (kg)
Time – Second (s)
Electric current – Ampere (A)
Temperature – Kelvin (K)
Amount of substance – Mole (mol)
Luminous intensity – Candela (cd)
1.7 Why is a prefix used? Name three sub-multiples and three multiples with their symbols.
Prefixes are used to represent very large or very small values conveniently.
Sub-multiples:
Milli (\(10^{-3}\), symbol: m)
Micro (\(10^{-6}\), symbol: μ)
Nano (\(10^{-9}\), symbol: n)
Multiples:
Kilo (\(10^{3}\), symbol: k)
Mega (\(10^{6}\), symbol: M)
Giga (\(10^{9}\), symbol: G)
1.8 What is meant by:
(a) 5 pm
(b) 15 ns
(c) 6 μm
(d) 5 fs
Answer
(a) 5 pm: 5 picometers, which is \(5 \times 10^{-12}\) meters (very small length).
(b) 15 ns: 15 nanoseconds, which is \(15 \times 10^{-9}\) seconds (very small time).
(c) 6 μm: 6 micrometers, which is \(6 \times 10^{-6}\) meters (small length).
(d) 5 fs: 5 femtoseconds, which is \(5 \times 10^{-15}\) seconds (extremely small time).
1.9 (a) For what purpose is a Vernier Callipers used?
A Vernier Callipers is used to measure the internal, external, and depth dimensions of objects accurately.
1.9 (b) Name its two main parts.
The two main parts of a Vernier Callipers are:
Main scale
Vernier scale
1.9 (c) How is least count found?
The least count of a Vernier Callipers is calculated as:
\[\text{Least Count} = \frac{\text{Value of 1 main scale division}}{\text{Total number of divisions on the Vernier scale}}\]
1.9 (d) What is meant by zero error?
Zero error occurs when the zero mark of the Vernier scale does not align with the zero mark of the main scale when the jaws of the Vernier Callipers are fully closed.
1.10 State least count and Vernier scale reading as shown in the figure and hence, find the length.
Least Count: \(0.1 \, \text{mm}\)
Vernier Scale Reading: The reading shown in the figure adds the main scale reading and the Vernier scale reading to calculate the total length:
\[ \text{Total Length} = 5.9 \, \text{cm} + 0.1 \, \text{cm} = 6.0 \, \text{cm} \]
1.11 Which reading out of A, B, and C shows the correct length and why?
Correct Reading: \(6.0 \, \text{cm} (B)\)
Reason: The reading \(6.0 \, \text{cm}\) (B) aligns with both the main scale and Vernier scale readings, making it accurate without any zero error or misalignment.
Constructed Response Questions
1.1 In what unit will you express each of the following?
(a) Thickness of a five-rupee coin
Unit: Millimeter (mm)
Reason: The thickness of a coin is very small and is measured in millimeters.
(b) Length of a book
Unit: Centimeter (cm)
Reason: The length of a book is typically measured in centimeters.
(c) Length of a football field
Unit: Meter (m)
Reason: The length of a football field is large and is commonly measured in meters.
(d) The distance between two cities
Unit Kilometer (km)
Reason: The distance between cities is very large and is measured in kilometers.
(e) Mass of a five-rupee coin
Unit: Gram (g)
Reason: The mass of a coin is small and is expressed in grams.
(f) Mass of your school bag
Unit: Kilogram (kg)
Reason: The mass of a school bag is larger and is measured in kilograms.
(g) Duration of your class period
Unit: Minute (min)
Reason: Class periods are typically measured in minutes.
(h) Volume of petrol filled in the tank of a car
Unit: Liter (L)
Reason: The volume of petrol is measured in liters.
(i) Time to boil one litre of milk
Unit: Minute (min)
Reason: The time taken to boil milk is generally expressed in minutes.
1.2 Why might a standard system of measurement be helpful to a tailor?
A standard system of measurement helps a tailor take accurate and consistent measurements for clothes, ensuring a proper fit for customers.
1.3 The minimum main scale reading of a micrometer screw gauge is 1 mm, and there are 100 divisions on the circular scale. What is the least count of the instrument? The reading for the thickness of a steel rod is shown in the figure. What is the thickness of the rod?
Least Count: \( \text{Least Count} = \frac{1 \, \text{mm}}{100} = 0.01 \, \text{mm} \)
Thickness of the rod: The main scale shows \(4 \, \text{mm}\), and the circular scale shows \(78\). Thus:
\[ \text{Thickness} = 4 \, \text{mm} + (78 \times 0.01) = 4.78 \, \text{mm} \]
1.4 You are provided a metre scale and a bundle of pencils; how can the diameter of a pencil be measured using the metre scale with the same precision as that of Vernier Callipers?
Bundle multiple pencils tightly together and measure their total width using the metre scale. Divide the total width by the number of pencils to estimate the diameter of one pencil with better precision.
1.5 The end of a metre scale is worn out. Where will you place a pencil to find the length?
Place the pencil starting at the \(1 \, \text{cm}\) mark instead of the worn-out edge. Subtract \(1 \, \text{cm}\) from the final reading to get the correct length.
1.6 Why is it better to place the object close to the metre scale?
Placing the object close to the scale reduces parallax error, ensuring more accurate measurements.
1.7 Why is a standard unit needed to measure a quantity correctly?
A standard unit ensures consistency and accuracy in measurements, making it easy to compare and communicate results globally.
1.8 Suggest some natural phenomena that could serve as a reasonably accurate time standard.
Natural phenomena like the oscillation of a pendulum, Earth’s rotation (a day), or the vibrations of a cesium atom (atomic clock) can serve as accurate time standards.
1.9 It is difficult to locate the meniscus in a wider vessel. Why?
In wider vessels, the meniscus is less curved and more spread out, making it harder to identify its lowest point.
1.10 Which instrument can be used to measure
(i) Internal diameter of a test tube: Vernier Callipers
(ii) Depth of a beaker: Vernier Callipers or a depth gauge
Comprehensive Questions
1.1 What is meant by base and derived quantities? Give the names and symbols of SI base units.
Base Quantities
These are the fundamental physical quantities that cannot be derived from other quantities. Examples include length, mass, and time.
Derived Quantities
These are physical quantities derived from base quantities using mathematical relationships, such as velocity, acceleration, and force.
Names and Symbols of SI Base Units
Length – Meter (m)
Mass – Kilogram (kg)
Time – Second (s)
Electric Current – Ampere (A)
Temperature – Kelvin (K)
Amount of Substance – Mole (mol)
Luminous Intensity – Candela (cd)
1.2 Give three examples of derived units in SI. How are they derived from base units? Describe briefly.
Velocity (m/s) Derived by dividing distance (meter) by time (second).
Formula: \( \text{Velocity} = \frac{\text{Length}}{\text{Time}} \).
Force (Newton, N) Derived using mass (kilogram), acceleration (meter per second squared).
Formula: \( F = m \cdot a \).
Energy (Joule, J): Derived using force (Newton) and displacement (meter).
Formula: \( \text{Energy} = \text{Force} \cdot \text{Displacement} \).
1.3 State the similarities and differences between Vernier Calipers and Micrometer Screw Gauge.
Similarities
Both are used to measure small dimensions accurately.
Both have a scale for measurement and provide readings in millimeters or smaller units.
Differences
Vernier Calipers
Measures internal, external dimensions, and depth of objects. Its least count is usually 0.01 cm.
Micrometer Screw Gauge
Measures very small thicknesses or diameters of objects. Its least count is more precise, usually 0.001 cm.
1.4 Identify and explain the reasons for human errors, random errors, and systematic errors in experiments.
Human Errors
Caused by mistakes in reading instruments or recording values. Example: Misreading a scale.
Random Errors
Occur due to unpredictable variations in experimental conditions, such as slight temperature fluctuations.
Systematic Errors
Caused by faulty equipment or incorrect calibration, leading to consistent deviations. Example: A misaligned measuring tool.
1.5 Differentiate between precision and accuracy of a measurement with examples.
Precision
Refers to how closely repeated measurements agree with each other.
Example
Measuring an object’s length three times and getting 10.1 cm, 10.1 cm, and 10.2 cm shows high precision.
Accuracy
Refers to how close a measurement is to the true value.
Example
Measuring a 10 cm object and getting a result of 10.0 cm shows high accuracy.
Numerical Problems
1.1 Calculate the number of seconds in a (a) day (b) week (c) month and state your answers using SI prefixes.
Given:
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
To Find:
(a) Seconds in a day
(b) Seconds in a week
(c) Seconds in a month
Formula:
\[\text{Seconds} = \text{Total time} \times 60 \times 60\]
Solution:
(a) Seconds in a day:
\[24 \, \text{hours} \times 60 \, \text{minutes/hour} \times 60 \, \text{seconds/minute} = 86,400 \, \text{seconds}\]
Using SI Prefix:
\[86,400 \, \text{seconds} = 86.4 \, \text{ks} \, (\text{kilo seconds})\]
(b) Seconds in a week:
\[7 \, \text{days} \times 86,400 \, \text{seconds/day} = 604,800 \, \text{seconds}\]
Using SI Prefix:
\[604,800 \, \text{seconds} = 604.8 \, \text{ks}\]
(c) Seconds in a month (30 days):
\[30 \, \text{days} \times 86,400 \, \text{seconds/day} = 2,592,000 \, \text{seconds}\]
Using SI Prefix:
\[2,592,000 \, \text{seconds} = 2.592 \, \text{Ms} \, (\text{Mega seconds})\]
Final Answers:
(a) \( 86.4 \, \text{ks} \)
(b) \( 604.8 \, \text{ks} \)
(c) \( 2.592 \, \text{Ms} \)
1.2 State the answers of problem 1.1 in scientific notation.
Solution:
(a) \( 86,400 = 8.64 \times 10^4 \, \text{seconds} \)
(b) \( 604,800 = 6.048 \times 10^5 \, \text{seconds} \)
(c) \( 2,592,000 = 2.592 \times 10^6 \, \text{seconds} \)
Final Answers:
(a) \( 8.64 \times 10^4 \, \text{s} \)
(b) \( 6.048 \times 10^5 \, \text{s} \)
(c) \( 2.592 \times 10^6 \, \text{s} \)
1.3 Solve the following addition or subtraction. State your answers in scientific notation.
(a) \( 4 \times 10^{-4} \, \text{kg} + 3 \times 10^{-5} \, \text{kg} \)
(b) \( 5.4 \times 10^{-6} \, \text{m} – 3.2 \times 10^{-5} \, \text{m} \)
Solution:
(a) Addition
\[(4 \times 10^{-4}) + (3 \times 10^{-5}) = (40 \times 10^{-5}) + (3 \times 10^{-5}) = 43 \times 10^{-5} = 4.3 \times 10^{-4} \, \text{kg}\]
(b) Subtraction
\[(5.4 \times 10^{-6}) – (3.2 \times 10^{-5}) = (-26.6 \times 10^{-6}) = -2.66 \times 10^{-5} \, \text{m}\]
Final Answers:
(a) \( 4.3 \times 10^{-4} \, \text{kg} \)
(b) \( -2.66 \times 10^{-5} \, \text{m} \)
1.4 Solve the following multiplication or division. State your answers in scientific notation.
(a) \( (5 \times 10^4 \, \text{m}) \times (3 \times 10^{-2} \, \text{m}) \)
(b) \( \frac{6 \times 10^8 \, \text{kg}}{3 \times 10^4 \, \text{m}^3} \)
Solution
(a) Multiplication
\[(5 \times 10^4) \times (3 \times 10^{-2}) = 15 \times 10^{2} = 1.5 \times 10^3 \, \text{m}^2\]
(b) Division
\[\frac{6 \times 10^8}{3 \times 10^4} = \frac{6}{3} \times 10^{8 – 4} = 2.0 \times 10^4 \, \text{kg/m}^3\]
Final Answers
(a) \( 1.5 \times 10^3 \, \text{m}^2 \)
(b) \( 2.0 \times 10^4 \, \text{kg/m}^3 \)
1.5 Calculate the following and state your answer in scientific notation.
Given
\[\frac{(3 \times 10^2 \, \text{kg}) \times (4.0 \, \text{km})}{5 \times 10^2 \, \text{s}^2}\]
Solution
Convert \( 4.0 \, \text{km} \) to meters
\[4.0 \, \text{km} = 4.0 \times 10^3 \, \text{m}\]
Substitute into the formula
\[\frac{(3 \times 10^2) \times (4.0 \times 10^3)}{5 \times 10^2} = \frac{12.0 \times 10^5}{5 \times 10^2} = 2.4 \times 10^3 \, \text{kg} \, \text{m/s}^2\]
Final Answer:
\( 2.4 \times 10^3 \, \text{kg} \, \text{m/s}^2 \)
1.6 State the number of significant digits in each measurement.
Given
(a) \( 0.0045 \, \text{m} \)
(b) \( 2.047 \, \text{m} \)
(c) \( 3.40 \, \text{m} \)
(d) \( 3.420 \times 10^4 \, \text{m} \)
To Find
Number of significant digits in each case.
Solution
(a) \( 0.0045 \): 2 significant digits (only non-zero digits count).
(b) \( 2.047 \): 4 significant digits (all non-zero digits and zeros between significant digits count).
(c) \( 3.40 \): 3 significant digits (trailing zero after decimal is significant).
(d) \( 3.420 \times 10^4 \): 4 significant digits (trailing zero after decimal is significant).
Final Answer:
(a) 2, (b) 4, (c) 3, (d) 4
1.7 Write in scientific notation.
(a) \( 0.0035 \, \text{m} \)
(b) \( 206.4 \times 10^2 \, \text{m} \)
Solution
(a) \( 0.0035 \)
\[0.0035 = 3.5 \times 10^{-3} \, \text{m}\]
(b) \( 206.4 \times 10^2 \)
\[206.4 \times 10^2 = 2.064 \times 10^4 \, \text{m}\]
Final Answer
(a) \( 3.5 \times 10^{-3} \, \text{m} \)
(b) \( 2.064 \times 10^4 \, \text{m} \)
1.8 Write using correct prefixes.
(a) \( 5.0 \times 10^4 \, \text{cm} \)
(b) \( 580 \times 10^2 \, \text{g} \)
(c) \( 45 \times 10^{-4} \, \text{s} \)
Solution
(a) \( 5.0 \times 10^4 \, \text{cm} \)
Convert \( \text{cm} \) to \( \text{m} \)
\[5.0 \times 10^4 \, \text{cm} = 5.0 \times 10^2 \, \text{m} = 500 \, \text{m} \, (\text{kilo prefix as 500 km})\]
(b) \( 580 \times 10^2 \, \text{g} \)
Convert \( \text{g} \) to \( \text{kg} \)
\[580 \times 10^2 \, \text{g} = 5.8 \times 10^4 \, \text{g} = 58 \, \text{kg}\]
(c) \( 45 \times 10^{-4} \, \text{s} \)
Convert \( \text{s} \) to correct SI prefixes
\[45 \times 10^{-4} \, \text{s} = 4.5 \times 10^{-3} \, \text{s} = 4.5 \, \text{ms} \, (\text{milli prefix})\]
Final Answers
(a) \( 500 \, \text{m} \) or \( 0.5 \, \text{km} \)
(b) \( 58 \, \text{kg} \)
(c) \( 4.5 \, \text{ms} \)
1.9 Light year is a unit of distance used in astronomy. It is the distance covered by light in one year. Taking the speed of light as \( 3.0 \times 10^8 \, \text{m/s} \), calculate the distance.
Calculate the distance covered in one light year.
Given
Speed of light \( = 3.0 \times 10^8 \, \text{m/s} \)
Time for 1 year \( = 365 \, \text{days} \times 24 \, \text{hours/day} \times 60 \, \text{minutes/hour} \times 60 \, \text{seconds/minute} \)
To Find
Distance covered in one light year.
Formula
\[\text{Distance} = \text{Speed} \times \text{Time}\]
Solution
1 year in seconds
\[365 \times 24 \times 60 \times 60 = 31,536,000 \, \text{seconds}\]
Distance
\[\text{Distance} = (3.0 \times 10^8) \times (31,536,000) = 9.46 \times 10^{15} \, \text{m}\]
Final Answer
The distance covered in one light year is \( 9.46 \times 10^{15} \, \text{m} \).
1.10 Express the density of mercury given as \( 13.6 \, \text{g/cm}^3 \) in \( \text{kg/m}^3 \).
Convert \( 13.6 \, \text{g/cm}^3 \) into \( \text{kg/m}^3 \).
Given
Density of mercury \( = 13.6 \, \text{g/cm}^3 \)
\( 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 \)
To Find
Density in \( \text{kg/m}^3 \).
Formula
\[\text{Density in } \text{kg/m}^3 = \text{Density in } \text{g/cm}^3 \times 1000\]
Solution
\[13.6 \, \text{g/cm}^3 \times 1000 = 13,600 \, \text{kg/m}^3\]
Expressed in scientific notation
\[13,600 = 1.36 \times 10^4 \, \text{kg/m}^3\]
Final Answer
The density of mercury is \( 1.36 \times 10^4 \, \text{kg/m}^3 \).
