Federal board 9th Class Physics Notes Chapter 1 , FBISE 9th Class new book notes of chapter 1
Short Response Question
1. How physics plays an important role in our life?
Ans: Physics in Science:
Physics is the most fundamental of all the sciences. In order to study biology, chemistry, or any other natural science, one should have a firm understanding of the principles of physics.
{Example:}
For example, biology uses the physics’ principles of fluid movement to understand how the blood flows through the heart, arteries, and veins. Chemistry relies on the physics of subatomic particles to understand why chemical reactions take place.
Physics and Technology:
Rockets and space shuttles, Magnetically levitating trains, and microscopic robots that fight cancer cells in our bodies. All of these technologies, whether common place or exciting, are based on the principles of physics.
Physics in Society:
Apart from playing a significant role in technology, professionals such as engineers, pilots, physicians, electricians, and computer programmers apply physics concepts in their daily work.
Example:
For example, a pilot must understand how wind forces affect a flight path.
Q2. Estimate your age in minutes and seconds?
Ans: Suppose my age =15years
$$ \begin{aligned} & =15 \times 365=5475 \text { days } ; \quad(\because 1 \text { year }=365 \text { days }) \\ & =5475 \times 24=131400 \text { hours } ; \quad(\because 1 \text { day }=24 \text { Hours }) \\ & =131400 \times 60=7884000 \text { minutes } ; \quad(\because 1 \text { hour }=60 \text { Minutes }) \\ & =7884000 \times 60=473040000 \text { seconds } \quad ; \quad(\because 1 \text { minute }=60 \text { second }) \end{aligned} $$So , if I am 15 years old, my age in minutes is approximately 7,884,000 minutes, and in seconds, it’s about 473,040,000 seconds.
Q3. What base quantities are involved in these derived physical quantities; force, pressure, power and charge?
(a) Force:
Force F=maSI. Unit of force is Newton $$(\mathrm{N})=\mathrm{kgms}^{-2}$$Base quantities involved in force are mass, length, and time.(b) Pressure:
Pressure $$=\frac{\text { force }}{\text { area }} \Rightarrow Pressure =\frac{F}{A}$$The SI unit of force $$(F=\mathrm{ma})$$ is Newton $$\mathrm{N}=\mathrm{kgms}^{-2}$$ and unit of area is $$\mathrm{m}^{2}$$. The SI Unit of Pressure is Pascal “Pa” $$\frac{\mathrm{kgms}^{-2}}{\mathrm{~m}^{2}}=\mathbf{k g m}^{-1} \mathrm{~s}^{-2}$$So, the Base quantities involved in Pressure are Mass, length and time.(c) Power:
Power $$=\frac{\text { Work }}{\text { time }} \quad ; \quad Power =\frac{w}{t}$$The SI unit of Work W=F . d is joule $$(J)=\mathrm{kgm}^{2} \mathrm{~s}^{-2}$$ and unit of time is second (s) So, the Base quantities involved in Power are Mass length and time.(d) Charge:
Q=I*t
SI Unit of charge is Coulomb (C) As Base quantities involved are Electric current, and time.
9th class physics notes chapter 1
Q4. Show that prefix micro is thousand times smaller than prefix milli ?
Ans: To show that the prefix “micro” is a thousand times smaller than the prefix “milli”, we can look at the definitions of these prefixes according to the International System of Units (SI).
The prefix “milli” (symbol: $$m=10^{-3}=\frac{1}{1,000}$$ of a unit.The prefix “micro” (symbol: $$\boldsymbol{\mu}=10^{-6}=\frac{1}{1,000,000}$$ of a unit.To find out how many times smaller “micro” is compared to “milli”, we can divide the value of “milli” by the value of “micro”:$$\frac{10^{-3}}{10^{-6}}$$When you divide fractions, you multiply by the reciprocal of the denominator:$$=\frac{1}{1,000} \times \frac{1,000,000}{1}$$ $$=\frac{1,000,000}{1,000}=1,000$$Therefore, the prefix “micro” is 1,000 times smaller than the prefix “milli”.Q5. Justify that displacement is a vector quantity while energy is a scalar quantity?
Ans: Displacement is a vector quantity because it has both magnitude and direction. When an object moves from one position to another, its displacement is the straight-line distance between the initial and final positions, along with the direction from the starting point to the ending point.
On the other hand, energy is a scalar quantity because it only has magnitude and no specific direction associated with it. For instance, the kinetic energy of an object depends only on its speed and mass, not on the direction in which it is moving. Therefore, energy is classified as a scalar quantity.
Q6. Screw gauge can give more precise length than vernier calipers. Briefly explain why?
Ans: Screw gauge can give more precise length than vernier calipers because of the smaller least count of screw gauge.
Explanationt
Smaller the least count of an instrument, more precise will be its reading. A typical screw gauge has a least count of 0.01mm OR 0.001cm while a typical vernier calipers has a least count of 0.1 mm OR 0.01cm . As least count of screw gauge is about 10 times smaller than that of vernier calipers that’s why screw gauge can give about 10 times more precise reading of longth than vernier calipers.
Q7. Differentiate between mechanical stop watch and digital ‘stopwatch?
| Mechanical Stopwatch | Digital Stopwatch |
| It has two circular dials. | It has no dials but a screen. |
| it has one knob which is used for starting and stopping the stopwatch. By long pressing this knob will reset the stopwatch. | It has two buttons. Pressing the left button starts the timer and by pressing it again the time stops, thus the elapsed time is shown. Right button is- used for reset. |
| Its accuracy is up to 1/100th second. | Its accuracy is up to 1/1000th of a second. |
| Generally the least count of analogue stop watch is 1s or 0.1s | Generally the least count of digital stop watch is 0.01s. |
Q8. How measuring cylinder is used to measure volume of an irregular shaped stone?
Ans: Measuring cylinder can be used for measuring the volume of an irregular solid body such as metallic bob as shown in figure. When the object is completely immersed the volume of the water is read again. The volume of the object is found by subtracting the first reading from the second.
Measuring cylinder can be used to find the volume of a small irregular shaped solid that sinks in water. Let us find the volume of a small stone. Take some water in a graduated measuring cylinder. Note the volume ‘ V’ of water in the cylinder. Tie the solid with a thread. Lower the solid into the cylinder till it is fully immersed in water. Note the volume ” Vi ” of water and the solid. Volume of the solid will be Vf – Vi
Q9. What precaution should be kept in mind while taking measurement using measuring cylinder?
1.While using a measuring cylinder, it must be kept vertical on a plane surface.
2.The correct method to note the level of a liquid in the cylinder is to keep the eye at the same level as the meniscus of the liquid.
3.Formation of bubbles inside the cylinder should be completely avoided. Any bubbles within leads to wrong measurements.
Q10. Why do we need to consider significant digits in measurements?
Ans: We need to consider significant digits in measurements to accurately convey the precision and reliability of the measured quantity. Significant digits help us understand the level of uncertainty associated with a measurement, facilitate error analysis, ensure accurate mathematical operations, and enable clear communication of scientific findings. By paying attention to significant digits, we can convey the magnitude of quantities effectively and maintain consistency in reporting measurements across different contexts.
9th Class Physics Numerical Chapter 1
Q1. Write the following numbers in scientific notations.?
$$a. \quad 1234 \mathrm{~m}$$ $$b. 0.000023 \mathrm{~s}$$ $$c. \quad 469.3 \times 10^{5} \mathrm{~m}$$ $$d. 0.00985 \times 10^{7} s$$Ans:a. $$\quad 1234 \mathrm{~m}$$The decimal should be in between 1 and 2 . So, we have to move 3 steps towards left that’s why the power will be positive 3 . The result will be:$$ 1234 \mathrm{~m}=1.234 \times 10^{3} \mathrm{~m} $$b. $$\quad 0.000023 \mathrm{~s}$$The decimal should be in between 2 and 3 . So, we have to move 5 steps towards right that’s why the power will be negative 5 . The result will be:$$ 0.000023 \mathrm{~s}=2.3 \times 10^{-5} \mathrm{~s} $$c. $$\quad 469.3 \times 10^{5} \mathrm{~m}$$The decimal should be in between 4 and 6 . So, we have to move 2 steps towards left that’s why the power will be positive 2. As the number has already a power of 5,2 will be added. The result will be:$$ 469.3 \times 10^{5} \mathrm{~m}=4.693 \times 10^{7} \mathrm{~m} $$d. $$\quad 0.00985 \times 10^{7} \mathrm{~s}$$The decimal should be in between 9 and 8 . So, we have to move 3 steps towards right that’s why the power will be negative 3 . As the number has already a power of 7,-3 will be added. The result will be:$$ 0.0098 \times 10^{7} \mathrm{~s}=9.85 \times 10^{4} \mathrm{~s} $$Q2. Express the followings measurements using prefixes?
a. $$\quad 27.5 \times 10^{-10} \mathrm{~m}$$ b. $$0.00023 \times 10^{-2} s$$Ans: a. $$\quad 27.5 \times 10^{-10} \mathrm{~m}$$We will convert the number first to scientific notation. For this purpose, we have to move the decimal to bring it in between 2 and 7 .$$27.5 \times 10^{-10} \mathrm{~m}=2.75 \times 10^{+1} \times 10^{-10} \mathrm{~m}=2.75 \times 10^{-9} \mathrm{~m}$$As $$10^{-9}$$ is equal to nano, so: $$27.5 \times 10^{-10} \mathrm{~m}=2.75 \mathrm{~nm}$$ b. $$0.00023 \times 10^{-2} \mathrm{~s}$$We will convert the number first to scientific notation. For this purpose, we have to move the decimal to bring it in between 2 and 3.$$ \begin{aligned} & \text { move the decimal to bring } \\ & 0.00023 \times 10^{-2} \mathrm{~s}=2.3 \times 10^{-4} \times 10^{-2} \mathrm{~s}=2.3 \times 10^{-6} \mathrm{~s} \end{aligned} $$As $$10^{-6}$$ is equal to micro, so: $$0.00023 \times 10^{-2} \mathrm{~m}=2.3 \mu \mathrm{m}$$Q3. If a boy has age of 15 . years 2 months and 10 days, convert his age in
a. seconds b. milliseconds c. mega seconds ?
Ans: First we will find number boy’s age in days.Days in 15 years $$=16 \times 365.25=5478.75$$ daysDays in 2 months $$=2 \times \mathbf{3 0}=\mathbf{6 0}$$ daysDays in 10 days $$=1 \times 10=10$$ daysBoys age in days $$=5548.75$$ daysa. As there are 24 hours in a day and 3600 seconds in an hour, so the age in seconds will be: $$5548.75 \times 24 \times 36005=4.79412 \times 10^{8} \mathrm{~s}$$b. As there are 1000 or $$10^{3}$$ milliseconds in a second, so the age in milliseconds will be: $$4.79412 \times 10^{8} \mathrm{~s}=4.79412 \times 10^{8} \times 10^{3} \mathrm{~ms}=4.79412 \times 10^{11} \mathrm{~ms}$$c. As there are $$\mathbf{1 0 0 , 0 0 0}$$ or $$\mathbf{1 0}$$ seconds in a mega second, so the age in mega seconds will be:$$4.79412 \times 10^{6} \mathrm{~s}=4.79412 \times 10^{8} \times 10^{3} \mathrm{~ms}=4.79412 \times 10^{2} \mathrm{Ms}$$Q4. How many kllometers are there in 25 micrometers?
Ans: As there are $$10^{3}$$ meters in a kilometer and $$10^{-6}$$ meters in a micrometer, so$$25 \mu \mathrm{m}=25 \times 10^{-6} \mathrm{~m}=25 \times 10^{-6} \times 10^{-3} \mathrm{~km}$$$$25 \mu \mathrm{m}=25 \times 10^{-1} \mathrm{~km}=2.5 \times 10^{-4} \mathrm{~km}$$Q5. What is pltcch and least count of:
a. Vernler callpers if smallest dilvision on main scale is 1mm and total divisions on vernier scale are 20.
b. Screw gauge if smallest division on its maln scale is 0.5mm and its movable scale has 50 divisions.
Ans:a. The least count of vernier calipers is given by:Least Count $$=\frac{\text { Smallest division on main scale }}{\text { Number of dividions on vernier scale }}$$Least Count $$=\frac{1 \mathrm{~mm}}{20}=0.05 \mathrm{~mm}=0.005 \mathrm{~cm}$$b. The least count of screw gauge is given by:$$ \begin{aligned} & \text { Least Count }=\frac{\text { Pitch of the screw }}{\text { Number of divisions on circular scal }} \\ & \text { Least Count }=\frac{0.5 \mathrm{~mm}}{50}=0.01 \mathrm{~mm}=0.001 \mathrm{~cm} \end{aligned} $$Q6. Look at the measurement of vernier callpers:
a. What is its main scale reading?
b. What is its coinciding division on vernier scale?
c. Calculate total reading on the vernier callipers?
Ans:
a. The main scale reading (reading to which zero of vernier scale coincides) is 2.5 cm
b. The coinciding division on vernier scale is 8th division.
c. The total reading: 2.5+8(0.01)cm=2.5+0.08 cm=2.58 cm
Q7. Look at the figure of screw gauge, let a small steel ball is place between its thimble and anvil then:
a. What is its main scale reading?
b. What is coinciding division of circular scale?
c. Calculate the total diameter of the ball?
Ans:
a. The main scale reading (reading which is visible) is 6.5mm.
b. The coinciding division on circular scale is 46th division.
c. The total reading:$$6.5+46(0.01) \mathrm{mm}=6.5+0.46 \mathrm{~cm}=6.96 \mathrm{~mm}$$9th class physics notes chapter 1, federal board 9th class physics notes new book 9th class notes for federal board
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