Physics class 9 Chapter 1 questions and answers Mardan board

KPK Mardan board Physics class 9 Chapter 1 questions and answers Mardan board, (Numerical, Short, Long, MCQ’s) These notes are great for learning.

Physics class 9 Chapter 1 short questions

Q.1) How technology is shaped by physics?

Answer:
Physics and technology are closely related however they are also different from each other. Physics is concerned with gathering knowledge and organizing it. Technology lets us use that knowledge for practical purposes.
Behind every technology a physical phenomenon exists, Physics has played its part in the process of humankind and in the improvement of the quality of living.

Q.2) Physics and biology are considered different branches of science, how physics links with biology?

Answer:
Physics has provided a basic understanding of the development of new instruments. In biology, we have developed CT scan machines, MRI and laser surgeries. These technologies work on the principle of Physics i.e. resonance, magnetism and optics respectively.

Q.3) Why are measurements important?

Answer:
Measurement is a comparison between an unknown quantity and a standard to see how many times as big it is.
Measurement is important because it let the scientist perform the experiments or form theories. It’s not only important in science but in various fields.

Q.4) Why area is called a derived quantity ?

Answer:
Generally, a derived physical quantity is a combination of more than one base physical quantities. The area is a derived physical quantity because a basic quantity (length) is occurring two times.

Q.5) Name any four derived units and write them as their base units ?

Q.6) Why in Physics we need to write in scientific notation?

Answer:
The need to write in scientific notation is that it is an internationally accepted way of writing numbers. With the help of it, we can write very large or very small numbers in the power of 10 with only one non zero digit before the decimal.

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Q.7) What is least count? How least count for vernier calliper and screw gauge are defined?

Answer:
The smallest value that can be measured accurately by an instrument is called its least count.
Least count for Vernier Calliper.

If the smallest main scale division is 1 mm and vernier scale has 10 division on it then the least count is
                                        Least count = 1 mm/ 10
                                                            = 0.1 mm
Least count for Screw Gauge:

If the pitch of the screw gauge is 0.5 mm and the number of divisions on the circular scale is 50 then
                                        Least count = 0.5 mm/ 50
                                                            = 0.01 mm

Q.8) How can we find the volume of a small pebble with the help of measuring cylinder?

Answer:
    We can find the volume of a small pebble with the help of measuring cylinder in the following steps
1.      In the first step, put some liquid in the measuring cylinder more than the height of the pebble.
2.      Now measure the level of the liquid and call it ‘V₁‘.
3.      In the next step, add the pebble into the cylinder and measure the level of the liquid again.
4.      Now measure the level of the liquid and call it “V₂.”
5.      The difference between the two levels will give the volume of the pebble
                                                            ∆V= V₂ – V₁.

Comprehensive Questions

Q.1) Define physics. How physics play a crucial role in science technology and society?

“Physics is the branch of science, which deals with the study of properties of matter and energy along with their mutual interaction”.

The word ‘physics’ comes from the Greek ‘knowledge of nature,’ and in general, the field aims to analyze and understand the natural phenomena of the universe. It covers a wide range of phenomena, from the smallest atomic particle to largest galaxies.

Physics and technology are closely related however they are also different from each other. Physics is concerned with gathering knowledge and organizing it. Technology lets us use that knowledge for practical purposes. Behind every technology a physical phenomenon exists, Physics has played its part in the process of humankind and in the improvement of the quality of living.

Physics has provided a basic understanding of the development of new instruments. In biology, we have developed CT scan machines, MRI and laser surgeries. These technologies work on the principle of Physics i.e. resonance, magnetism and optics.

The use of Physics in information technology has improved the standard of communication. Mobile cell phones are commonly used even by illiterates. Hologram technology is a three-dimensional image, created with photographic projection. The hologram technology is also incorporated in cell phones.

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Q.2) What is SI? Name SI base quantities and their units.

Answer:
        The complete set of units for all physical quantities is called the international system of units. SI stands for “System International“. SI units are based on seven base units which are as follows.

Q.3) What are physical quantities? Distinguish between base and derived physical quantities.

Answer:
            “All quantities that can be measured are called physical quantities”
Examples of physical quantities are mass, amount of substance, length, time, temperature, electric current, light intensity, force, velocity, density, and many others. A physical quantity is always measured of natural non-living objects. The foundation of physics rests upon physical quantities in term of which the laws of physics are expressed. Therefore, these quantities have to be measured accurately.

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Types of physical quantities
i.            Base physical quantities
ii.           Derived physical quantities
Base physical quantities
        “Those physical quantities in terms of which other physical quantities can be expressed are called base physical quantities”. There are seven base quantities which are given below

Derived physical quantities
        “Those physical quantities which are derived from base quantities are called derived physical quantities”. There are several derived quantities few of which are given below

Q.4) What is standard form or scientific notation?

Answer:
Scientific notationis an internationally accepted way of writing numbers. With the help of it, we can write very large or very small numbers in the power of 10 with only one non zero digit before the decimal.
        In physics, we deal with very large or very small numbers. For example the mass of moon is 70,000,000,000,000,000,000,000 kilograms. Unisg this number occasionally will create a mess. To avoid this mess we use a standard form or scientific notation. It represents a number as the product of a number greater than 1 and less than 10 (mantissa) and a power of 10 (exponent).
                            number (N) = mantissa (M) x 10^{exponent (n)}
        The mass of the moon can thus be written compactly as 7 x 10²² kg, where 7 is the mantissa and 22 is the exponent.

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Q.5) What are prefixes? Explain with examples.

Answer:
A mechanism in which a very small or very large number is expressed as in the form of power of ten by giving a proper name to it is called prefix to the power of ten.
        Prefix makes standard form to be written even more easily. Large  number are simple written in more convenient prefix with units.
        The thickness of a paper can be written conveniently in smaller units of millimeters instead of meter. Similarly the long distance between two cities can be expressed in a bigger unit, i.e., kilometer. A useful set of prefixes in SI to replace powers of 10.

Q.6) Describe the construction and use for measurement of the following instruments:

a) Venire Calipers               b) Screw gauge

Answer:
a) Venire Calipers
Definition
“A device used to measure a fraction of smallest scale division by sliding another scale over it is called vernier caliper”. 
   Vernier Callipers is used to measure small lengths, thickness, an internal and external diameter of a hollow cylinder, depth of an object with accuracy up to 0.1 mm or 0.01cm.
There are two scales on vernier calipers.
 Main (fixed) Scale:
            The main scale which has markings of usually of 1 mm each and it contains jaw A on its left end.
Vernier (sliding) Scale: 
            A vernier (sliding) scale which has markings of some multiple of the marking on the main scale. The vernier scale usually has the length of 9 mm and is divided equally into 10 divisions (thus the separation between two lines on vernier scale is 9/10 mm = 0.9 mm).

Suppose we want to measure the diameter of a small solid cylinder with vernier caliper we will use the following method.

1.    Note the least count of the vernier, (it is usually written on vernier caliper, otherwise we can find it out by a method described previously) which in our case is 0.01 cm (0.1 mm).

2.    Close the jaws fully with no object between the jaws. If the zero line of the vernier scale coincides with the zero lines of the main scale then there is no zero error.

3.    Now fix the solid cylinder in between the two jaws and tighten the vernier with the help of screw ‘S’ suppose the zero of the vernier scale is to the right of 1.9 cm the mark and to the left of the 2.0 cm (20 mm) mark as shown in the figure1 shown below. Thus the required length is somewhat greater than 1.9 cm (19 mm).

4.     To find the fraction to be added. We see that division of the vernier scale which coincides (inline) with any division of the main scale. As the 6th division of the vernier scale coincides with one of the main scale divisions as in the figure1 shown below.

5.    Multiply the least count 0.01 cm (0.1 mm) by 6 which gives 0.06 cm (0.6 mm) and add to 1.9 cm (19 mm). The measured length is 1.9 cm + 0.06 cm = 1.96 cm or 19 mm+ 0.6 mm = 19.6 mm Hence the diameter of solid cylinder is 1.96 cm (19.6 mm)

Hence the diameter of a solid cylinder is 1.96 cm (19.6 mm).

b) Screw gauge
Definition
    ” A device used to measure a fraction of smallest scale division by a rotatory motion of circular scale over it is known as screw gauge”.
    The main objective of the Screw Gauge is to measure the diameter of a given object and its volume.
Parts of Screw Gauge (SG)

• Datum line / Central line
• Anvil
• Spindle
• Thimble (circular scale)
• Ratchet knob
• Sphere

If the pitch of the screw is 0.5 mm and the number of divisions on the circular scale is 50 then the diameter of a small sphere is measured by using a screw gauge by the following method.
1.    Place the object between thimble and anvil. Now turn the thimble until the anvil and spindle gently grip the object. Then turn the ratchet until it starts to click. The ratchet prevents the user from exerting too much pressure on the object.
2.    Read the main scale reading. For example, in the figure2 shown above the edge of circular scale is lying between 5.5 mm and 6.0 mm, i.e. the diameter of the sphere is more than 5.5 mm and less than 6.0 mm.
3.    To know the part more than 5.5 mm. We look for the division of circular scale (Thimble scale) which is in front of the datum line. In the figure2 shown above, it is 28.
4.    Now multiplying 28 by the least count which is 0.01 mm we get 0.28 mm. Add this product to 5.5 mm. 5.5 mm + 0.28 mm = 5.78 mm Thus 5.78 mm is the diameter of the sphere.

Q.7) What is meant by the significant figures of measurement? What are the main points to be kept in mind while determining the significant figures of a measurement?

Answer:
    It is defined as accurately known digits and first doubtful digit in a measurement is known as a significant figure.
Explanation
    Consider we want to measure the length of a rod with the help of meter rod. We assume that three students measured the length as 9.63 cm, 9.62 cm and 9.64 cm. In this case, we agree with 10.6 cm, which we called accurately known digits. However, there is doubt about the third digit, which is 3, 2 and 4. These digits are called doubtful digits. Out of three digits, two are accurately known while the third one is the doubtful digit.

Rules for determining significant figures
    Following rules should be kept in mind while dealing with significant figures in a given data.
         i.            All non-zero numbers (1,2,3,4,5,6,7,8,and 9) are always significant.
        ii.            All zeroes between non-zero numbers are always significant.
      iii.            All zeroes, which are simultaneous to the right of the decimal point and at the end of the number, are always significant.
     iv.        All zeroes which are to the left of a written decimal point and are in a number ≥ 10 are always significant.

Rules for rounding of non-significant figures
We have following rules for dropping the non-significant figures
         i.    If the last digit is less than 5 then it will be ignored e.g. 84.74 is rounded to 84.7.
        ii.    If terminating digit is greater than 5 then the last digit increase by 1. e.g. 87.76 is rounded to 87.8.
      iii.    In case if dropping digit is 5 and the last retained digit is even then the last digit i.e. 5 will be dropped without affecting the next one. e.g. 77.65 is rounded to 77.6.
      iv.     And if the last digit is 5 and the 2nd last is an odd digit then the 2nd last digit is increased by1 in order to round off 5 e.g. 87.35 is rounded to 87.4.

Numerical Questions

Q.1) Write the number in prefix to power of ten 

a) Mechanical nano-oscillators can detect a mass change as small as 10^-21 kg. b) The nearest neutron star (a collapsed star made primarily of neutrons) is about 3.00×10¹⁸ m away from earth. 
c) Earth to sun distance is 149.6 million km

Answer:
a) Mass = 10^-21 kg                 = 10^-21 ×10³g             ∴k = 10³
                 =10^-21+3 g
                 =10^-18 g  
                 =1 atto g                                ∴atto = 10^-18
                 =1 ag

b) Distance     =3.00 × 10¹⁸ m
                         =3.0 Em                         ∴Exa = 10¹⁸

c) Distance = 149.6 million km
                      =149.6 x 10⁶ x 10³ m
                      =149.6 x 10⁹ m
                      =149.6 Gm                          ∴Giga = 10⁹ 

Q.2) An angstrom (symbol A^o) is a unit of length (commonly used in atomic physics), defined as 10^-10 m which is of the order of the diameter of an atom.

a) How many nanometers are in 1.0 angstrom?
b) How many femtometers or fermis (the common unit of length in nuclear physics) are in 1.0 angstrom?
c) How many angstroms are in 1.0 m?

Answer:
a) 1.0 angstrom  = 10^-10 m
                            = 10^-9 x 10^-1 m
                            = 10^-1 nm                                      ∴nano = 10^-9
                            = 0.1 nm

b) 1.0 angstrom   = 10^-10 m
                               = 10⁵ x 10^-15 m
                               = 10⁵ fm                                          ∴femto = 10^-15
                               = 100,000 fm

c) 1.0 angstrom = 10^-10 m

                        1 m = 1 /10^-10 A⁰

                              =10¹⁰ A⁰           

Q.3) The speed of light is c = 299,792,458 m/s.

a) Write this value in scientific notation.
b) Express the speed of light to 
i) Five significant figures,  ii. Three significant figures.

Answer:
a) The speed of light is c = 299, 792, 458 m/s
                                           = 2.99792458 × 10⁸ m/s

b) Rounding to significant figures 

i. Five significant figures.
The speed of light is     c = 2.9979 × 10⁸ m/s

ii. Three significant figures.
The speed of light is     c = 2.9979 × 10⁸ m/s
                                       c = 2.998 × 10⁸ m/s
                                       c = 3.00 × 10⁸ m/s

Q.4)  Express the following in terms of powers of 10.

a)  7 nanometre           b) 96 megawatt         c) 2 gigabite
d)  43 picofarad           e) 2 millimetre.

Answer:
a)       7 nanometer
                                                               We now that  1 n = 10^-9,
                                                                So, 7 nanometer = 7 x­ 10^-9 m.
b)       96 megawatt
                                                               We now that 1 M = 10⁶
                                                               So, 96 MWatt =96 x­ 10⁶ W = 9.6 x­ 10⁷ W.
c)      2 gigabits
                                                                We now that 1 G = 10⁹
                                                                So, 2 gigabits = 2 x­ 10⁹ bits.
d)       43 picofarad
                                                                We now that 1 pico = 10^-12
                                                          So, 43 picofarad = 43 x­ 10^-12 farad = 4.3 x­ 10^-11 farad.
e)       2 millimeter
                                                                We now that 1 mm = 10^-3 m
                                                                So, 2 millimeter = 2 x­ 10^-3 m.

Q.5) Write the following numbers in standard form:

a) Mass of bacterial cell: 0.000,000,000,005 kg 
b) Diameter of sun: 1,390,000,000 m

Answer:
a) Mass of Bacterial cell    = 0.000,000,000,005 kg
                                             = 0.000,000,000,005×10⁰ kg
                                             = 5 × 10^-12 kg
b) Diameter of Sun            = 1,390,000,000 m
                                            = 1,390,000,000.0 m
                                            = 1.39 × 10⁹ m