Selina ICSE Solutions for Class 9 Physics Chapter - 1 Measurements and Experimentation
Exercise 1(A)
1. What is meant by measurement?
Solution 1.
Measurement is the process of comparing a
given physical quantity with a known standard quantity of the same nature.
2. What do you understand by the term unit?
Solution 2.
Unit is a quantity of constant magnitude which is used to measure the magnitudes of other quantities of the same manner.
3. What are the three requirements for selecting a unit of a physical quantity?
Solution 3.
The three requirements for selecting a unit of a physical quantity are
1. It should be possible to define the unit without ambiguity.
2. The unit should be reproducible.
3. The value of units should not change with
space and time.
4. Define the three fundamental quantities.
4. Define the three fundamental quantities.
Solution 4.
Definitions of three fundamental quantities:
1. S.I. unit of length (m): A metre
was originally defined in 1889 as the distance between two marks drawn on
a platinum-iridium (an alloy made of 90% platinum and 10% iridium) rod kept
at 0°C in the International Bureau of Weights and Measures at Serves near
Paris.
2. S.I. unit of mass (kg): In 1889, one kilogramme was defined as the mass of a cylindrical piece of a platinum-iridium alloy kept at International Bureau of Weights and Measures at Serves near Paris.
3. S.I. unit of time (s): A second is defined as 1/86400th part of a mean solar day, i.e.
5. Name the three systems of unit and state the various fundamental units in them.
Solution 5.
Three systems of unit and their fundamental units:
1. C.G.S. system (or French system): In this system, the unit of length is centimeter (cm), unit of mass is gramme (g) and unit of time is second (s).
2. F.P.S. system (or British system): In this system, the unit of length is foot (ft), unit of mass is pound (lb) and unit of time is second (s).
3. M.K.S. system (or metric system): In this system, the unit of length is metre (m), unit of mass is kilogramme (kg) and unit of time is second (s).
6. Define a fundamental unit.
Solution 6.
A fundamental (or basic) unit is that which is independent of any other unit or which can neither be changed nor can be related to any other fundamental unit.
7. What are the fundamental units in S.I. system? Name them along with their symbols.
Solution 7.
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
8. Explain the meaning of a derived unit with the help of one example.
Solution 8.
The units of quantities other than those measured in fundamental units can be obtained in terms of the fundamental units, and thus the units so obtained are called derived units.
Example:
Speed = Distance/time
Hence, the unit of speed = fundamental unit of distance/fundamental unit of time
Or, the unit of speed = metre/second or ms-1.
As the unit of speed is derived from the fundamental units of distance and time, it is a derived unit.
9. Define standard metre.
Solution 9.
A metre was originally defined in 1889 as the distance between two marks drawn on a platinum-iridium (an alloy with 90% platinum and 10% iridium) rod kept at 0o C in the International Bureau of Weights and Measures at Serves near Paris.
10. Name two units of length which are bigger than a metre. How are they related to the metre?
Solution 10.
Astronomical unit (A.U.) and kilometer (km) are units of length which are bigger than a metre.
1 km = 1000 m
1 A.U. = 1.496 × 1011 m
11. Write the names of two units of length smaller than a metre. Express their relationship with metre.
Solution 11.
Centimeter (cm) and millimeter (mm) are units of length smaller than a metre.
1 cm = 10-2 m
1 mm = 10-3 m
12. How is nanometer related to Angstrom?
12. How is nanometer related to Angstrom?
Solution 12.
1 nm = 10 Å
13. Name three convenient units used to measure lengths ranging from very short to very long value. How are they related to SI unit?
Solution 13.
Three convenient units of length and their relation with the S.I. unit of length:
1. 1 Angstrom (Å) = 10-10 m
2. 1 kilometre (km) = 103 m
3. 1 light year (ly) = 9.46 × 1015 m
14. Name the S.I. unit of mass and define it.
Solution 14.
S.I. unit of mass is ‘kilogramme’.
In 1889, one kilogramme was defined as the mass of a cylindrical piece of a platinum-iridium alloy kept at the International Bureau of Weights and Measures at Serves near Paris.
(a) 1 light year = ...........m
(b) 1 m = ...........Å
(c) 1 m = ...........µ (micron)
(d) 1 micron = ...........Å
(e) 1 fermi = ...........m
Solution 15.
(a) 1 light year = 9.46 × 1015 m
(b) 1 m = 1010 Å
(c) 1 m = 106 µ (micron)
(d) 1 micron = 104 Å
(e) 1 fermi = 10-15 m
16. State two units of mass smaller than a kilogram. How are they related to the kilogramme?
Solution 16.
The units ‘gramme’ (g) and ‘milligramme’ (mg) are two units of mass smaller than ‘kilogramme’.
1 g = 10-3 kg
1 mg = 10-6 kg
17. State two units of mass bigger than a kilogram. Give their relationship with the kilogramme.
Solution 17.
The units ‘quintal’ and ‘metric tonne’ are two units of mass bigger than ‘kilogramme’.
1 quintal = 100 kg
1 metric tonne = 1000 kg
18. Complete the following
(a) 1 g = .......... kg
(b) 1 mg = .......... kg
(c) 1 quintal = .......... kg
(d) 1 a.m.u (or u) = .......... kg
Solution 18.
(a) 1 g = 10-3 kg
(b) 1 mg = 10-6 kg
(c) 1 quintal = 100 kg
(d) 1 a.m.u (or u) = 1.66 x 10-27 kg
19. Name the S.I. unit of time and define it.
Solution 19.
The S.I. unit of time is second (s).
A second is defined as 1/86400th part of a mean solar day, i.e.
20. Name two units of time bigger than a second. How are they related to second?
Solution 20.
The units ‘minute’ (min) and ‘year’ (yr) are two units of time bigger than second(s).
1 min = 60 s
1 yr = 3.1536 × 107 s
21. What is a leap year?
Solution 21.
A leap year is the year in which the month of February has 29 days.
22. 'The year 2016 will have February of 29 days'. Is this statement true?
Solution 22.
Yes, the given statement is true.
23. What is a lunar month?
Solution 23.
One lunar month is the time in which the moon completes one revolution around the earth. A lunar month is made of nearly 4 weeks.
24. Complete the following
(a) 1 nanosecond = .........s.
(b) 1 µs = .........s.
(c) 1 mean solar day = .........s.
(d) 1 year = .........s.
(a) 1 nanosecond = .........s.
(b) 1 µs = .........s.
(c) 1 mean solar day = .........s.
(d) 1 year = .........s.
Solution 24.
(a) 1 nanosecond = 10-9 s
(b) 1 µs = 10-6 s
(c) 1 mean solar day = 86400 s
(d) 1 year = 3.15 × 107 s
25. Name the physical quantities which are measured in the following units
(a) u
(b) ly
(c) ns
(d) nm
Solution 25.
(a) Mass
(b) Distance (or length)
(c) Time
(d) Length
(b) Distance (or length)
(c) Time
(d) Length
26. Write the derived units of the following
(a) Speed
(b) Force
(c) Work
(d) Pressure.
Solution 26.
(a) ms-1
(b) kg ms-2
(c) kg m2 s-2
(d) kg m-1s-2
(b) kg ms-2
(c) kg m2 s-2
(d) kg m-1s-2
27. How are the following derived units related to the fundamental units?
(a) Newton
(b) Watt
(c) Joule
(d) Pascal
Solution 27.
(a) kg m s-2
(b) kg m2s-3
(c) kg m2s-2(b) kg m2s-3
(d) kg m-1s-2
28. Name the physical quantities related to the following units:
(a) km2 (b) Newton (c) Joule (d) Pascal (e) Watt
Solution 28.
(a) Area
(b) Force
(c) Energy
(d) Pressure (b) Force
(c) Energy
(e) Power
MULTIPLE CHOICE TYPE
1. The fundamental unit present in the list mentioned below is
1. Newton
2. Pascal
3. Hertz
4. Second
Solution 1.
4. Second
2. Which of the following unit is not a fundamental unit:
1. Metre
2. Litre
3. Second
4. Kilogramme
1. Metre
2. Litre
3. Second
4. Kilogramme
Solution 2.
2. Litre
3. The unit of time is
1. Light year
2. Parsec
3. Leap year
4. Angstrom.
1. Light year
2. Parsec
3. Leap year
4. Angstrom.
Solution 3.
3. Leap year
4. 1 Å is equal to
1. 0.1 nm
2. 10-10cm
3. 10-8m
4. 104 µ.
1. 0.1 nm
2. 10-10cm
3. 10-8m
4. 104 µ.
Solution 4.
1. 0.1 nm
5. Light year (ly) is the unit of
1. Time
2. Length
3. Mass
4. None of these.
Solution 5.
2. Length
NUMERICALS
1. The wavelength of light of a particular colour is 5800 Å.(a) Express it in nanometre and (b) metre.
Solution 1.
Wavelength of light of particular colour = 5800 Å
(a) (i) 1 Å = 10-1 nm
5800 Å = 5800 × 10-1nm
= 580 nm
(ii) 1 Å = 10-10 m
5800 Å = 5800 × 10-10 m
= 5.8 × 10-7m
2. The size of bacteria is 1 µ. Find the number of bacteria present in 1 m length.
Solution 2.
Size of a bacteria = 1 µ
Since 1 µ = 10-6 m
Number of the particle = Total length/size of
one bacteria
= 1 m/10-6 m
= 106
3. The distance of a galaxy is 5·6 × 1025 m.
Assuming the speed of light to be 3 × 108 m s -1 . Find the time taken by light to travel this distance
[Hint : Time taken = distance travelled/speed]
Solution 3.
Distance of galaxy = 5.6 × 1025 m
Speed of light = 3 × 108 m/s
(a) Time taken by light = Distance travelled/speed of light
= (5.6 × 1025 / 3 × 108) s
= 1.87 × 1017 s
4. Wavelength of light is 589 nm. What is its wavelength in Å
Solution 4.
Wavelength of light = 589 nm
= 589 × 10-9 m
= 5.89 × 10-7m
Order of magnitude = 101 × 10-7 m
= 10-6 m
(This is because the numerical value of 5.89 is more than the numerical value 3.2)
5. The mass of an oxygen atom is 16.00µ. Find it's mass in kg.
Solution 5.
Mass of an oxygen atom = 16.00 u
Now, 1 u = 1.66 × 10-27 kg
Hence, mass of oxygen in kg = 16 × 1.66 × 10-27 kg
= 26.56 × 10-27 kg
Because the numerical value of 26.56 is greater than the numerical value of 3.2, the order of magnitude of mass of oxygen in kg
= 101 × 10-27 kg
= 10-27 kg
6. It takes time 8 min for light to reach from the sun to the earth surface. If speed of light taken to be 3 × 108 m/s, find the distance from the sun to the earth in km.
Solution 6.
Time taken by light to reach from the Sun to the Earth = 8 min = 480 s.
Speed of light = 3 × 108 m/s
Distance from the Sun to the Earth = Speed × time
= 3 × 108 × 480 m
= 1440 × 108 m
= 1440 × 108 × 10-3 km
= 1440 × 105 km
= 1.44 × 108 km
Because the numerical value of 1.44 is less than the numerical value of 3.2, the order of magnitude of distance from the Sun to the Earth in km = 100 × 108 km
= 108 km
7. If the distance of a star from the earth is 8.33 light minutes. What do you mean by this statement? Express the distance in metre.
Solution 7.
The statement ‘the distance of a star from the Earth is 8.33 light minutes’ means that the light from the star takes 8.33 minutes to reach Earth.
Exercise 1(B)
1. Explain the meaning of the term 'least count of an instrument' by taking a suitable example.
Solution 1.
The least count of an instrument is the smallest measurement that can be taken accurately with it. For example, if an ammeter has 5 divisions between the marks 0 and 1A, then its least count is 1/5 = 0.2 A or it can measure current up to the value 0.2 accurately.
2. A boy makes a ruler with graduations in cm on it, i.e. 100 divisions in 1 m. To what accuracy can this ruler measure? How can this accuracy be increased?
Solution 2.
Total length of the ruler = 1 m = 100 cm
No. of divisions = 100
Length of each division = Total length/total no. of divisions
= 100 cm/100
= 1 cm
Thus, this ruler can measure with an accuracy of 1 cm.
To increase the accuracy, the total number of divisions on the ruler must be increased.
3. A boy measures the length of a pencil and expresses it to be 2.6 cm. What is the accuracy of hies measurement? Can he write it as 2.60 cm?
Solution 3.
The least count of a metre rule is 1 cm.
The length cannot be expressed as 2.60 cm because a metre scale measures length correctly only up to one decimal place of a centimeter.4. Define the least count of vernier callipers. How do you determine it?
Solution 4.
The least count of vernier callipers is equal to the difference between the values of one main scale division and one vernier scale division.
Let n divisions on vernier callipers be of length equal to that of (n – 1) divisions on the main scale and the value of 1 main scale division be x. Then,
Value of n divisions on vernier = (n – 1) x
5. Define the term 'Vernier constant'.
Solution 5.
Vernier constant is equal to the difference between the values of one main scale division and one vernier scale division. It is the least count of vernier callipers.
6. When is a vernier calipers said to be free from zero error?
Solution 6.
A vernier calipers is said to be free from zero error, if the zero mark of the vernier scale coincides with the zero mark of the main scale.
7. What is meant by zero error of vernier callipers? How is it determined? Draw neat diagrams to explain it. How is it taken in account to get the correct measurement?
Solution 7.
Due to mechanical errors, sometimes the zero mark of the vernier scale does not coincide with the zero mark of the main scale, the vernier callipers is said to have zero error.
It is determined by measuring the distance between the zero mark of the main scale and the zero mark of the vernier scale.
The zero error is of two kinds
1. Positive zero error
2. Negative zero error
1. Positive zero error: On bringing the two jaws together, if the zero mark of the vernier scale is on the right of the zero mark of the main scale, the error is said to be positive.
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
To find this error, we note the division of the vernier scale, which coincides with any division of the main scale. The number of this vernier division when multiplied by the least count of the vernier callipers, gives the zero error.
For example, for the scales shown, the least count is 0.01 cm and the 6th division of the vernier scale coincides with a main scale division.
Zero error = +6 × L.C. = +6 × 0.01 cm
= +0.06 cm
2. Negative zero error: On bringing the two jaws together, if the zero mark of the vernier scale is on the left of the zero mark of the main scale, then the error is said to be negative.
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
To find this error, we note the division of the vernier scale coinciding with any division of the main scale. The number of this vernier division is subtracted from the total number of divisions on the vernier scale and then the difference is multiplied by the least count.
For example, for the scales shown, the least count is 0.01 cm and the sixth division of the vernier scale coincides with a certain division of the main scale. The total number of divisions on vernier callipers is 10.
Zero error = – (10 – 6) × L.C.
= – 4 × 0.01 cm = – 0.04 cm
Correction:
To get correct measurement with vernier callipers having a zero error, the zero error with its proper sign is always subtracted from the observed reading.
Correct reading = Observed reading – zero error (with sign)
8. A Vernier callipers has a zero error of + 0.06 cm. Draw a neat labelled diagram to represent it.
8. A Vernier callipers has a zero error of + 0.06 cm. Draw a neat labelled diagram to represent it.
Solution 8.
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
9. Draw a neat and labelled diagram of a vernier callipers. Name its main parts and state their functions.
Solution 9.
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
Main parts and their functions:
• Main scale: It is used to measure length correct up to 1 mm.
• Vernier scale: It helps to measure length correct up to 0.1 mm.
• Outside jaws: It helps to measure length of a rod, diameter of a sphere, external diameter of a hollow cylinder.
• Inside jaws: It helps to measure the internal diameter of a hollow cylinder or pipe.
• Strip: It helps to measure the depth of a beaker or a bottle.
10. State three uses of the vernier callipers.
10. State three uses of the vernier callipers.
Solution 10.
Three uses of vernier callipers are
1. Measuring the internal diameter of a tube or a cylinder.
2. Measuring the length of an object.
3. Measuring the depth of a beaker or a bottle.
11. Name the two scales of a vernier callipers and explain how it is used to measure length correct up to 0.01 cm.
11. Name the two scales of a vernier callipers and explain how it is used to measure length correct up to 0.01 cm.
Solution 11.
Two scales of vernier calipers are
1. Main scale
2. Vernier scale
The main scale is graduated to read up to 1 mm and on vernier scale, the length of 10 divisions is equal to the length of 9 divisions on the main scale.
Value of 1 division on the main scale= 1 mm
Total no. of divisions on the vernier scale = 10
Thus, L.C. = 1 mm /10 = 0.1 mm = 0.01 cm.
Hence, a vernier callipers can measure length correct up to 0.01 cm.
12. Describe in steps, how would you use a vernier callipers to measure the length of a small rod?
12. Describe in steps, how would you use a vernier callipers to measure the length of a small rod?
Solution 12.
Measuring the length of a small rod using vernier calipers:
The rod whose length is to be measured is placed in between the fixed end and the vernier scale as shown in the figure.
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
In this position, the zero mark of the vernier scale is ahead of 1.2 cm mark on main scale. Thus the actual length of the rod is 1.2 cm plus the length ab (i.e., the length between the 1.2 cm mark on the main scale and 0 mark on vernier scale).
To measure the length ab, we note the pth division of the vernier scale, which coincides with any division of main scale.
Now, ab + length of p divisions on vernier scale = length of p divisions on main scale
Alternatively, ab = length of p divisions on the main scale – length of p divisions on the vernier scale.
= p (length of 1 division on main scale – length of 1 division on vernier scale)
= p × L.C.
Therefore, total reading = main scale reading + vernier scale reading
= 1.2 cm + (p × L.C.)
13. Name the part of the vernier callipers which is used to measure the following
(a) External diameter of a tube,
(b) Internal diameter of a mug,
(c) Depth of a small bottle,
(d) Thickness of a pencil.
13. Name the part of the vernier callipers which is used to measure the following
(a) External diameter of a tube,
(b) Internal diameter of a mug,
(c) Depth of a small bottle,
(d) Thickness of a pencil.
Solution 13.
(a) Outside jaws
(b) Inside jaws
(c) Strip
(d) Outer jaws
14. Explain the terms (i) Pitch and (ii) Least count of a screw gauge. How are they determined?
14. Explain the terms (i) Pitch and (ii) Least count of a screw gauge. How are they determined?
Solution 14.
(i) Pitch: The pitch of a screw gauge is the distance moved by the screw along its axis in one complete rotation.
(ii) Least count (L.C.) of a screw gauge: The L.C. of a screw gauge is the distance moved by it in rotating the circular scale by one division.
Thus, L.C. = Pitch of the screw gauge/total no. of divisions on its circular scale.
If a screw moves by 1 mm in one rotation and it has 100 divisions on its circular scale, then pitch of screw = 1 mm.
Thus, L.C. = 1 mm / 100 = 0.01 mm = 0.001 cm
15. How can the least count of a screw gauge be decreased?
15. How can the least count of a screw gauge be decreased?
Solution 15.
The least count of a screw gauge can be increased by decreasing the pitch and increasing the total number of divisions on the circular scale.
16. Draw a neat and labelled diagram of a screw gauge. Name its main parts and state their functions.
16. Draw a neat and labelled diagram of a screw gauge. Name its main parts and state their functions.
Solution 16.
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
Main parts and their functions:
1. Ratchet: It advances the screw by turning it until the object is gently held between the stud and spindle of screw.
2. Sleeve:It marks the main scale and base line.
3. Thimble: It marks the circular scale.
4. Main scale:It helps to read the length correct up to 1 mm.
5. Circular scale: It helps to read length correct up to 0.01 mm.
17. State one use of a screw gauge.
17. State one use of a screw gauge.
Solution 17.
A screw gauge is used for measuring diameter of circular objects mostly wires with an accuracy of 0.001 cm.
18. State the purpose of ratchet in a screw gauge.
Solution 18.
Ratchet helps to advance the screw by turning it until the object is gently held between the stud and spindle of the screw.
19. What do you mean by zero error of a screw gauge? How is it accounted for?
Solution 19.
Due to mechanical errors, sometimes when the anvil and spindle end are brought in contact, the zero mark of the circular scale does not coincide with the base line of main scale. It is either above or below the base line of the main scale, in which case the screw gauge is said to have a zero error. It can be both positive and negative.
It is accounted by subtracting the zero error (with sign) from the observed reading in order to get the correct reading.
Correct reading = Observed reading – zero error (with sign)]
20. A screw gauge has a least count 0.001 cm and zero error +0.007 cm. Draw a neat diagram to represent it.
20. A screw gauge has a least count 0.001 cm and zero error +0.007 cm. Draw a neat diagram to represent it.
Solution 20.
Diagram of a screw gauge with L.C. 0.001 cm and zero error +0.007 cm.
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
21. What is backlash error? Why is it caused? How is it avoided?
Solution 21.
Backlash error: If by reversing the direction of rotation of the thimble, the tip of the screw does not start moving in the opposite direction immediately but remains stationary for a part of rotation; it is called backlash error.
Reason: It happens due to wear and tear of the screw threads.
To avoid the backlash error, while taking the measurements the screw should be rotated in one direction only. If the direction of rotation of the screw needs to be changed, then it should be stopped for a while and then rotated in the reverse direction.
22. Describe the procedure to measure the diameter of a wire with the help of a screw gauge.
Solution 22.
Measurement of diameter of wire with a screw gauge:
The wire whose thickness is to be determined is placed between the anvil and spindle end, the thimble is rotated until the wire is firmly held between the anvil and spindle. The rachet is provided to avoid excessive pressure on the wire. It prevents the spindle from further movement. The thickness of the wire could be determined from the reading as shown in the figure below.
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
The pitch of the screw = 1 mm
L.C. of screw gauge = 0.01 mm
Main scale reading = 2.5 mm
46th division of circular scale coincides with the base line.
Therefore, circular scale reading = 46 × 0.01 = 0.46 mm
Total reading = Main scale reading + circular scale reading
= (2.5 + 0.46) mm
= 2.96 mm
23. Name the instrument which can accurately measure the following
(a) The diameter of a needle,
(b) The thickness of a paper,
(c) The internal diameter of the neck of a water bottle,
(d) The diameter of a pencil.
23. Name the instrument which can accurately measure the following
(a) The diameter of a needle,
(b) The thickness of a paper,
(c) The internal diameter of the neck of a water bottle,
(d) The diameter of a pencil.
Solution 23.
(a) Screw gauge
(b) Screw gauge
(c) Vernier calipers
(d) Screw gauge
24.Which of the following measures a small length with high accuracy: metre rule, vernier callipers or screw gauge?
24.Which of the following measures a small length with high accuracy: metre rule, vernier callipers or screw gauge?
Solution 24.
Screw gauge measures a length to a high accuracy.
25. Name the instrument which has the least count
(a) 0.1 mm (b) 1 mm (c) 0.01 mm.
25. Name the instrument which has the least count
(a) 0.1 mm (b) 1 mm (c) 0.01 mm.
Solution 25.
(a) Vernier callipers (b) Metre scale (c) Screw gauge.
1. The least count of a vernier calipers is :MULTIPLE CHOICE TYPE
a. 1 cm
b. 0.001 cm
c. 0.1 cm
d. 0.01 cm
Solution 1.
d. The least count of a vernier calipers is 0.01 cm
2. A microscope has its main scale with 20 divisions in 1 cm and vernier scale with 25 divisions, the length of which is equal to the length of 24 divisions of main scale. The least count of microscope is
1. 0.002 cm
2. 0.001 cm
3. 0.02 cm
4. 0.01 cm
1. 0.002 cm
2. 0.001 cm
3. 0.02 cm
4. 0.01 cm
Solution 2.
1. 0.002 cm
3. The diameter of a thin wire can be measured by
1. A vernier calipers
2. A metre rule
3. A screw gauge
4. None of these.
1. A vernier calipers
2. A metre rule
3. A screw gauge
4. None of these.
Solution 3.
A screw gauge
NUMERICALS
1. A stopwatch has 10 divisions graduated between the 0 and 5 s marks. What is its least count?
Solution 1.
Range of the stop watch = 5s
Total number of divisions = 10
L.C. = 5/10 = 0.5 s
2. A vernier has 10 divisions and they are equal to 9 divisions of the main scale in length. If the main scale is calibrated in mm, what is its least count?
2. A vernier has 10 divisions and they are equal to 9 divisions of the main scale in length. If the main scale is calibrated in mm, what is its least count?
Solution 2.
Value of 1 m.s.d. = 1 mm
10 vernier divisions = 9 m.s.d.
L.C. = Value of 1 m.s.d./number of divisions on vernier scale
= 1 mm/10
= 0.1 mm or 0.01 cm
3. A microscopic is provided with a main scale graduated with 20 devision in 1 cm and a Vernier scale with 50 divisions on it of length same as of 49 divisions of main scale. Find the least count of the microscope.
Solution 3.
There are 20 divisions in 1 cm on the main scale.
Therefore, the value of 1 m.s.d. (x) = 1/20 cm = 0.05 cm
No. of divisions on the vernier scale (n) = 25
Hence, the L.C. of the microscope = x/n = (0.05 / 25) cm
= 0.002 cm
4. A boy uses a vernier callipers to measure the thickness of his pencil. He measures it to be 1.4 mm. If the zero error of vernier callipers is +0.02 cm, what is the correct thickness of the pencil?
Solution 4.
Thickness of the pencil (observed reading) = 1.4 mm
Zero error = + 0.02 cm = + 0.2 mm
Correct reading = observed reading – zero error (with sign)
= 1.4 mm – 0.2 mm
= 1.2 mm
5. A vernier callipers has its main scale graduated in mm and 10 divisions on its vernier scale are equal in length to 9 mm. When the two jaws are in contact, the zero of the vernier scale is ahead of the zero of the main scale and the 3rd division of the vernier scale coincides with a main scale division.
Find : (i) The least count and
(ii) The zero error of the vernier callipers.
5. A vernier callipers has its main scale graduated in mm and 10 divisions on its vernier scale are equal in length to 9 mm. When the two jaws are in contact, the zero of the vernier scale is ahead of the zero of the main scale and the 3rd division of the vernier scale coincides with a main scale division.
Find : (i) The least count and
(ii) The zero error of the vernier callipers.
Solution 5.
(i) Value of 1 m.s.d. =1 mm
10 vernier divisions = 9 m.s.d.
L.C. = Value of 1 m.s.d./number of divisions on the vernier scale
= 1 mm/10
= 0.1 mm or 0.01 cm
(ii) On bringing the jaws together, the zero of the vernier scale is ahead of zero of the main scale, the error is positive.
3rd vernier division coincides with a main scale division.
Total no. of vernier divisions = 10
Zero error = +3 × L.C.
= +3 × 0.01 cm
= +0.03 cm
6. The main scale of a vernier callipers is calibrated in mm and 19 divisions of main scale are equal in length to 20 divisions of the vernier scale. In measuring the diameter of a cylinder by this instrument, the main scale reads 35 divisions and 4th division of vernier scale coincides with a main scale division. Find
(i) Least count and (ii) Radius of a cylinder.
Solution 6.
(i) Value of 1 m.s.d = 1 mm = 0.1 cm
20 vernier divisions = 19 m.s.d.
L.C. = Value of 1 m.s.d./number of divisions on the vernier scale
= 1mm/20
= (0.1/20) cm
= 0.005 cm
(ii) Main scale reading = 35 mm = 3.5 cm
Since 4th division of the main scale coincides with the main scale, i.e. p = 4.
Therefore, the vernier scale reading = 4 × 0.005 cm = 0.02 cm
Total reading = Main scale reading + vernier scale reading
= (3.5 + 0.02) cm
= 3.52 cm
Radius of the cylinder = Diameter (Total reading) / 2
= (3.52/2) cm
= 1.76 cm
7. In the vernier callipers, there are 10 divisions on the vernier scale and 1 cm on the main scale is divided into 10 parts. While measuring the length, the zero of the vernier lies just ahead of the 1.8 cm mark and the 4th division of vernier coincides with a main scale division.
(a) Find the length.
(b) If zero error of vernier callipers is -0.02 cm,
What is the correct length ?
Solution 7.
(a) L.C. of vernier callipers = 0.01 cm
Main scale reading = 1.8 cm
Since 4th division of the main scale coincides with the main scale, i.e. p = 4.
Therefore, the vernier scale reading = 4 × 0.01 cm = 0.04 cm
Total reading = Main scale reading + vernier scale reading
= (1.8 + 0.04) cm
= 1.84 cm
(b) Observed reading = 1.84 cm
Zero error = -0.02 cm
Correct reading = Observed reading – Zero error (with sign)
= [1.84 – (-0.02)] cm
= 1.86 cm
8. While measuring the length of a rod with a vernier callipers, Fig. 2.22 shows the position of its scales. What is the length of the rod?
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
Solution 8.
L.C. of vernier callipers = 0.01 cm
In the shown scale,
Main scale reading = 3.3 mm
6th vernier division coincides with an m.s.d.
Therefore, vernier scale reading = 6 × 0.01 cm = 0.06 cm
Total reading = m.s.r. + v.s.r.
= 3.3 + 0.06
= 3.36 cm
9. The pitch of a screw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of a screw gauge?
Solution 9.
Pitch of a screw gauge = 0.5 mm
No. of divisions on the circular scale = 100
L.C. = (0.5/100) mm
= 0.005 mm or 0.0005 cm
10. The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions.
(i) What is the pitch of the screw gauge?
(ii) What is the least count of the screw gauge?
Solution 10.
No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/2 = 0.5 mm
(ii) L.C. = Pitch/No. of divisions on the circular head
= (0.5/50) mm
= 0.01 mm
11. The pitch of a screw gauge is 1 mm and the circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on the circular scale coincides with the base line. Find
(i) The least count and
(ii) The diameter of the wire.
Solution 11.
Pitch of the screw gauge = 1mm
No. of divisions on the circular scale = 100
(i) L.C. = Pitch/No. of divisions on the circular head
= (1/100) mm
= 0.01 mm or 0.001 cm
(ii) Main scale reading = 2mm = 0.2 cm
No. of division of circular head in line with the base line (p) = 45
Circular scale reading = (p) × L.C.
= 45 x 0.001 cm
= 0.045 cm
Total reading = M.s.r. + circular scale reading
= (0.2 + 0.045) cm
= 0.245 cm
12. When a screw gauge with a least count of 0.01 mm is used to measure the diameter of a wire, the reading on the sleeve is found to be 1 mm and the reading on the thimble is found to be 27 divisions.
(i) What is the diameter of the wire in cm?
(ii) If the zero error is +0.005 cm, what is the correct diameter?
Solution 12.
(i) L.C. of screw gauge = 0.01 mm or 0.001 cm
Main scale reading = 1 mm or 0.1 cm
No. of division of circular head in line with the base line (p) = 27
Circular scale reading = (p) × L.C.
= 27 × 0.001 cm
= 0.027 cm
Diameter (Total reading) = M.s.r. + circular scale reading
= (0.1 + 0.027) cm
= 0.127 cm
(ii) Zero error = 0.005 cm
Correct reading = Observed reading – zero error (with sign)
= [0.127 – (+0.005)] cm
= 0.122 cm
13. A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two revolutions. When the flat end of the screw is in contact with the stud, the zero of the circular scale lies below the base line and 4th division of the circular scale is in line with the base line. Find
(i) The pitch,
(ii) The least count and
(iii) The zero error of the screw gauge
Solution 13.
No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/2 = 0.5 mm
(ii) L.C. = Pitch/No. of divisions on the circular head
= (0.5/50) mm
= 0.01 mm
(iii) Because the zero of the circular scale lies below the base line, when the flat end of the screw is in contact with the stud, the error is positive.
No. of circular division coinciding with m.s.d. = 4
Zero error = + (4 × L.C.)
= + (4 × 0.01) mm
= + 0.04 mm
14. Fig., below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on the main scale when circular head is rotated once.
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
Find:
(i) Pitch of the screw gauge,
(ii) Least count of the screw gauge and
(iii) The diameter of the wire.
Solution 14.
No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/1 = 1 mm.
(ii) L.C. = Pitch/No. of divisions on the circular head
= (1/50) mm
= 0.02 mm
(iii) Main scale reading = 4 mm
No. of circular division coinciding with m.s.d. (p) = 47
Circular scale reading = p × L.C.
= (47 × 0.02) mm
= 0.94 mm
Diameter (Total reading) = M.s.r. + circular scale reading
= (4 + 0.94) mm
= 4.94 mm
15. A screw has a pitch equal to 0.5 mm. What should be the number of divisions on its head in order to read correctly up to 0.001 mm with its help?
Solution 15.
Pitch of the screw gauge = 0.5 mm
L.C. of the screw gauge = 0.001 mm
No. of divisions on circular scale = Pitch / L.C.
= 0.5 / 0.001
= 500
Exercise 1(C)
1. What is a simple pendulum? Is the pendulum used in a pendulum clock as a simple pendulum? Give reason to your answer.
Solution 1.
A simple pendulum is a heavy point mass (known as bob) suspended from a rigid support by a massless and inextensible string.
No, the pendulum used in pendulum clock is not a simple pendulum because the simple pendulum is an ideal case. We cannot have a heavy mass having the size of a point and string having no mass.
2. Define the terms: (i) oscillation, (ii) amplitude, (iii) frequency and (iv) time period as related to a simple pendulum.
Solution 2.
(i) Oscillation: One complete to and fro motion of the pendulum is called one oscillation.
(ii) Amplitude: The maximum displacement of the bob from its mean position on either side is called the amplitude of oscillation. It is measured in metres (m).
(iii) Frequency: It is the number of oscillations made in one second. Its unit is hertz (Hz).
(iv) Time period: This is the time taken to complete one oscillation. It is measured in second (s).
3. Draw a neat diagram of a simple pendulum.
Show the effective length and one oscillation of the pendulum.
Solution 3.
Simple Pendulum:
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
4. Name two factors on which the time period of a simple pendulum depends. Write the relation for the time period in terms of the above named factors.
Solution 4.
Two factors on which the time period of a simple pendulum depends are
1. Length of pendulum (l)
2. Acceleration due to gravity (g)
5. Name two factors on which the time period of a simple pendulum does not depend.
Solution 5.
Two factors on which the time period of a simple pendulum does not depend are
1. Material of the bob
2. Amplitude
6. How is the time period of a simple pendulum affected, if at all, in the following situations:
(a) The length is quadrupled.
(b) The acceleration due to gravity is reduced to one-fourth.
Solution 6.
Therefore, the time period is doubled.
(b) If the acceleration due to gravity is reduced to one-fourth,
Therefore, the time period is doubled
7. How is the time period T and frequency f of a simple pendulum related to each other?
Solution 7.
8. How do you measure the time period of a given pendulum? Why do you note the time for more than one oscillation?
Solution 8.
Measurement of time period of a simple pendulum:
1. To measure the time period of a simple pendulum, the bob is slightly displaced from its mean position and is then released. This gives a to and fro motion about the mean position to the pendulum.
2. The time ‘t’ for 20 complete oscillations is measured with the help of a stop watch.
3. Time period ‘T’ can be found by dividing ‘t’ by 20.
To find the time period, the time for the number of oscillations more than 1 is noted because the least count of stop watch is either 1 s or 0.5 s. It cannot record the time period in fractions such as 1.2 or 1.4 and so on.
9. How does the time period (T) of a simple pendulum depend on its length (l)? Draw a graph showing the variation of T2 with l. How will you use this graph to determine the value of g (acceleration due to gravity)?
Solution 9.
The time period of a simple pendulum is directly proportional to the square root of its effective length.
 |
| Selina ICSE Solutions for Class 9 Physics Chapter 1 Measurements and Experimentation |
From this graph, the value of acceleration due to gravity (g) can be calculated as follows.
The slope of the straight line can be found by taking two points P and Q on the straight line and drawing normals from these points on the X- and Y-axis, respectively. Then, the value of T2 is to be noted at a and b, the value of l at c and d. Then,
where g is the acceleration due to gravity at that place. Thus, g can be determined at a place from these measurements using the following relation:
10. Two simple pendulums A and B have equal lengths, but their bobs weigh 50 gf and 100 gf, respectively. What would be the ratio of their time periods? Give reasons for your answer.
Solution 10.
The ratio of their time periods would be 1:1 because the time period does not depend on the weight of the bob.
11. Two simple pendulums A and B have lengths 1.0 m and 4.0 m, respectively, at a certain place. Which pendulum will make more oscillations in 1 min? Explain your answer.
Solution 11.
Pendulum A will take more time (twice) in a given time because the time period of oscillation is directly proportional to the square root of the length of the pendulum. Therefore, the pendulum B will have a greater time period and thus making lesser oscillations.
12. State how does the time period of a simple pendulum depend on (a) length of pendulum, (b) mass of bob, (c) amplitude of oscillation and (d) acceleration due to gravity.
Solution 12.
(a) The time period of oscillations is directly proportional to the square root of the length of the pendulum.
(b) The time period of oscillations of simple pendulum does not depend on the mass of the bob.
(c) The time period of oscillations of simple pendulum does not depend on the amplitude of oscillations.
(d) The time period of oscillations of simple pendulum is inversely proportional to the square root of acceleration due to gravity.
13. What is a second's pendulum?
Solution 13.
A pendulum with the time period of oscillation equal to two seconds is known as a seconds pendulum.
14S.State the numerical value of the frequency of oscillation of a second's pendulum. Does it depend on the amplitude of oscillations?
Solution 14.
The frequency of oscillation of a seconds’ pendulum is 0.5 s-1. It does not depend on the amplitude of oscillation.
1. If the length of a simple pendulum is made one-fourth, then its time period becomes
1. Four times
2. One-fourth
3. Double
4. Half.
1. Four times
2. One-fourth
3. Double
4. Half.
Solution 1.
4. Half
2. The time period of a pendulum clock is
1. 1 s
2. 2 s
3. 1 min
4. 12 h
1. 1 s
2. 2 s
3. 1 min
4. 12 h
Solution 2.
2.2 s
3. The length of a second's pendulum clock is :
a. 0.5 m
b. 9.8 m
c. 1.0m
d. 2.0 m
Solution 3.
1. A simple pendulum completes 40 oscillations in a minute.
Find its (a) Frequency and (b) Time period.
Solution 1N.
(a) Frequency = Oscillations per second
= (40/60) s-1
= 0.67 s-1
(b) Time period = 1/frequency
= (1/0.67) s
= 1.5 s
2. The time period of a simple pendulum is 2 s. What is its frequency?
Solution 2.
Time period = 2 s
Frequency = 1/time period
= (½)s-1
= 0.5 s-1
Such a pendulum is called the seconds’ pendulum.
3. A seconds pendulum is taken to a place where acceleration due to the gravity falls to one-fourth. How is the time period of the pendulum affected, if at all? Give reasons. What will be its new time period?
Solution 3.
4. Find the length of a second's pendulum at a place where g = 10 m s-2 (Take p = 3.14)
Solution 4.
5. Compare the time periods of the two pendulums of lengths 1 m and 9 m, respectively.
Solution 5.
6. A pendulum completes two oscillations in 5 s. What is its time period? If g = 9.8 m s-2, find its length.
Solution 6.
7. The time periods of two simple pendulums at a place are in the ratio 2:1. What will be the corresponding ratio of their lengths?
Solution 7.
Let T1 and T2 be the time periods of the two pendulums of lengths l1 and l2, respectively.
Then, we know that the time period is directly proportional to the square root of the length of the pendulum.
8. It takes 0.2 s for a pendulum bob to move from the mean position to one end. What is the time period of the pendulum?
Solution 8.
Time period = Time taken to complete 1 oscillation
= (4 × 0.2) s
= 0.8 s
9. How much time does the bob of a second's pendulum take to move from one extreme to the other extreme of its oscillation?
Solution 9.
Time period of a seconds’ pendulum = 2 s
Time taken to complete half oscillation, i.e. from one extreme to the other extreme = 1 s.
28 / 3 / 4
28 / 3 / 4