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PHYSICS CLASS 9TH CHAPTER-1 All Exercise |
PHYSICS CLASS 9TH CHAPTER-1 NOTES
QUICK QUIZ
QUICK QUIZ QUICK QUIZ 1.1:
1. Why do we study physics?
Ans: We study physics to understand the physical phenomena happening around us.
2. Name five branches of physics?
Name of five branches of physics are i. Mechanics ii. Heat iii. Sound iv. Nuclear physics v. Atomic physics
QUICK QUIZ 1.2:
1. How can you differentiate between base and derived quantities? Ans:
Base Quantities Derived Quantities Base quantities are the quantities on the The quantities that are expressed in basis of which other quantities are terms of base quantities are derived expressed. quantities. There are seven base quantities. These There are more than seven derived are length, mass, time, electric current, quantities. These include area, volume, temperature, intensity of light and the speed, force, work, energy, power, amount of a substance. electric charge, electric potential etc.
2. Identify the base quantities in the following:
a) Speed b) Area c) Force d) Distance
Ans: Here distance is the base quantity.
3. Identify the following as base or derived quantity:
Density, force, mass, speed, time, length, temperature and volume Ans:
Base Quantities Derived Quantities: Mass Density Time Force Length Speed Temperature Volume
QUICK QUIZ 1.3:
1. Name five prefixes most commonly used. Ans:
Prefixes Symbol Multiplier: mega M 106 kilo k 103 centi c 10–2 milli m 10–3 micro 10–6
2. The Sun is one hundred and fifty million kilometers away from the Earth. Write this a) As an ordinary whole number b) In scientific notation
Ans: a) As an ordinary whole number: 150 x 106 km as 1 million = 106 150 x 106 x 103 m 1 kilo = 103 150 x 109 m 150,000,000,000 m
b) In scientific notation: 150,000,000,000 m = 1.5 x 1011 m
3. Write the numbers given below in scientific notation?
a) 3000000000 ms–1 b) 6400000 m c) 0.0000000016 g d) 0.0000548 s Ans:
Numbers Scientific Notation 3000000000 ms–1 3 x 109ms–1 6400000 m 6.4 x 106 m 0.0000000016 g 1.6 x10–9g 0.0000548 s 5.48 x10–5s
QUICK QUIZ 1.4:
1. What is the least count of vernier callipers?
Ans: The least count of vernier callipers is 0.1 mm or 0.01 cm.
2. What is the range of vernier callipers used in your physics laboratory?
Ans: The range of vernier callipers used in physics laboratory is from 0.01cm to 12 cm.
3. How many divisions are there on its vernier scale?
Ans: Vernier scale has 10 divisions over it such that each of its division is 0.9 mm.
4. Why do we use zero correction?
Ans: To apply zero is known as zero correction. Zero correction is applied to get accurate result.
PHYSICS CLASS 9TH CHAPTER-1 NOTES
MINI EXERCISE
MINI EXERCISE MINI EXERCISE 1.1:
1. Express 1 cm3 in millilitres?
Ans: Volume is a derived quantity. As 1 L = 1000 mL Then 1 L = 1 dm3 = (10 cm)3 = 1000 cm3 1 cm3 = 1 mL
2. Express 1 m3 in litres?
Ans: As 1 L = 1000 cm3 = 1000 100 x 100 x 100 m3 1 L= 1 1000 m3 1 m3 =1000 L
MINI EXERCISE 1.2:
Cut a strip of paper sheet. Fold it along its length. Now mark centimeters and half centimeters along its length using a ruler. Answer the following questions:
1. What is the range of your paper scale?
Ans: The range of my paper scale is from 0.5 cm to 30 cm.
2. What is its least count?
Ans: The least count of my paper scale is 0.5 cm.
3. Measure the length of pencil using your paper scale and with meter ruler. Which one is more accurate and why?
Ans: The measurement of pencil measured by paper scale is 5.5 cm while the measurement of pencil measured by meter ruler is 5.2 cm.
The measurement of pencil measured by the meter ruler is more accurate because it even can measure the length in millimetres.
MINI EXERCISE 1.3:
1. What is the least count of screw gauge?
Ans: The least count of screw gauge is 0.01 mm or 0.001 cm.
2. What is the pitch of your laboratory screw gauge?
Ans: The pitch of laboratory screw gauge is 1 mm.
3. What is the range of school laboratory screw gauge?
Ans: The range of laboratory screw gauge is 0.01 mm to 100 mm.
4. Which one of the two instruments is more precise and why? a) Vernier callipers b) Screw gauge
Ans: The instrument screw gauge is more precise than verniercallipers because the least count of vernier callipers is 0.01 cm while the least count of screw gauge is 0.001 cm.
MINI EXERCISE 1.4:
1. What is the function of balancing screws in a physical balance?
Ans: Balancing screws in a physical balance is used to bring the pointer at zero position.
2. On what pan we place the object?
Ans: We place the object to be measured in the left pan and the varying weights or masses in the right pan. It is due to the reason that we feel convenience to change weights with our right hand when required.
PHYSICS CLASS 9TH CHAPTER-1 NOTES
DO YOU KNOW
1. What is saying of Lord Kelvin about measurements?
Ans: When you can measure what you are speaking about and express it in numbers, you know something about it.
When you cannot measure what you are speaking about or you cannot express it in numbers, your knowledge is of a meager and of unsatisfactory kind.
2. What is Andromeda?
Ans: Andromeda is one of the billions of galaxies of known universe.
3. How can we produce pollution free electricity?
Ans: Wind turbines are used to produce pollution free electricity.
4. What is the function of Hubble Telescope?
Ans: Hubble Space Telescope orbits around the earth. It provides information about stars.
5. Which one of the instruments is more precise and why? a) meter ruler b) Vernier callipers c) Screw gauge
Ans: Least count of ruler is 1 mm. It is 0.1 mm for vernier callipers and 0.01 mm for screw gauge. Thus measurements taken by micrometer screw gauge are the most precise than the other two.
6. What do you know about more precise balance?
Ans: The precision of a balance in measuring mass of an object is different for different balances. e.g. a sensitive balance cannot measure large masses. Similarly a balance that measures large masses cannot be sensitive.
Some digital balances measure even smaller difference of the order of 0.0001 g or 0.1 mg. such balances are considered the most precise balance.
PHYSICS CLASS 9TH CHAPTER-1 NOTES
MULTIPLE CHOICE QUESTIONS
MULTIPLE CHOICE EXERCISE
Q-1.1: Multiple Choice Questions:
1. The number of base units in SI are …………………….. a) 3 b) 6 c) 7 d) 9
2. Which one of the following unit is not a derived unit? a) pascal b) kilogram c) newton d) watt
3. Amount of substance in terms of numbers is measured in ……………………… a) gram b) kilogramme c) newton d) mole
4. An interval of 200 s is equivalent to …………………. a) 0.2 s b) 0.02 s c) 2 x 10–4 s d)2 x 10–6 s
5. Which one of the following is the smallest quantity? a) 0.01 g b) 2 mg c) 100 μg d) 5000 ng
6. Which instrument is most suitable to measure the internal diameter of a test tube? a) metre rule b) verniercallipers c) measuring tape d) screw gauge
7. A student claims the diameter of a wire as 1.032 cm using verniercallipers. Upto what extent do you agree with it?
a) 1 cm b) 1.0 cm c) 1.03 cm d) 1.032 cm
8. A measuring cylinder is used to measure ……………….. a) mass b) area c) volume d) level of a liquid
9. A student noted the thickness of a glass sheet using a screw gauge. On the main scale, it reads 3 divisions while 8th division on the circular scale coincides with index line. It thickness is …………………… a) 3.8 cm b) 3.08 mm c) 3.8 mm d)3.08 m
10. Significant figures in an expression are ………………… a) all the digits b) all the accurately known digits c) all the accurately known digits and the first doubtful digit d) all the accurately known digits and all the doubtful digits
MCQS ANSWERS 1 2 3 4 5 6 7 8 9 10 c b d c d b c c b c
PHYSICS CLASS 9TH CHAPTER-1 NOTES
REMAINING EXERCISE
Q-1.2: What is the difference between base quantities and derived quantities? Give three examples in each case.
Base Quantities Derived Quantities: Base quantities are the quantities on the The quantities that are expressed in basis of which other quantities are terms of base quantities are derived expressed. quantities. There are seven base quantities. These There are more than seven derived are length, mass, time, electric current, quantities. These include area, volume, temperature, intensity of light and the speed, force, work, energy, power, amount of a substance. electric charge, electric potential etc.
Q-1.3: Pick out the base units in the following: joule, newton, kilograms, hertz, mole, ampere, meter, Kelvin, coulomb and watt.
Ans: Here Kilogram, mole, ampere, meter, and Kelvin are the base units.
Q-1.4: Find the base quantities involved in each of the following derived quantities? a) speed b) volume c) force d) work
Ans: a) Speed: The formula of speed is Speed = DISTANCE time Unit of speed = unit of distance unit of time = m s = ms–1
Conclusion:
Unit of speed shows that speed is derived quantity and it is derived from base quantities length and time. Moreover unit of length is mere and unit of time is second.
b) Volume: The formula of volume of cube is Volume = length x width x height Unit of volume = unit of length x unit of width x unit of height = m x m x m = m3
Conclusion: Unit of volume shows that volume is derived quantity and it is derived from base quantity length. Moreover unit of length is meter.
c) Force: The formula of force is Force = mass x acceleration
Unit of force = unit of mass x unit of acceleration = kg x ms–2 = kgms–2 = N
Conclusion: Unit of force shows that force is derived quantity and it is derived from base quantities mass, length and time. Moreover unit of mass is kilogram unit of length is mere and unit of time is second.
d) Work: The formula of work is Work = force x distance
Unit of work = unit of force x unit of distance = kgms–2 x m = kgm2s–2 = J
Conclusion: Unit of work shows that work is derived quantity and it is derived from base quantities mass, length and time. Moreover unit of mass is kilograms unit of length is meter and unit of time is second.
Q-1.5: Estimate your age in seconds?
Ans: Suppose age of person = 14 years
Since Total days in one year = 365 Total hours in one day = 24 hours Total minutes in one hour = 60 min Total seconds in minutes = 60 sec Therefore, Age of person in seconds = 14 x 365 x 24 x 60 x 60 = 441,504,000 sec
Q-1.6: What role SI units have played in the development of science?
Ans: The SI units have brought consistency and uniformity in measurements, calculations. SI is very helpful to exchange scientific and technical information at the international level. It has also facility to convert one unit into other.
Q-1.7: What is meant by vernier constant?
Ans: Least count of vernier callipers is also called vernier constant.
“The difference between one small division on main scale division and one vernier scale division is 0.1 mm. It is called least count (LC) or vernier constant of the VernierCallipers”.
Least count or vernier constant of the Vernier Callipers can also be found as given below:
Least count of vernier calipers =smallest reading on main scale no. of division on vernier scale 1mm 10 div = 0.1 mm = 0.01 cm
Q-1.8: What do you understand by the zero error of measuring instrument?
Ans: Zero error is basically the systematic error which exists in the measuring instrument. This is the error in instrument due to which it shows measurement more than actual measurement or less than actual measurement. This error can be recovered by adding or subtracting from the observed measurement.
Q-1.9: Why is the use of zero error necessary in measuring instruments?
Ans: Zero error is necessary in measuring instrument to obtain an extreme correct value.
Q-1.10: What is a stopwatch? What is the least count of a mechanical stopwatch you have used in the laboratories?
Ans: STOPWATCH: “A stopwatch is used to measure the time interval of an event”. A mechanical stopwatch can measure a time interval up to a minimum 0.1 second.
Q-1.11: Why do we need to measure extremely small interval of times? Ans: We need to measure extremely small interval of times for greater accuracy in obtaining results.
For Example: Time period of simple pendulum and in free fall experiments time taken in falling the bob etc, these time intervals must be measured with perfect accuracy.
Q-1.12: What is meant by significant figures of measurement? Ans: SIGNIFICANT FIGURES:
“All the accurately known digits and the first doubtful digit in an expression are called significant figures”.
For Example: i. 0.055 has 2 significant digits ii.2.907 has 4 significant digits.
Q-1.13: How is precision related to the significant figures in a measured quantity?
Ans: More significant means more precision. Thus a measured quantity having more significant figures will be more precise or accurate.
For Example: If length of rod is measured using ruler and it is 5 cm. When the same length is measured by using vernier callipers it becomes 5.02 cm. In first case significant figure is 1 while in the second case it becomes 3. Since second measurement is more precise because the number of significant figures are increased. Hence greater number of significant figures means greater precision.
EXAMPLES
Example 1.1: Find the diameter of a cylinder placed between the outer jaws of vernier callipers as shown in figure:
Sol: Zero Correction:
On closing the jaws of vernier callipers, the position of vernier scale as shown in fig is Vernier division coinciding with main scale = 7 div Zero error (Z.E) = 7 x 0.01 cm = +0.07 cm Zero correction (Z.C) = – 0.07 cm
Diameter of the Cylinder: Main scale reading Vernier division coinciding with main scale Vernier Scale reading Observed diameter of the cylinder Correct diameter of the cylinder = 2.2 cm = 6 div = 6 x 0.01 cm = 0.06 cm = 2.2 cm + 0.06 cm = 2.26 cm = 2.26 – 0.07 = 2.19 cm Thus the correct diameter of the given cylinder as found by vernier calipers is 2.19 cm.
Example 1.2: Find the diameter of a wire using screw gauge as shown in figure:
Sol: Zero Correction:
On closing the gap of the screw gauge, the position of circular scale is
Circular division coinciding with index line Zero error (Z.E)
Zero correction (Z.C) = 24 div = 24 x 0.01 mm = +0.24 mm = –0.24 mm = 1 mm = 85 div = 85 x 0.01 mm = 0.85 mm = 1 mm + 0.85 mm = 1.85 mm = 1.85 – 0.24 = 1.61 mm Thus the correct diameter of the given wire as found by screw gauge is 1.61 mm.
Example 1.3: Find the mass of a small stone by a physical balance.
Sol: Follow the steps to measure the mass of a given object.
i. Adjusting the leveling screws with the help of plumb line to level the platform of physical balance. ii. Raise the beam gently by turning the arresting knob clockwise. Using balancing screws at the ends of its beam, bring the pointer at zero position. iii. Turn the arresting knob to bring the beam back on its supports. Place the given object (stone) on its left pan. iv. Place suitable standard masses from the weight box on the right pan. Raise the beam. Lower the beam if its pointer is not at zero. v. Repeat adding or removing suitable standard masses in the right pan till the pointer rests at zero on raising the beam. vi. Note the standard masses on the right pan. Their sum is the mass of the object on the left pan.
Example 1.4: Find the number of significant figures in each of the following values. Also express them in scientific notations.
a) 100.8s b) 0.00580 km c) 210.0 g
Sol: a) All the four digits are significant. The zeros between the two significant figures 1 and 8 are significant. To write the quantity in scientific notation, we move the decimal point two places to the left, thus 100.8s = 1.008 x102 s
b) The first two zeros are not significant. They are used to space the decimal point. The digit 5, 8 and the final zero are significant. Thus there are three significant figures. In scientific notation, it can be written as 5.80×10–3 km.
c) The final zero is significant since it comes after the decimal point. The zero between last zero and 1 is also significant because it comes between the significant figures. Thus the number of significant figures in this case is four. In scientific notation, it can be written as 210.0 g = 2.100 x 102g. Main scale reading Circular division coinciding with index line Circular Scale reading Observed diameter of the wire Correct diameter of the cylinder.
PROBLEMS
Problem Sol: a)5000 a) b) c)52 d) 1.1: 5000 2000 225 x 10g Express x g 000 –10 10–8 b) kg W s the 2000 following = = = = = 000 5 2 5.2 5.2 2.25 W x x 1010x x quantities x 10 1036c) 10W g 1 x 2 – 1052 x 10 10–10 = = x + –8 5 2 using 103 x kg MW g s
–10 10= 35.2 prefixes. kg g
d) 225 x 10–8 s x 10–6g = 5.2 μg (as (as 10103 6 = = k) M)
(as 10–6= μ)
= 2.25 x 102 – 8 s = 2.25 x 10–6s = 2.25 μs (as 10–6= μ)
Problem 1.2: How do the prefixes micro, nano and pico relate to each other? Sol: As we know
μ = n = p = Relation of micro with nano:
1010–6 –9
10–12
One n = = = Relation of micro with pico: One p = = = Relation of nano with pico: One p = = =
10–9 10–3 x 10–6 10–3 μ
10–12 1010–6 –6 x μ
10–6
10–12 10–3 x 10–9 10–3 n Problem 1.3: Your hair grows at the rate of 1 mm per day. Find their growth rate in nms–1. Sol: Given data:
Growth rate = 1 mm/day Required:
Growth rate in nms–1= ? As we know
1 milli 1 nano = = m n = = 1010–3 –9
In one day = 24 x 60 x 60 = 86400sec Now:
Growth rate in m/day
Growth rate in nm/day
Growth rate in nm/sec
= 1 x 10–3m/day 1 x 10–3 = 10–9 nm/day
= 1 1 x x 10106 6 nm/day = nm/sec
= 0.00001157 x 106 = 11.57 nm/sec
Problem a) 1168 1.4: Rewrite x 10–27 the b) following 32 x 105 in standard c) 725 x 10form.
–5d) 0.02 x 10–8 Sol: a) 1168 x 10–27 = 1168 x 10–27 + 3 = 1.168 x 10–24
b) c) 32 725 x x 10105 –5 = = 32 725 x x 10105 –5 + 1 + 2 = = 7.25 3.2 x x 10106
–3
d) 0.02 x 10–8 = 0.02 x 10–8 – 2 = 2 x 10–10
Problem Sol: a) a) b) c) 1.5: 6400 6400 380 300 Write 000 000 km km km 000 the b) msfollowing 380 –1 000 km = =3.8 = quantities 6.4 3 x x 10x 10108 5 msc) 3 km km in –1 300 standard 000 000 form.
ms–1 d) seconds in a day
d) seconds in a day = 24 x 60 x 60 s = 86400 s = 8.64 x 104 s
Problem to its main 1.6: scale On closing such that the 4th jaws division of a Vernier of its vernier callipers, scale zero coincides of the vernier with one scale of the is on main the scale right
division. Find its zero error and zero correction. Sol: Vernier division coinciding with main scale = 4 div
Least count of vernier calipers = 0.01 cm Now Zero error (Z.E) = 4 x 0.01 cm
= +0.04 cm Zero correction (Z.C) = – 0.04 cm
Problem 1.7: A screw gauge has 50 divisions on its circular scale. The pitch of the screw gauge is 0.5 mm. What is its least count? Sol: No of divisions on circular scale = 50 div
Pitch of Screw Gauge = 0.5 mm Least count of a screw gauge can be found as given below:
Least count = =
Pitch Number divisionson circular scale 0.5 mm 50 = 0.01 mm = 0.001 cm Thus least count of the screw gauge is 0.01 mm or 0.001 cm.
Problem 1.8: Which of the following a) 3.0066 m b) 0.00309 kg quantities c) 5.05 have x 10three –27 kg significant d) figures?
301.0 s Sol: a) 3.0066 m
The significant figures in this quantity are five. Because the zeros between significant figures are also significant.
b) 0.00309 kg The significant figures in this quantity are three. Because the zeros used for spacing the decimal point are not significant.
c) 5.05 x 10–27 kg The significant figures in this quantity are three. Because the zeros between significant figures are also significant.
d) 301.0 s The significant figures in this quantity are four. Because the zeros after decimal are also significant.
Hence (b) and (c) have three numbers of significant figures.
Problem 1.9: What are the significant a) 1.009 mb) 0.00450 kg figures c) 1.66 d) 2001 s in the x 10 following –27 kg measurements?
Sol: a) 1.009 m The significant figures in this quantity are four. Because the zeros between significant figures are also significant.
b) 0.00450 kg The significant figures in this quantity are three. Because the zeros used for spacing the decimal point are not significant.
c) 1.66 x 10–27 kg The significant figures in this quantity are three. Because all the digits before the power of 10 are significant.
d) 2001 s The significant figures in this quantity are four. Because the zeros between significant figures are also significant.
Problem 1.10: A chocolate wrapper is 6.7 cm long and 5.4 cm wide. Calculate its area upto reasonable number of significant figures. Sol: Length= 6.7 cm
Width Area = 5.4 cm = ? Now
Area = Length x Width
= 6.7 x 5.4 = 36.18 cm2 = 36 cm2
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