9th Class Physics New Book 2025 Chapter 1 Solved Exercise
9th Class New Book Physics 2025 Chapter 1 Solved Exercise
We will Solve the exercise for the 9th class physics revised smart syllabus 2025.
9th Class Physics New Book 2025 Chapter 1 MCQs
1.1 The instrument that is most suitable for measuring the thickness of a few sheets of cardboard is
(a) meter rule (b) Measuring tape
(c) Vernier Callipers (d) micermeter screw gauge
Option C is correct.
Reason:- Vernier callipers precisely measures the thickness of cardboard sheets due to its flat jaw.
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1.2 One femtometer is equal to
$$(a)\;10^{-9}m\;(b)\;10^{-15}m\;(c)\;10^9m\;(d)\;10^{15}m$$
Option B is correct.
Reason:- Femto means 10 to raised power -15. In Nuclear physics, we use this unit.
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1.4 Which one is a non-physical quantity ?
(a) distance (b) density
(c) colour (d) temperature
Option C is correct.
Reason:-Colour depends on observer.
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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 bottom of the meniscus
Option D is correct.
Reason:- Correct position of eye will almost eliminate the chances of human error.
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1.7 A displacement can is used to measure:
(a) mass of a liquid (b) mass of a solid
(c) volume of a liquid (d) volume of a solid
Option D is correct.
Reason:- Displacement can can measure volume of regular as well as irregular shaped solid by displacement method.
We can directly measure mass of liquid and solid with the help of balance and measuring cylinder can measure the volume of a liquid.
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1.9 Four students measure the diameter of a cylinder with vernier callipers. which of the following reading is correct ?
(a) 3.4 cm (b) 3.475 cm (c) 3.47 cm (d) 3.5 cm
Option C is correct.
Reason:- Least count of vernier calliper is 0.01 cm which is 2 decimal places and there are also 2 digits after the decimal in 3.47.
1.10 Which of the following measures are likely to present the thickness of a sheet of this book ?
$$(a)\;\;6\times10^{-25}\;m\;\;\;(b)\;1\times10^{-4}\;m\\\\(c)\;1.2\times10^{-15}\;m\;\;(d)\;4\times10^{-2}\;m$$
Option B is correct.
Reason:- Measurements in option a and c are very small and measurement in option d is 4 cm. Option B is 0.1 mm so, its correct.
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9th Class Physics New Book 2025 Chapter 1 Short Questions
Q 1.1 Can a non-physical quantity be measured ? If yes, then how ?
No. With the help of using tools and instruments, we cannot measure a non-physical quantity because it does not have a defined physical unit. Love, beauty, honesty and trust are non-physical quantities.
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Q 1.2 What is measurement ? Name its two parts
Measurement is a process of comparison of an unknown quantity with a standard quantity (unit).
Measurement consists of two parts, a number (magnitude or numerical value) and a unit. In 5 meters, 5 is magnitude and meter is unit.
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Q 1.3 Why do we need a standard unit for measurement ?
Standard unit for measurement can be for efficient working and growth of trade, business and share scientific information.
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Q 1.4 Write the name of 3 base quantities and 3 derived quantities.
Base quantities are length, mass and time.
Derived quantities are area, volume and speed.
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Q 1.6 Write the name and symbols of all SI base units.
Page 08
| Name of Unit | Symbol of Unit |
|---|---|
| Meter | m |
| Kilogram | kg |
| Second | s |
| Kelvin | K |
| Ampere | A |
| Candela | cd |
| Mole | mol |
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Q 1.7 Why prefix is used ? Name three sub-multiples and three multiple prefixes with their symbols.
We use prefix to make unit simple and easy. Three sub-multiples are deci, centi and milli denoted by d, c and m respectively.
Three multiples are kilo, mega and giga denoted by k, M and G respectively.
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9th Class Physics New Book 2025 Chapter 1 Constructed Response Questions
1.3 The minimum main scale reading of a micrometre screw gauge is 1 mm and there are 100 divisions on the circular scale. When thimble is rotated once, 1 mm is its measurement on the main scale. What is the least count of the instrument? The reading for thickness of a steel rod shown in the figure. What is the thickness of the rod?
Least count of micrometre screw gauge is 0.01 mm or 0.001 cm.
The edge of the thimble in the figure is just pass the 3mm mark on the main scale. The reference line coincides with 70 divisions on the circular scale.
Thickness = 3+70(0.01)
=3+0.70= 3.70 mm
Hence thickness of steel rod is 3.70 mm.
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1.5 The end of a metre scale is worn out. Where will you place a pencil to find the length?
We will avoid the worn out mark and place the pencil at any clear mark. We will read the scale at the outer end and subtract the starting reading to get the true length.
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1.6 Why is it better to place the object close to the metre scale?
It is better to place the object close to the meter scale to read the marking correctly. This will reduce the chance of error.
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1.7 Why a standard unit is needed to measure a quantity correctly?
A standard unit will ensure accurate, uniform and universally understood measurement.
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1.9 It is difficult to locate the meniscus in a wider vessel. Why?
It is difficult to locate the meniscus ( curved surface of liquid inside a container ) in a wider vessel because the liquid surface becomes flat. It makes the meniscus less curved and hard to see.
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9th Class Physics New Book 2025 Chapter 1 Comprehensive Questions
Q 1.1 What is meant by base and derived quantities ? Give the names and symbols of SI base units.
The quantities which are independent of the other physical quantities are called base quantities. We cannot define any base quantity in terms of any other quantity.
Derived quantities are the quantities which can be defined with reference to base quantities.
There are seven base physical quantities and hence seven base units. Symbol of meter is m, kilogram is kg, second is s, kelvin is K, ampere is A, candela is cd and mole is mol.
| Name of Unit | Symbol of Unit |
|---|---|
| Meter | m |
| Kilogram | kg |
| Second | s |
| Kelvin | K |
| Ampere | A |
| Candela | cd |
| Mole | mol |
Q 1.2 Give three examples of derived units in SI. How are they derived from base units ? Describe briefly.
Some of the derived physical quantities are area, volume and speed and there units are square meter, cubic meter and meter per second respectively.
When we multiply or divide two or more than two base units, we get derived units. Remember that by adding or subtracting base units, we cannot get a derived unit.
$$Area\;=\;Length\;\times\;breath\\=\;meter\;\times\;meter\\=\;Square\;meter\;=\;m^2$$
$$Volume\mathit=\mathit(Length\mathit)\mathit(breath\mathit)\mathit(height\mathit)\\\mathit=\mathit\;meter\mathit\times meter\mathit\times meter\\\mathit=\mathit\;Cubic\mathit\;meter\mathit\;\mathit=\mathit\;m^{\mathit3}$$
$$Speed\;=\;\frac{dis\tan ce}{time}\\=\frac{meter}{second}=\;m\;s^{-1}$$
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9th Class Physics New Book 2025 Chapter 1 Numerical
1.1 Calculate the number of seconds in a (a) day (b) week (c) month and state you answer using SI Prefixes.
Number of minutes in 1 hour = 60
Seconds in 1 hour = (60) (60) = 3600
Seconds in 1 day = (3600) (24) = 86,400
$$=(\frac{86400}{1000})\;(1000)s\\\\=(86.4)\;(1000)s\\\\=86.4\;Ks\\$$
Number of seconds in a week = (86,400) (7)
= 604,800
$$=(\frac{604800}{1000})\;(1000)s\\\\=(604.8)\;(1000)s\\\\=604.8\;Ks\\$$
Number of seconds in 1 month = (86,400) (30)
= 2,592,000
$$=(\frac{2592000}{1000000})\;(1000000)s\\\\=(2.592)\;(1000000)s\\\\=2.592\;Ms\\$$
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1.3 Solve the following addition or subtraction. State your answer in scientific notation
$$(a)\;\;4\times10^{-4}kg\;+3\times10^{-5}kg\\\\=\;10^{-4}(4\;+\;3\times10^{-1}\;)\;kg\\\\=\;10^{-4}(4\;+\;\frac3{10}\;)\;kg\\\\=\;10^{-4}(4\;+\;0.3)\;kg\\\\=\;10^{-4}(4.3\;)\;kg\\\\=\;4.3\;\times10^{-4}\;kg\\\\\\(b)\;\;5.4\times10^{-6}\;m\;-\;3.2\times10^{-5}\;m\\\\=\;10^{-5}(5.4\;\times10^{-1}-3.2)m\\\\=\;10^{-5}(\frac{5.4}{10}-3.2)m\\\\=\;10^{-5}(0.54-3.2)m\\\\=\;10^{-5}(-2.66)m\\\\=\;-2.66\;\times10^{-5}m\\\\\\\\\\$$
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1.4 Solve the following multiplication or division. State your answer in scientific notation.
$$(a)\;\;\;(5\times10^4m)\;\times\;(3\times10^{-2}m)\\(b)\;\;\;\frac{6\times10^8kg}{3\times10^4\;m^3}\;\\$$
$$(a)\;\;\;(5\times10^4m)\;\times\;(3\times10^{-2}m)\\\\=\;\;5\times3\times10^4\times10^{-2}\times m\times m\\\\=\;15\times10^{4-2}\;\times m^2\\\\=\;15\times10^2\;m^2\\\\=\;1.5\;\times10^1\;\times10^2\;m^2\\\\=\;1.5\;\times10^{1+2}m^2\\\\=\;1.5\;\times10^3m^2\\\\\\(b)\;\;\;\frac{6\times10^8kg}{3\times10^4\;m^3}\;\\=\;2\times10^8\times10^{-4}\;kg\;m^{-3}\\\\=\;2\times{(10)}^{8-4}\;kg\;m^{-3}\\\\=\;2\times10^4\;kg\;m^{-3}\\\\$$
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1.6 State the number of significant digits in each measurement.
$$(a)\;0.0045m\\(b)\;2.047m\;\\(c)\;3.40m\\(d)\;3.420\;\times\;10^4m\\\\(a)\;0.0045m\\All\mathit\;non\mathit\;zero\mathit\;digits\mathit\;are\\significant\mathit.\\Zeros\;are\;not\;significant\\here.\;\\4\;and\;5\;are\;significant.\\Hence,\;there\;are\;only\;2\\significant\;figures.\\\\(b)\;2.047m\\All\mathit\;non\mathit\;zero\mathit\;digits\mathit\;are\\significant\mathit.\;\\Zero\;between\;non\;zeros\\are\;significant.\;\\All\;the\;digits\;are\;significant\\here.\\Hence,\;there\;are\;4\\significant\;figures.\\\\c)\;3.40m\\All\mathit\;non\mathit\;zero\mathit\;digits\mathit\;are\\significant\mathit.\\Zero\;after\;the\;decimical\\point\;is\;significant.\\Hence,\;there\;are\;3\\significant\;figures.\\\\(d)\;3.420\;\times\;10^4m\\All\mathit\;non\mathit\;zero\mathit\;digits\\are\mathit\;significant\mathit.\\Zero\;after\;the\;decimical\\point\;is\;significant.\\Hence,\;there\;are\;4\\significant\;figures.$$
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1.8 Write using correct prefixes
$$\boldsymbol(\boldsymbol a\boldsymbol)\boldsymbol\;\boldsymbol\;\mathbf5\boldsymbol.\mathbf0\boldsymbol\;\boldsymbol\times\boldsymbol\;\mathbf{10}^{\mathbf4}\boldsymbol\;\boldsymbol c\boldsymbol m\\\\=\;5.0\;\times\;10^4\;\times\;10^{-2}m\\\\=\;5.0\;\times\;10^{4-2}m\\\\=5.0\;\times\;10^2m\\\\=\;\frac{5.0\;\times\;10^2km}{1000}\\\\=\;\frac{5.0\;\times\;100\;km}{1000}\\\\=\;\frac{500\;km}{1000}\\\\=\;0.5\;km\\\\\boldsymbol(\boldsymbol b\boldsymbol)\boldsymbol\;\boldsymbol\;\mathbf{580}\boldsymbol\;\boldsymbol\times\boldsymbol\;\mathbf{10}^{\mathbf2}\boldsymbol\;\boldsymbol g\\\\=\frac{\;580\;\times\;10^2\;kg}{1000}\\\\=\;\frac{\;580\;\times\;100\;kg}{1000}\\\\=\;58\;kg\\\\\boldsymbol(\boldsymbol c\boldsymbol)\boldsymbol\;\boldsymbol\;\mathbf{45}\boldsymbol\;\boldsymbol\times\boldsymbol\;\mathbf{10}^{\boldsymbol-\mathbf4}\boldsymbol\;\boldsymbol s\\\\=\;45\;\times\;10^{-4}\;\times\;1000\;\;ms\\\\=\;\frac{45\;\;\times\;1000\;\;ms}{10^4}\\\\=\;\frac{\;45000\;\;ms}{10000}\\\\=\;4.5\;ms\\\\$$
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1.10
$$Express\;the\;density\\of\;mercury\;given\;as\\13.6\;g\;c\;m^{-3}\;in\;kg\;m^{-3}\\\\\\$$
$$13.6\;g\;c\;m^{-3}\\\\=\;\frac{13.6\;\;kg\;c\;m^{-3}}{1000}\\\\=\;0.0136\;\;kg\;c\;m^{-3}\\\\\mathbf1\boldsymbol\;\boldsymbol c\boldsymbol m\boldsymbol\;\boldsymbol=\boldsymbol\;\mathbf{10}^{\boldsymbol-\mathbf2}\boldsymbol\;\boldsymbol m\\\\\mathbf1\boldsymbol\;\boldsymbol c\boldsymbol\;\boldsymbol m^{\mathbf3}\boldsymbol\;\boldsymbol=\boldsymbol\;\mathbf{10}^{\boldsymbol-\mathbf6}\boldsymbol\;\boldsymbol m^{\mathbf3}\\\\\;0.0136\;\;kg\;c\;m^{-3}\\\\=\frac{0.0136\;kg}{\;c\;m^3}\\\\=\;\frac{0.0136\;kg}{\;10^{-6}\;m^3}\\\\=0.0136\;\times10^6\;\;kg\;\;\;m^{-3}\\\\=\;1.36\;\times\;10^{-2\;}\times10^6\;\;kg\;\;\;m^{-3}\\\\=\;1.36\;\times10^{-2+6}\;\;kg\;\;\;m^{-3}\\\\=\;1.36\;\times10^4\;\;kg\;\;\;m^{-3}\\\\$$
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