MEASUREMENTS AND EXPERIMENTATION - Q&A
EXERCISE 1 (A)
1. What is meant by measurement ?
Measurement is the process of comparison of the given physical quantity with the known standard quantity of the same nature.
2. What do you understand by the term unit ?
The standard quantity used to measure the given physical quantity is called the unit. It is a constant magnitude used to measure the magnitudes of other quantities of the same nature.
3. What are the three requirements for selecting a unit of a physical quantity ?
The requirements for selecting a unit are:
1. The unit should be of convenient size.
2. It should be possible to define the unit without ambiguity.
3. The unit should be reproducible.
4. The value of unit should not change with space and time.
4. Name the three fundamental quantities.
The three fundamental quantities are Length, Mass, and Time.
5. Name the three systems of unit and state the various fundamental units in them.
The three systems of units are:
1. C.G.S. system (French system): Length (centimetre), Mass (gram), Time (second).
2. F.P.S. system (British system): Length (foot), Mass (pound), Time (second).
3. M.K.S. system (Metric system): Length (metre), Mass (kilogram), Time (second).
6. Define a fundamental unit.
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.
There are seven fundamental units in the S.I. system:
1. Length - metre (m)
2. Mass - kilogram (kg)
3. Time - second (s)
4. Temperature - kelvin (K)
5. Luminous intensity - candela (cd)
6. Electric current - ampere (A)
7. Amount of substance - mole (mol)
8. Explain the meaning of derived unit with the help of one example.
Derived units are those which depend on the fundamental units or which can be expressed in terms of the fundamental units.
Example: The unit of Area is derived from the fundamental unit of length. Area = length × breadth. Since both are measured in metres, the unit of area is metre × metre = m2.
9. Define standard metre.
Originally, a standard metre was defined as the distance between two marks drawn on a platinum-iridium rod kept at 0°C at Sevres near Paris.
In 1983, it was re-defined in terms of the speed of light: One metre is the distance travelled by light in air (or vacuum) in 1/299,792,458 of a second.
10. Name two units of length which are bigger than a metre. How are they related to the metre?
Two units of length bigger than a metre are:
1. Kilometre (km): 1 km = 1000 m = 103 m.
2. Astronomical Unit (A.U.): 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.
Two units of length smaller than a metre are:
1. Centimetre (cm): 1 cm = 10-2 m.
2. Millimetre (mm): 1 mm = 10-3 m.
12. How is nanometre related to Angstrom?
1 nanometre (nm) = 10-9 m.
1 Angstrom (Å) = 10-10 m.
Therefore, 1 nm = 10 Å.
13. Name the three convenient units used to measure length ranging from very short to very long value. How are they related to the S.I. unit ?
1. For very short lengths: Angstrom (Å). 1 Å = 10-10 m.
2. For common lengths: Metre (m). (This is the S.I. unit itself).
3. For very long lengths (astronomical distances): Light year (ly). 1 ly = 9.46 × 1015 m.
14. Name the S.I. unit of mass and define it.
The S.I. unit of mass is the kilogram (kg).
Definition: One kilogram is defined as the mass of a specific cylindrical piece of platinum-iridium alloy kept at the International Bureau of Weights and Measures at Sevres near Paris. Alternatively, it is the mass of 1 litre of water at 4°C.
15. Complete the following:
(a) 1 light year = 9.46 × 1015 m
(b) 1 m = 1010 Å
(c) 1 m = 106 μ
(d) 1 micron = 10,000 Å
(e) 1 fermi = 10-15 m
16. State two units of mass smaller than a kilogram. How are they related to the kilogram ?
1. Gram (g): 1 g = 10-3 kg.
2. Milligram (mg): 1 mg = 10-6 kg.
17. State two units of mass bigger than a kilogram. Give their relationship with the kilogram.
1. Quintal: 1 quintal = 100 kg.
2. Metric tonne: 1 metric tonne = 1000 kg.
18. Complete the following:
(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 × 10-27 kg
19. Name the S.I. unit of time and define it.
The S.I. unit of time is the second (s).
Definition: One second is defined as the time interval of 9,192,631,770 vibrations of radiation corresponding to the transition between two hyperfine levels of the ground state of the cesium-133 atom.
20. Name two units of time bigger than a second. How are they related to the second ?
1. Minute (min): 1 min = 60 s.
2. Hour (h): 1 h = 3600 s.
21. What is a leap year ?
A leap year is a year in which the month of February has 29 days instead of 28. It has a total of 366 days. It occurs every fourth year (except for century years not divisible by 400).
22. 'The year 2016 will have February of 29 days'. Is this statement true ?
Yes, this statement is true. 2016 is divisible by 4, so it is a leap year.
23. What is a lunar month?
One lunar month is the time of one lunar cycle (phases of the moon) as seen from the earth. It is approximately 29.5 days.
24. Complete the following:
(a) 1 nano second = 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
(a) u (atomic mass unit): Mass
(b) ly (light year): Distance (or Length)
(c) ns (nanosecond): Time
(d) nm (nanometre): Length
26. Write the derived units of the following:
(a) speed (b) force (c) work (d) pressure.
(a) Speed: m s-1
(b) Force: kg m s-2 (or Newton)
(c) Work: kg m2 s-2 (or Joule)
(d) Pressure: kg m-1 s-2 (or Pascal)
27. How are the following derived units related to the fundamental units ?
(a) newton (b) watt (c) joule (d) pascal.
(a) Newton (Force) = mass × acceleration = kg × m s-2 = kg m s-2
(b) Watt (Power) = work / time = (kg m2 s-2) / s = kg m2 s-3
(c) Joule (Energy/Work) = force × distance = (kg m s-2) × m = kg m2 s-2
(d) Pascal (Pressure) = force / area = (kg m s-2) / m2 = kg m-1 s-2
28. Name the physical quantities related to the following units :
(a) km2 (b) newton (c) joule (d) pascal (e) watt
(a) km2: Area
(b) newton: Force
(c) joule: Energy (or Work)
(d) pascal: Pressure
(e) watt: Power
Numericals :
1. The wavelength of light of a particular colour is 5800 Å. Express it in (a) nanometre and (b) metre.
Given: Wavelength = 5800 Å
(a) In nanometre (nm):
1 Å = 0.1 nm (or 1 nm = 10 Å)
5800 Å = 5800 × 0.1 nm = 580 nm.
(b) In metre (m):
1 Å = 10-10 m
5800 Å = 5800 × 10-10 m = 5.8 × 103 × 10-10 m = 5.8 × 10-7 m.
2. The size of a bacteria is 1 μ. Find the number of bacteria in 1 m length.
Given: Size of 1 bacteria = 1 μ (micron) = 10-6 m.
Total length = 1 m.
Number of bacteria = Total length / Size of 1 bacteria
Number = 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.
Given: Distance = 5.6 × 1025 m
Speed of light = 3 × 108 m s-1
Time taken = Distance / Speed
Time = (5.6 × 1025) / (3 × 108)
Time = 1.87 × 1017 s.
4. The wavelength of light is 589 nm. What is its wavelength in Å?
Given: Wavelength = 589 nm.
Relation: 1 nm = 10 Å.
Wavelength in Å = 589 × 10 = 5890 Å.
5. The mass of an oxygen atom is 16.00 u. Find its mass in kg.
Given: Mass = 16.00 u.
Relation: 1 u = 1.66 × 10-27 kg.
Mass in kg = 16.00 × 1.66 × 10-27 kg
Mass = 26.56 × 10-27 kg = 2.656 × 10-26 kg.
6. It takes time 8 min for light to reach from the sun to the earth surface. If speed of light is taken to be 3 × 108 m s-1, find the distance from the sun to the earth in km.
Given: Time = 8 min = 8 × 60 s = 480 s.
Speed of light = 3 × 108 m s-1.
Distance = Speed × Time
Distance = (3 × 108) × 480 = 1440 × 108 m = 1.44 × 1011 m.
To convert to km (divide by 1000 or 103):
Distance = 1.44 × 1011 / 103 km = 1.44 × 108 km.
7. '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.
Meaning: It means that light from the star takes 8.33 minutes to reach the earth.
Calculation:
Time = 8.33 minutes = 8.33 × 60 seconds = 499.8 seconds (approx 500 s).
Speed of light = 3 × 108 m s-1.
Distance = Speed × Time
Distance = 3 × 108 × 499.8
Distance ≈ 1.5 × 1011 m.
EXERCISE 1 (B)
1. Explain the meaning of the term 'least count of an instrument' by taking a suitable example.
The least count of an instrument is the smallest measurement that can be taken accurately with it.
Example: An ordinary metre rule has divisions marked in cm and mm. The smallest division is 1 mm. Therefore, the least count of a metre rule is 1 mm or 0.1 cm.
2. A boy makes a ruler with graduations in cm on it (i.e., 100 divisions in 1 m). To what accuracy this ruler can measure? How can this accuracy be increased ?
There are 100 divisions in 1 m, so 1 division = 1/100 m = 1 cm.
The accuracy (least count) of this ruler is 1 cm.
To increase the accuracy, the size of the divisions must be decreased, for example by dividing each cm into mm (10 subdivisions).
3. A boy measures the length of a pencil and expresses it to be 2.6 cm. What is the accuracy of his measurement ? Can he write it as 2.60 cm?
The measurement is 2.6 cm, which has one decimal place. This implies the instrument used has a least count (accuracy) of 0.1 cm (or 1 mm).
No, he cannot write it as 2.60 cm because 2.60 cm implies an accuracy up to the second decimal place (0.01 cm), which his instrument does not possess.
4. Define least count of a vernier callipers. How do you determine it ?
The least count of a vernier callipers is the difference between the value of one main scale division and one vernier scale division.
Determination: L.C. = (Value of 1 main scale division) / (Total number of divisions on vernier scale).
5. Define the term 'Vernier constant'.
Vernier constant is another name for the least count of a vernier. It is the smallest length that can be measured accurately using the vernier scale.
6. When is a vernier callipers said to be free from zero error ?
A vernier callipers is said to be free from zero error if the zero mark of the vernier scale coincides exactly with the zero mark of the main scale when the two jaws are brought in contact.
7. What is meant by zero error of a vernier callipers? How is it determined ? Draw neat diagrams to explain it. How is it taken in account to get the correct measurement ?
Zero error: When the jaws of the vernier callipers are in contact, if the zero of the vernier scale does not coincide with the zero of the main scale, the instrument has a zero error.
Determination:
1. Positive Zero Error: If vernier zero is to the right of main scale zero. Error = + (coinciding vernier division × L.C.).
2. Negative Zero Error: If vernier zero is to the left of main scale zero. Error = - (Total vernier divisions - coinciding division) × L.C.
Correction: Correct reading = Observed reading - Zero error (with proper sign).
8. A vernier callipers has a zero error + 0.06 cm. Draw a neat labelled diagram to represent it.
(For the diagram: The zero of the vernier scale should be shown shifted to the right of the zero of the main scale such that the 6th division of the vernier coincides with a main scale division, assuming L.C. is 0.01 cm).
9. Draw a neat labelled diagram of a vernier callipers. Name its main parts and state their functions.
Main parts and functions:
1. Outside jaws: To measure external dimensions (length, diameter).
2. Inside jaws: To measure internal dimensions (inner diameter).
3. Strip: To measure depth.
4. Main Scale: To measure length up to 1 mm.
5. Vernier Scale: To measure length accurately up to 0.1 mm (or 0.01 cm).
10. State three uses of a vernier callipers.
1. To measure the length of a rod.
2. To measure the diameter of a sphere.
3. To measure the internal diameter of a hollow cylinder.
11. Name the two scales of a vernier callipers and explain, how is it used to measure a length correct up to 0.01 cm.
The two scales are the Main Scale and the Vernier Scale.
Method:
1. Note the main scale reading (a) to the left of the vernier zero.
2. Note the vernier division (p) that coincides with any main scale division.
3. Vernier reading = p × Least Count (0.01 cm).
4. Total length = Main scale reading + Vernier reading.
12. Describe in steps, how would you use a vernier callipers to measure the length of a small rod ?
1. Find the least count and zero error of the instrument.
2. Place the rod between the fixed and movable jaws.
3. Tighten the screw to hold the rod gently.
4. Note the main scale reading (division just before vernier zero).
5. Note the coinciding vernier scale division.
6. Calculate observed length = Main scale reading + (Vernier division × L.C.).
7. Subtract zero error (with sign) to get true length.
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.
(a) Outside jaws (J1, J2)
(b) Inside jaws
(c) Strip (T)
(d) Outside jaws
14. Explain the terms (i) pitch, and (ii) least count of a screw gauge. How are they determined ?
(i) Pitch: The distance moved by the screw along its axis in one complete rotation of its head. Determined by dividing the distance moved on the main scale by the number of rotations.
(ii) Least Count: The distance moved by the screw along the axis when the circular scale is rotated by one division. Determined by: L.C. = Pitch / Total number of divisions on circular scale.
15. How can the least count of a screw gauge be decreased ?
The least count can be decreased (accuracy increased) by:
1. Decreasing the pitch of the screw.
2. 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.
Main parts: Ratchet, Sleeve (Main scale), Thimble (Circular scale), Stud, Spindle, U-frame.
Functions: Ratchet holds the object gently. Sleeve carries the main scale. Thimble carries the circular scale. Stud and spindle hold the object.
17. State one use of a screw gauge.
To measure the diameter of a thin wire or the thickness of a paper.
18. State the purpose of ratchet in a screw gauge.
The ratchet helps in holding the object gently between the stud and the spindle. It prevents tightening the screw too much, which could deform the object or damage the threads.
19. What do you mean by zero error of a screw gauge? How is it accounted for?
Zero error occurs if the zero of the circular scale does not coincide with the base line of the main scale when the stud and spindle are in contact.
It is accounted for by subtracting the zero error (with proper sign) from the observed reading. True reading = Observed reading - Zero error.
20. A screw gauge has a least count 0.001 cm and zero error + 0.007 cm. Draw a neat diagram to represent it.
(For the diagram: The zero of the circular scale should be shown below the base line of the main scale such that the 7th division coincides with the base line).
21. What is backlash error? Why is it caused ? How is it avoided ?
Backlash error is the error where the screw does not move immediately on reversing the direction of rotation.
Cause: Wear and tear of the screw threads.
Avoidance: Rotate the screw in one direction only while taking measurements.
22. Describe the procedure to measure the diameter of a wire with the help of a screw gauge.
1. Determine Pitch and L.C. and note the zero error.
2. Place the wire between the stud and the spindle.
3. Turn the ratchet until it clicks.
4. Note Main scale reading (visible divisions on sleeve).
5. Note Circular scale reading (division coinciding with base line).
6. Observed Diameter = M.S.R. + (C.S.R. × L.C.).
7. Correct Diameter = Observed Diameter - Zero Error.
23. Name the instrument which can measure accurately 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.
(a) Screw gauge
(b) Screw gauge
(c) Vernier callipers
(d) Screw gauge (or Vernier callipers)
24. Which of the following measures a small length to a high accuracy: metre rule, vernier callipers, screw gauge ?
Screw gauge.
25. Name the instrument which has the least count:
(a) 0.1 mm (b) 1 mm (c) 0.01 mm.
(a) Vernier callipers
(b) Metre rule
(c) Screw gauge
Numericals :
1. A stop watch has 10 divisions graduated between the 0 and 5 s marks. What is its least count?
Total time interval = 5 s.
Number of divisions = 10.
Least Count = Total time / Number of divisions = 5 / 10 = 0.5 s.
2. A vernier has 10 divisions and they are equal to 9 divisions of main scale in length. If the main scale is calibrated in mm, what is its least count?
Value of 1 main scale division (x) = 1 mm.
Number of vernier divisions (n) = 10.
Least Count = x/n = 1 mm / 10 = 0.1 mm = 0.01 cm.
3. A microsocpe is provided with a main scale graduated with 20 divisions 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.
Value of 1 main scale division (x) = 1 cm / 20 = 0.05 cm.
Number of vernier divisions (n) = 50.
Least Count = x/n = 0.05 cm / 50 = 0.001 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 pencil ?
Observed reading = 1.4 mm = 0.14 cm.
Zero error = + 0.02 cm.
Correct reading = Observed reading - Zero error
Correct reading = 0.14 cm - (+ 0.02 cm) = 0.12 cm = 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 vernier scale is ahead of zero of main scale and 3rd division of vernier scale coincides with a main scale division. Find: (i) the least count and (ii) the zero error of the vernier callipers.
(i) Least Count:
1 M.S.D. = 1 mm. n = 10.
L.C. = 1 mm / 10 = 0.1 mm = 0.01 cm.
(ii) Zero Error:
Zero is ahead (right), so it's a positive error.
Coinciding division = 3.
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 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 cylinder.
(i) Least Count:
1 M.S.D. = 1 mm. n = 20.
L.C. = 1 mm / 20 = 0.05 mm = 0.005 cm.
(ii) Radius:
Main scale reading = 35 divisions = 35 mm = 3.5 cm.
Vernier reading = 4 × L.C. = 4 × 0.005 cm = 0.020 cm.
Diameter = 3.5 cm + 0.020 cm = 3.52 cm.
Radius = Diameter / 2 = 3.52 / 2 = 1.76 cm.
7. In a vernier callipers, there are 10 divisions on the vernier scale and 1 cm on the main scale is divided in 10 parts. While measuring a length, the zero of the vernier lies just ahead of 1.8 cm mark and 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 ?
1 M.S.D = 1/10 cm = 0.1 cm. n = 10.
L.C. = 0.1 / 10 = 0.01 cm.
(a) Observed Length:
Main scale reading = 1.8 cm.
Vernier reading = 4 × 0.01 = 0.04 cm.
Length = 1.8 + 0.04 = 1.84 cm.
(b) Correct Length:
Zero error = -0.02 cm.
Correct length = Observed - Error = 1.84 - (-0.02) = 1.84 + 0.02 = 1.86 cm.
8. While measuring the length of a rod with a vernier callipers, Fig. 1.14 below shows the position of its scales. What is the length of the rod ?
(Based on standard reading of such diagrams):
Main scale reading (before vernier zero) = 3.3 cm.
Coinciding vernier division = 6th.
Least count (standard) = 0.01 cm.
Reading = 3.3 + (6 × 0.01) = 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 screw gauge ?
Pitch = 0.5 mm.
Number of divisions = 100.
L.C. = Pitch / Divisions = 0.5 / 100 = 0.005 mm = 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 screw gauge ? (ii) What is the least count of the screw gauge?
(i) Pitch = Distance moved / Number of revolutions = 1 mm / 2 = 0.5 mm.
(ii) Least Count = Pitch / Number of divisions = 0.5 mm / 50 = 0.01 mm.
11. The pitch of a screw gauge is 1 mm and its circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on circular scale coincides with the base line. Find: (i) the least count, and (ii) the diameter of the wire.
(i) L.C. = 1 mm / 100 = 0.01 mm = 0.001 cm.
(ii) Main scale reading = 2 mm.
Circular scale reading = 45 × 0.01 mm = 0.45 mm.
Diameter = 2 mm + 0.45 mm = 2.45 mm = 0.245 cm.
12. When a screw gauge of least count 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 ?
(i) Observed Reading:
M.S.R = 1 mm.
C.S.R = 27 × 0.01 mm = 0.27 mm.
Diameter = 1.27 mm = 0.127 cm.
(ii) Correct Reading:
Zero error = + 0.005 cm.
Correct Dia = 0.127 cm - 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 rotations. When the flat end of the screw is in contact with the stud, the zero of circular scale lies below the base line and 4th division of 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.
(i) Pitch = 1 mm / 2 = 0.5 mm.
(ii) L.C. = 0.5 mm / 50 = 0.01 mm.
(iii) Zero Error:
Zero is below the base line, so positive error.
Coinciding division = 4.
Error = + (4 × 0.01 mm) = + 0.04 mm.
14. Fig. 1.15 below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on main scale when circular head is rotated once. Find: (i) pitch of the screw gauge, (ii) least count of the screw gauge, and (iii) the diameter of the wire.
(i) Pitch: Main scale division is usually 1 mm. If screw advances 1 div per rotation, Pitch = 1 mm.
(ii) L.C.: Assuming 50 divisions on circular scale (based on diagram showing 45, 40 etc). L.C. = 1/50 = 0.02 mm.
(iii) Diameter:
Main scale reading (visible) = 4 mm.
Circular scale coincidence = 47.
Reading = 4 + (47 × 0.02) = 4 + 0.94 = 4.94 mm.
15. A screw has a pitch equal to 0.5 mm. What should be the number of divisions on its head so as to read correct up to 0.001 mm with its help?
L.C. = Pitch / Number of divisions.
0.001 mm = 0.5 mm / n
n = 0.5 / 0.001 = 500 divisions.
EXERCISE 1 (C)
1. What is a simple pendulum ? Is the pendulum used in a pendulum clock simple pendulum ? Give reason to your answer.
A simple pendulum is a heavy point mass (bob) suspended from a rigid support by a massless and inextensible string.
No, the pendulum used in a clock is not a simple pendulum; it is a compound pendulum because it is a rigid body capable of oscillating about a horizontal axis, and the mass is distributed (not a point mass).
2. Define the terms: (i) oscillation, (ii) amplitude, (iii) frequency, and (iv) time period as related to a simple pendulum.
(i) Oscillation: One complete to and fro motion of the bob.
(ii) Amplitude: The maximum displacement of the bob from its mean position on either side.
(iii) Frequency: The number of oscillations made in one second.
(iv) Time period: The time taken to complete one oscillation.
3. Draw a neat diagram of a simple pendulum. Show on it the effective length of the pendulum and its one oscillation.
(Diagram should show a support, a string, and a bob. Effective length 'l' is from support to the center of the bob. One oscillation is the path A to B and back to A).
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.
Factors:
1. Effective length of the pendulum (l).
2. Acceleration due to gravity (g).
Relation: T = 2π√(l/g).
5. Name two factors on which the time period of a simple pendulum does not depend.
1. Mass of the bob.
2. Amplitude of oscillation (provided it is small).
6. How is the time period of a simple pendulum affected, if at all, in the following situations: (a) the length is made four times, (b) the acceleration due to gravity is reduced to one-fourth,
Formula: T ∝ √l and T ∝ 1/√g.
(a) If length becomes 4 times: T' ∝ √4l ⇒ T' = 2T. Time period gets doubled.
(b) If g becomes 1/4: T' ∝ 1/√(g/4) ⇒ T' ∝ 1/(0.5√g) ⇒ T' = 2T. Time period gets doubled.
7. How are the time period T and frequency f of an oscillation of a simple pendulum related ?
They are inversely related: f = 1/T.
8. How do you measure the time period of a given pendulum ? Why do you note the time for more than one oscillation ?
To measure the time period, we measure the total time 't' for a large number of oscillations (say 20) using a stopwatch. Then Time Period T = t/20.
We measure time for multiple oscillations to reduce the percentage error in starting and stopping the stopwatch, which improves accuracy since the time for one oscillation is very small.
9. How does the time period (T) of a simple pendulum depends 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)?
Dependence: T is proportional to the square root of length (T ∝ √l), or T2 ∝ l.
Graph: A graph of T2 (y-axis) vs l (x-axis) is a straight line passing through the origin.
Determining g: The slope of the T2 vs l graph is 4π2/g. Therefore, g = 4π2 / Slope.
10. Two simple pendulums A and B have equal length, but their bobs weigh 50 gf and 100 gf respectively. What would be the ratio of their time periods? Give reason for your answer.
Ratio: 1:1.
Reason: Time period does not depend on the mass or weight of the bob. Since lengths are equal, time periods are equal.
11. Two simple pendulums A and B have length 1.0 m and 4.0 m respectively at a certain place. Which pendulum will make more oscillations in 1 minute? Explain your answer.
Pendulum A will make more oscillations.
Reason: T ∝ √l. Length of B is 4 times length of A, so Time period of B is 2 times Time period of A (TB = 2TA). Since B takes longer to complete one oscillation, A will complete more oscillations in the same time.
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.
(a) Directly proportional to square root of length (T ∝ √l).
(b) Does not depend on mass.
(c) Does not depend on amplitude (for small swings).
(d) Inversely proportional to square root of g (T ∝ 1/√g).
13. What is a seconds' pendulum ?
A seconds' pendulum is a pendulum whose time period is exactly 2 seconds. It takes 1 second to move from one extreme to the other.
14. State the numerical value of the frequency of oscillation of a seconds' pendulum. Does it depend on the amplitude of oscillation ?
Time period T = 2 s.
Frequency f = 1/T = 1/2 = 0.5 Hz (or 0.5 s-1).
No, it does not depend on the amplitude.
Numericals:
1. A simple pendulum completes 40 oscillations in one minute. Find its (a) frequency, (b) time period.
(a) Frequency = Oscillations / Time = 40 / 60 s = 2/3 = 0.67 s-1.
(b) Time period T = 1/f = 1 / (2/3) = 1.5 s.
2. The time period of a simple pendulum is 2 s. What is its frequency? What name is given to such a pendulum ?
Frequency f = 1/T = 1/2 = 0.5 s-1.
Name: Seconds' pendulum.
3. A seconds' pendulum is taken to a place where acceleration due to gravity falls to one-fourth. How is the time period of the pendulum affected, if at all? Give reason. What will be its new time period ?
Effect: Time period increases (doubles).
Reason: T ∝ 1/√g. If g becomes g/4, T becomes proportional to 1/√(g/4) = 2/√g.
New Time period: Original T = 2 s. New T = 2 × 2 = 4 s.
4. Find the length of a seconds' pendulum at a place where g = 10 m s-2 (Take π = 3.14).
Formula: T = 2π√(l/g) ⇒ l = (T2 × g) / 4π2.
Given T = 2 s, g = 10 m s-2.
l = (22 × 10) / (4 × 3.142)
l = 40 / (4 × 9.8596) = 10 / 9.8596 ≈ 1.0142 m.
5. Compare the time periods of two pendulums of length 1 m and 9 m.
Ratio T1/T2 = √(l1/l2)
T1/T2 = √(1/9) = 1/3.
Ratio is 1:3.
6. A pendulum completes 2 oscillations in 5 s. (a) What is its time period? (b) If g = 9.8 m s-2, find its length.
(a) Time period T = Total time / oscillations = 5 / 2 = 2.5 s.
(b) Length l = (T2 × g) / 4π2
l = (2.52 × 9.8) / (4 × 3.142)
l = (6.25 × 9.8) / 39.438
l = 61.25 / 39.438 ≈ 1.55 m.
7. The time periods of two simple pendulums at a place are in ratio 2: 1. What will be the ratio of their length ?
T ∝ √l ⇒ T2 ∝ l.
l1/l2 = (T1/T2)2
l1/l2 = (2/1)2 = 4/1.
Ratio is 4:1.
8. It takes 0.2 s for a pendulum bob to move from mean position to one end. What is the time period of pendulum ?
Movement from mean to one end is 1/4th of an oscillation.
Time for 1/4 oscillation = 0.2 s.
Time Period (full oscillation) = 4 × 0.2 s = 0.8 s.
9. How much time does the bob of a seconds' pendulum take to move from one extreme of its oscillation to the other extreme ?
Time Period of seconds' pendulum = 2 s.
Moving from one extreme to other is half an oscillation.
Time = T/2 = 2/2 = 1 s.