Selina Chapter 1 Measurements and Experimentation Questions Answers Class 9 Physics

ICSE Solutions for Chapter 1 Measurements and Experimentation Class 9 Physics Selina Publishers

Exercise-1(A)

1. What is meant by measurement?

Solution

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

Unit is the quantity of a constant magnitude which is 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?

Solution

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. Name the three fundamental quantities.

Solution

The three fundamental quantities are:

1. Length
2. Mass
3. Time

 

5. Name the three systems of unit and state the various fundamental units in them.

Solution

The three systems of unit and the corresponding fundamental units are:

1. CGS system (French System): unit of length is centimeter (cm), unit of mass is Gram(g), unit of time is Second(s)
2. F.P.S system (British system): unit of length is Foot(ft), unit of mass is Pound(Ib), unit of ime is Second(s)
3. M.K.S system (Metric system): unit of length is Metre(m), unit of mass is Kilogramme(kg), unit of time is Second(s)

 

6. Define a fundamental unit.

Solution

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

Quantity	Unit	Symbol
Length	metre	m
Mass	kilogram	Kg
Time	second	S
Temperature	kelvin	K
Luminous intensity	candela	cd
Electric current	ampere	A
Amount of substance	mole	mol
Angle	radian	rd
Solid angle	steradian	St-rd

 

8. Explain the meaning of derived unit with the help of one example.

Solution

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

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 0° 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

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 x 10¹¹ m

 

11. Write the name of two units of length smaller than a metre. How are they related to the metre?

Solution

The two units of length smaller than a metre are:

• Angstrom (Å)
• Fermi (f)

Relation between metre (m) and Angstrom (Å) is:

1 Angstrom (Å) = 10^-10 metre

Relation between metre (m) and Fermi is:

1 fermi (f) = 10^-15 m

 

12. How is nanometer related to Angstrom?

Solution

Relation between nanometer (nm) and Angstrom (A):

1 nanometer = 10Å

 

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?

Solution

The 3 convenient units used to measure length ranging from very short to very long value are:

• Centimeter (cm)
• Metre (m)
• Kilometer (km)

Relation between meter (m) and the units are:

1 m = 100cm

1 km = 1000m

 

14. Name the S.I. unit of mass and define it.

Solution

S.I. unit of mass is ‘kilogram’.

In 1889, one kilogram 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.

 

15. Complete the following:

(a) 1 light year = ___ m

(b) 1 m = ___ Å

(c) 1m = ___ μ

(d) 1 micron = ___ Å

(e) 1 fermi ___ m

Solution

(a) 1 light year = 9.46 x10¹⁵ m

(b) 1 m = 10¹⁰ Å

(c) 1m = 10⁶ μ

(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 kilogram?

Solution

The units ‘gramme’ (q) 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 kilogram.

Solution

The units ‘quintal’ and ‘metric tonne’ are two units of mass Digger 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 μ) = ___ kg

Solution

(a) 1g = 10^-3kg

(b) 1 mg = 10^-6 kg

(c) 1 quintal = 100 kg

(d) 1 a.m.u (or μ) = 1.66 x 10^-27 kg

 

19. Name the S.I. unit of time and define it.

Solution

The S.I. unit of time is second (s).

A second is defined as 1/86400th part of a mean solar day, i.e.

1s = 1/86400 x one mean solar day.

 

20. Name two units of time bigger than a second. How are they related to second?

Solution

The units ‘minute’ (min) and ‘year’ (yr) are two units of time bigger than second(s).

1 min = 60s

1 yr = 3.1536 x 10⁷s

 

21. What is a leap year?

Solution

A leap year is the year in which the month of February has 29 days and total number of days in leap year is 366 days.

 

22. ‘The year 2020 will have February of 29 days’. Is this statement true?

Solution

Yes, the given statement is true.

 

23. What is a lunar month?

Solution

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 nano second = ___ s

(b) 1 μs = ___ s

(c) 1 mean solar day = ____ s

(d) 1 year = ____ s

Solution

(a) 1 nanosecond = 10^-9 s

(b) 1 μs = 10^-6 s

(c) 1 mean solar day = 86400 s

(d) 1 year = 3.15 x 10⁷s

 

25. Name the physical quantities which are measured in the following units:

(a) μ

(b) ly

(c) ns

(d) nm

Solution

(a) Mass

(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

The derived units for the following:

(a) Speed: ms^-1

(b) Force: kg ms^-2

(c) Work: kg m²s^-2

(d) Pressure: 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

(a) Newton: kg ms^-2

(b) Watt: kg m²s^-3

(c) Joule: kg m² s^-2

(d) Pascal: 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

The physical quantities related to the following units are:

(a) km² - area

(b) newton - force

(c) joule - energy

(d) pascal - pressure

(e) watt – power

 

Multiple Choice Type

1. The fundamental unit is

(a) newton

(b) pascal

(c) hertz

(d) second

Solution

(d) second

Second is a fundamental unit, Some other fundamental units are meter (m), kilogram(kg).

 

2. Which of the following unit is not a fundamental unit:

(a) metre

(b) litre

(c) second

(d) kilogram

Solution

(b) litre

Litre is a unit of volume, which is a derived physical quantity.

 

3. The unit of time is:

(a) light year

(b) parsec

(c) leap year

(d) angstrom

Solution

(c) leap year

A leap year is the year in which the month of February is of 29 days.

 

4. 1 Å is equal to:

(a) 0.1 nm

(b) 10^-10 cm

(c) 10^-8 m

(d) 10^-4 μ

Solution

(a) 0.1 nm

1m=10 Å

 

5. ly is the unit of:

(a) time

(b) length

(c) mass

(d) none of these

Solution

(b) length

ly is the short for light year, which is a unit of distance or length.

 

Numericals

1. The wavelength of light of a particular colour is 5800 A. Express it in (a) nanometer and (b) metre

Solution

The wavelength of light of a particular colour is 5800 A.

(a) 1 nm=10 Å

5800 Å = 5800/10 = 580 nm

(b) 1 m=10¹⁰ Å

5800 A = 5800/10 = 5.8 x 10^-7 m

 

2. The size of a bacteria is 1 μ. Find the number of bacteria in 1m length.

Solution

Size of bacteria is 1 μ

1 μ =10^-6m

Number of bacteria in 1 m = 1/10^-6

Number of bacteria in 1 m length = 10⁶ bacteria

 

3. The distance of a galaxy from the earth is 5.6 x 10²⁵m. Assuming the speed of light to be 3 x 10⁸ ms^-1 find the time taken by light to travel this distance.

[Hint: Time taken = distance travelled/speed]

Solution

Given:

distance = 5.6 x 10²⁵m

speed = 3 x 10⁸ ms^-1

Time =?

 

4. The wavelength of light is 589nm. What is its wavelength in Å?

Solution

Wavelength of light = 589 nm

= 589 x 10^-9m

= 5.89 x 10^-7m

Order of magnitude = 10¹ x 10^-7 m

=10^-6m

 

5. The mass of an oxygen atom is 16.00 u, Find its mass in kg.

Solution

Mass of an oxygen atom = 16.00 μ

Now, 1 μ = 1.66 x 10^-27 kg

Hence, mass of oxygen in kg = 16 x 1.66 x 10^-27 kg

= 26.56 x 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.

= 10 x 10^-27 kg

= 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 x 10⁸ ms^-1, find the distance from the sun to the earth in km.

Solution

Time taken by light to reach from the Sun to the Earth = 8 min = 480 s.

Speed of light = 3 x 10⁸ m/s

Distance from the Sun to the Earth = Speed x time

=3 x 10⁸ x 480m

= 1440 x 10⁸m

= 1440 x 10⁸ x 10^-3 km

= 1440 x 10⁵ km

= 1.44 x 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 x 10⁸ km = 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.

Solution

The distance of a star from the earth is 8.33 light minutes means that it takes 8.33 minutes for light to reach the earth from the ultimate source of light — the Sun. The distance is large, hence light year is used.

Speed of light = 3 x 10⁸ m/s

Time = 8.33 min = 499.8s

Distance = speed x time

= 3x 10⁸ x 499.8 =1.5 x 10¹¹ m

Exercise-1(B)

1. Explain the meaning of the term ‘least count of an instrument’ by taking a suitable example.

Solution

The least count of an instrument is the smallest measurement which can be accurately taken from that particular instrument.

Example- if there are 10 divisions between 0 and 5s mark of a stop watch, the least count of that stop watch is 0.5s.

 

2. A boy makes a ruler with graduation in cm on it (i.e., 100 divisions in 1 m). To what accuracy this ruler can measure? How can this accuracy be increased?

Solution

Given: ruler has 100 divisions — centimeter scaling system

100 cm = 1 m

Hence, the accuracy the ruler can measure up to is the centimeter division.

∴ the ruler can be used to measure the length up to the accuracy of centimeter

In order to increase the accuracy, the scale must further be able to measure the next unit in-line, i.e., the millimeter division. By doing so the accuracy of the ruler can increase from centimeter to millimeter.

Hence,

1 m = 100 cm = 10mm

1 m= 1000 mm

 

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?

Solution

Given: The length of the pencil is 2.6cm.

We know that length can be expressed in different units such as metre, centimeter, millimeter etc.

Hence, we can say that the measurement may be accurate but not precise enough,

The boy can write 2.6cm or can express it as 2.60cm, both are the same as the value of zero here is not significant.

 

4. Define least count of a vernier calipers. How do you determine it?

Solution

Vernier constant or least count of a Vernier calipers 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 calipers.

It can be determined using the formula:

Least count (L.C.) = Value of one main scale division/total number of divisions on Vernier

 

5. Define the term ‘Vernier constant’.

Solution

Vernier constant can be defined as the difference between the values of one main scale division and one vernier scale division.

 

6. When is a vernier calipers said to be free from zero error?

Solution

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 a vernier calipers? How is it determined? Draw neat diagrams to explain it. How is it taken in account to get the correct measurement?

Solution

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.

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 calipers, gives the zero error.

For example, for the scales shown, the least count is 0.01 cm and the 6^th division of the vernier scale coincides with a main scale division,

Zero error = +6 x L.C, = +6 x 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.

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 calipers is 10.

Zero error = = (10 - 6) x L.C.
= -4x 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 calipers has a zero error +0.06cm. Draw a neat labelled diagram to represent it.

Solution

L.C. = 0.01cm

As per the scale readings, main scale reading = 3.3mm

The 6^th vernier division coincides with a main scale division

Vernier scale reading = 6 x 0.01 cm = 0.06cm

Total reading = m.s.r + v.s.r = 3.3 + 0.06 = 3.36cm

9. Draw a neat labelled diagram of a vernier calipers. Name its main parts and state their functions.

Solution

 

Main parts of Vernier calipers 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 a vernier calipers.

Solution

Three uses of vernier calipers 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 calipers and explain, how it is used to measure a length correct up to 0.01em.

Solution

Two scales of a vernier calipers are:

• Main scale - can read up till 1 mm
• Vernier scale - length of 10 divisions is equal to length of 9 divisions on the main scale

The value of one division on the main scale is 1 mm.

Total number of divisions on the vernier scale is 10

∴ Least count = 1 mm/10 = 0.1mm = 0.01 cm

Consequently, a vernier calipers can be used to measure a length accurately up to 0.01cm.

 

12. Describe in steps, how would you use a vernier calipers to measure the length of a small rod?

Solution

Measuring the length of a small rod using vernier calipers:

1. The rad whose length is to be measured is placed in between the fixed end and the vernier scale as shown in the figure.
2. 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).
3. To measure the length ab, we note the pth division of the vernier scale, which coincides with any division of main scale.
4. Now, ab + length of p divisions on vernier scale = length of p divisions on main scale
5. 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 x Lc.

Therefore, total reading = main scale reading + vernier scale reading = 1.2 cm + (p x L.C.)

 

13. Name the part of the vernier calipers 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

(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?

Solution

(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?

Solution

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 labelled diagram of a screw gauge. Name its main parts and state their functions.

Solution

The main parts of screw gauge and their functions are:

1. Ratchet: advances the screw by turning it till the object to be measured in held gently in between the spindle of the screw and the stud.
2. Sleeve: it notes the base line and the base line
3. Thimble: circular scale is marked by the thimble
4. Circular scale: reads length correctly to 0.01mm
5. Main scale: reads length to 1 mm

 

17. State one use of a screw gauge.

Solution

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

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

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 ta get the correct reading.

Correct reading = Observed reading - zero error (with sign)

 

20. A screw gauge has a least count 0.001cm and zero error + 0.007cm. Draw a neat diagram to represent it.

Solution

Diagram of a screw gauge with L.C. 0.001 cm and zero error +0.007 cm.

21. What is backlash error? Why is it caused? How is it avoided?

Solution

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

Measurement of diameter of wire with a screw gauge:

1. 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.
2. The rachet is provided to avoid excessive pressure on the wire. It prevents the spindle from further movement.
3. The thickness of the wire could be determined from the reading as shown in the figure below.

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 x 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 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

Solution

(a) Screw gauge

(b) Screw gauge

(c) Vernier calipers

(d) Screw gauge

24. Which of the following measures a small length to a high accuracy: metre rule, Vernier calipers, screw gauge?

Solution

The screw gauge measures a small length to a high accuracy.

 

25. Name the instrument which has the least count:

(a) 0.1 mm

(b) 1 mm

(c) 0.01mm

Solution

(a) 0.1 mm: vernier callipers

(b) 1 mm: metre rule

(c) 0.01mm: screw gauge

 

Multiple Choice Type

1. The least count of a vernier calipers is:

(a) 1 cm

(b) 0.001 cm

(c) 0.1 cm

(d) 0.01 cm

Solution

(c) 0.01 cm

Least count is the smallest measurement that can be accurately taken with the instrument

 

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:

(a) 0.002 cm

(b) 0.001 cm

(c) 0.02 cm

(d) 0.01 cm

Solution

(a) 0.002 cm

The least count of any instrument is the smallest measurement that can be taken accurately

 

3. The diameter of a thin wire can be measured by:

(a) A vernier calipers

(b) A metre rule

(c) A screw gauge

(d) None of these

Solution

(c) A screw gauge

 

Numericals

1. A stop watch has 10 divisions graduated between the 0 and 5s marks. What is its least count?

Solution

Least count is the smallest value that can be measured by an instrument.

L.C = (5-0)/10 = 0.5s

 

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?

Solution

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
= 1mm/10
= 0.1 mm or 0.01 cm

 

3. A microscope 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.

Solution

1 main scale division = 1/20 cm
50 divisions of vernier scale = 49 divisions of main scale
1 division of vernier scale = 49 ÷ 50 main scale divisions = 0.98 main scale division
Vernier calipers:
L.C. = 1 main scale division - 1 vernier scale division
= 1 main scale division - 0.98 main scale division
= 0.02 main scale division
Microscope:
1 main scale division = 1/20 cm
L.C = 0.02 × 1/20 = 0.001 cm
∴ Least count of the microscope is 0.001cm

 

4. A boy uses a vernier calipers to measure the thickness of his pencil. He measures it to be 1.4mm. If the zero error of vernier calipers is +0.02cm, what is the correct thickness of pencil?

Solution

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)

=1.4mm-0.2 mm

=1.2mm

5. A vernier calipers has its main scale graduated in mm and 10 divisions on its vernier scale are equal in length to 9mm. When the two jaws are in contact, the zero of vernier scale is ahead of zero of main scale and 3^rd division of vernier scale coincides with a main scale division. Find: (i) the least count and (ii) the zero error of the vernier calipers.

Solution

(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
= 1mm/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 x L.C. = +3 x 0.01cm = +0.03 cm

 

6. The main scale of a vernier calipers 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 4" division of vernier scale coincides with a main scale division. Find: (i) least count and (ii) radius of cylinder.

Solution

(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
= Imm/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 x 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 a vernier calipers, 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 1.8cm mark and 4^th division of vernier coincides with a main scale division.

(a) Find the length
(b) If zero of vernier calipers is -0.02cm, what is the correct length?

Solution

(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 x 0.01 cm = 0.04 cm
Total reading = Main scale reading + vernier scale reading
= (1.8 + 0.04) cm
= 1.84cm

(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 calipers, Figure below shows the position of its scales. What is the length of the rod?

Solution

L.C. of vernier callipers = 0.01 cm

In the shown scale,

Main scale reading = 3.3 mm

6^th vernier division coincides with an m.s.d.

Therefore, vernier scale reading = 6 x 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 0.5mm and the head scale is divided in 100 parts. What is the least count of screw gauge?

Solution

Pitch of a screw gauge = 0.5mm

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 screw gauge?

(ii) What is the least count of the screw gauge?

Solution

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 its circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2mm and 45" mark on circular scale coincides with the base line. Find:

(i) The least count, and
(ii) The diameter of the wire

Solution

Pitch of the screw gauge = 1 mm
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) x 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 of least count 0.01mm is used to measure the diameter of a wire, the reading on the sleeve is found to be Imm 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 +0.005cm, what is the correct diameter?

Solution

(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) x L.C.
= 27 x 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 or 1.22 mm

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 4^th 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.

Solution

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 x L.C.)
= +(4 x 0.01) mm
= +0.04 mm

 

14. Figure 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.

Solution

No. of divisions on the circular scale = 50

(i) Pitch = Distance moved ahead in one revolution
=1mm/1= 1mm.

(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 x L.C.
= (47 x 0.02) mm
= 0.94 mm

Diameter (Total reading) = M.s.r. + circular scale reading
= (4+ 0.94) mm
=4.94mm

 

15. A screw has a pitch equal to 0.5mm. What should be the number of division on its head so as to read correct up to 0.001mm with its help?

Solution

Given: Pitch = 0.5mm

Number of divisions = 0.5mm/0.001mm = 500

 

Exercise-1(C)

1. What is a simple pendulum? Is the pendulum used in a pendulum clock simple pendulum? Give reason to your answer.

Solution

A simple pendulum is a heavy point mass (known as bob) suspended from a rigid support by a massless and inextensible string.

No, the pendulurn 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 (iv) time period as related to a simple pendulum

Solution

(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 on it the effective length of the pendulum and its one oscillation.

Solution

 

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

Two factors on which the time period of a simple pendulum depends are:

1. Length of pendulum (l)
2. Acceleration due to gravity (g)

Time period in terms of the above named factors are:

Time period is directly dependent on length with square root and inversely proportional to ‘g’ i.e. acceleration due to gravity with square root.

5. Name two factors on which the time period of a simple pendulum does not spend.

Solution

The time period of a simple pendulum does not spend on the following factors:

• Mass of the bob
• Material of the pendulum

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.

Solution

We know that,

(a)  If length quadruples then,

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 are the time period T and frequency f of an oscillation of a simple pendulum related?

Solution

The time period and frequency of an oscillation of a simple pendulum are related as stated below:

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?

Solution

In order to measure the time period of pendulum, total time taken by the pendulum should be divided by the number of oscillations.

 

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

The time period of a simple pendulum is directly proportional to the square root of its effective length.
T∝√l

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,

This slope is found to be constant at a place and is equal to 4π²/g

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 pendulum A and B have equal lengths, but heir bobs weigh 50 gf and 100 gf respectively. What would be the ratio of their time periods? Give reason for your answer.

Solution

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 1.0m and 4.0m respectively at a certain place. Which pendulum will make more oscillations in 1 minute? Explain your answer.

Solution

Pendulum A will take more time (twice) in a given time because the time period of oscillation is directly proportianal to the square root of the length of the pendulum ie. T∝√l. 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

(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 seconds’ pendulum?

Solution

A pendulum with the time period of oscillation equal to two seconds is known as a seconds’ pendulum.

 

14. State the numerical value of the frequency of oscillation of a seconds’ pendulum. Does it depend on the amplitude of oscillation?

Solution

The frequency of oscillation of a seconds’ pendulum is 0.5 s^-1. It does not depend on the amplitude of oscillation.

 

Multiple Choice Type

1. The length of a simple pendulum is made one-fourth. Its time period becomes:

(a) Four times
(b) One-fourth
(c) Double
(d) Half

Solution

(d) Half
Time period is directly proportional to the square root of the length of the pendulum.

 

2. The time period of a pendulum clock is:

(a) 1s
(b) 2s
(c) 1 min
(d) 12h

Solution

(b) 2s
Time period is found using the number of oscillations more than once as least count of stop watch is either is or 0.5s.

 

3. The length of a seconds’ pendulum is nearly:

(a) 0.5m
(b) 9.8m
(c) 1.0m
(d) 2.0m

Solution

(c) 1.0m
The time period of a second’s pendulum is T = 2s.
The time period is related to the length as

Numericals

1. A simple pendulum completes 40 oscillations in one minute. Find its (a) frequency, (b) time period.

Solution

(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 2s. What is its frequency? What name is given to such a pendulum?

Solution

Time period = 2s

Frequency = 1/time period
= (1/2) 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 gravity falls to one-forth. How is the time period of the pendulum affected, if at all? Give reason. What will be its new time period?

Solution

Time period of ‘a’ is inversely proportional to the square root of acceleration due to gravity.
i.e. 

Now, if the acceleration due to gravity falls to one-fourth, the time period will be doubled.

Let the new time period be ‘T’ and let g’ be the acceleration due to gravity.

Then,

4. Find the length of a seconds’ pendulum at a place where g=10ms^-2 (Take π =3.14).

Solution

Given, g= 10 m/s² and time period (T) = 2s

Let 'T’ be the length of the seconds’ pendulum,

We know that,

or, l = 1.0142 s

5. Compare the time periods of two pendulums of length 1m and 9m.

Solution

Let T₁ and T₂ be the time periods of the two pendulums of lengths 1m and 9m, respectively.

6. A Pendulum completes 2 oscillations in 5s.
(a) What is its time period?
(b) If g=9.8 ms^-2, find its length.

Solution

Given,

oscillations = 2

(a) Time period = Total time/total no. of oscillations
= (5/2)s
=2.55s

(b) Let ‘l’ be the length. Then,

or, l = 1.55 m

7. The time periods of two simple pendulums at a place are in the ratio 2:1. What will be the ratio of their lengths?

Solution

We know that time period of pendulum is:

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 0.2s for a pendulum bob to move from mean position to one end. What is the time period of pendulum?

Solution

Time taken to complete one oscillation is the time period

= (4 x 0.2)s
= 0.85

9. How much time does the bob of a seconds’ pendulum take to move from one extreme of its oscillation to the other extreme?

Solution

We know that the time period of a seconds’ pendulum is 2seconds

∴ The time taken for a seconds’ pendulum to make half oscillation is 1/2 x 2 seconds = 1 second.