Measurements and Experimentation Class-9 ICSE Concise Selina Publishers

Chapter-1 Measurements and Experimentation Revised Concise Physics ICSE Class-9

ICSEHELP On Youtube Free Update Click Here 

Measurements and Experimentation Class-9 ICSE Concise Selina Publishers Solutions Chapter-1. Step By Step ICSE Selina Concise Solutions of Chapter-1 Measurements and Experimentation  with Exercise-1(A), Exercise-1(B) and Exercise-1(C) including Numerical and MCQ Questions Solved . Visit official Website CISCE for detail information about ICSE Board Class-9.

Measurements and Experimentation Class-9 ICSE Concise Selina Publishers Chapter-1


Select Topics 

 Exercise-1(A),  

MCQ-1(A),

  Numericals-1(A),

 Exercise-1(B)

MCQ-1(B)

Numericals-1(B),

 Exercise-1(C),

MCQ-1(C) , 

Numericals-1(C),

  Note :-  Before Viewing Selina Concise Physics Solutions of Chapter-1 Measurements and Experimentation Class-9 . Read the whole chapter carefully and Solved all example with Numerical s of Exercise-1 Measurements and Experimentation Class-9. 

Important Point for Measurement and Experimentation 

International system of unit in  . Other commonly used system of unit- FPS and CGS ,Measurement using common instrument Vernier caliper ,  Micrometer and screw gauge for  length and simple pendulum for time .

Measurement of length using vernier caliper and micrometer screw gauge . least count leads count least to an increasing an accuracy least count in  Measurement and Experimentation , Vernier caliper  Screw gauge in Measurement and Experimentation.


Exe-1(A) Measurements and Experimentation Class-9 Selina Physics Solutions  

  Question 1

What is meant by measurement?

Answer 1

Measurement is the process of comparing a given physical quantity with a known standard quantity of the same nature.

Question 2

What do you understand by the term unit?

Answer 2

Unit is a quantity of constant magnitude which is used to measure the magnitudes of other quantities of the same manner.

Question 3

What are the three requirements for selecting a unit of a physical quantity?

Answer 3

The three requirements for selecting a unit of a physical quantity are

(i) It should be possible to define the unit without ambiguity.

(ii) The unit should be reproducible.

(iii) The value of units should not change with space and time.

Question 4

Define the three fundamental quantities.

Answer 4

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.

aand 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.

while S.I. unit of time (s): A second is defined as 1/86400th part of a mean solar day, i.e.

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

Question 5

Name the three systems of unit and state their various fundamental units.

Answer 5

Three systems of unit and their fundamental units:

(i) 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).

(ii) 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).

(iii) M.K.S. system (or metric system): In this system, the unit of length is meter (m), unit of mass is kilogramme (kg) and unit of time is second (s).

Question 6

Define a fundamental unit.

Answer 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.

Question 7

What are the fundamental units in S.I. system? Name them along with their symbols.

Answer 7

Fundamental quantities, units and symbols in S.I. system are

Quantity Unit Symbol
Length metre m
Mass kilogramme 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

Question 8

Explain the meaning of a derived unit with the help of one example

Answer 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.

Question 9

Define standard metre.

Answer 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 0oC in the International Bureau of Weights and Measures at serves near Paris.

Question 10

Name two units of length which are bigger than a metre. How are they related to the unit metre?

Answer 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

Question 11

Write the names of two units of length smaller than a metre. Express their relationship with metre.

Answer 11

Centimeter (cm) and millimeter (mm) are units of length smaller than a metre.

1 cm = 10-2 m

1 mm = 10-3 m

Question 12

How is nanometer related to Angstrom?

Answer 12

1 nm = 10

Question 13

Name three convenient units used to measure lengths ranging from very short to very long value. How are they related to each other?

Answer 13

Three convenient units of length and their relation with the S.I. unit of length:

(i) 1 Angstrom (Å) = 10-10 m

(ii) 1 kilometre (km) = 103 m

(iii) 1 light year (ly) = 9.46 × 1015 m

Question 14

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

Answer 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.

Question 15

Complete the following

(a) 1 light year = ………..m

(b) 1 m   = ………..Å

(c) 1 m   = ………..µ (micron)

(d) 1 micron = ………..Å

(e) 1 fermi = ………..m

Answer 15

(a) 1 light year = 9.46 × 1015 m

(b) 1 m = 1010

(c) 1 m = 106 µ (micron)

(d) 1 micron = 104

(e) 1 fermi = 10-15 m

Question 16

State two units of mass smaller than a kilogram. How are they related to the unit kilogramme?

Answer 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

Question 17

State two units of mass bigger than a kilogram. Give their relationship with the unit kilogramme.

Answer 17

The units ‘quintal’ and ‘metric tonne’ are two units of mass bigger than ‘kilogramme’.

1 quintal = 100 kg

1 metric tonne = 1000 kg

Question 18

Complete the following

(a) 1 g    = ……….kg

(b) 1 mg    = ……….kg

(c) 1 quintal  = ……….kg

(d) 1 a.m.u (or u) = ……….kg

Answer 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

Question 19

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

Answer 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.

1 s =   × one mean solar day

Question 20

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

Answer 20

Minute and  hours

Question 21

What is a leap year?

Answer 21

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

Question 22

The year 2020  will have February of 29 days’.

Is this statement true?

Answer 22

Yes, the given statement is true.

Question 23

What is a lunar month?

Answer 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

Question 24

Complete the following

(a) 1 nanosecond =………s.

(b) 1 µs   =………s.

(c) 1 mean solar day =………s.

(d) 1 year = ………s.

Answer 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

Question 25

Name the physical quantities which are measured in the following units

(a) u  (b) ly  (c) ns  (d) nm

Answer 25

(a) Mass (b) Distance (or length) (c) Time (d) Length

Question 26

Write the derived units of the following

(a) Speed  (b) Force

(c) Work  (d) Pressure.

Answer 26

(a) ms-1 (b) kg ms-2 (c) kg m2s-2 (d) kg m-1s-2

Question 27

How are the following derived units related to the fundamental units?

(a) Newton   (b) Watt

(c) Joule   (d) Pascal.

Answer 27

(a) kg ms-2 (b) kg m2s-3

(c) kg m2s-2 (d) kg m-1s-2

Question 28

Name the physical quantities related to the following units:

(a) km2 (b) Newton (c) Joule

(d) Pascal (e) Watt

Answer 28

(a) Area (b) Force (c) Energy

(d) Pressure (f) Power


MULTIPLE CHOICE TYPE (EXERCISE 1-A)

Concise Selina Publishers Chapter-1 Measurements and Experimentation Class-9 Solutions 

 Question 1

The fundamental unit present in the list mentioned below is

(a) Newton

(b) Pascal

(c) Hertz

(d) Second

Answer  1

(d) Second

Question 2

Which of the following unit is not a fundamental unit:

(a) Metre

(b) Litre

(c) Second

(d) Kilogramme

Answer 2

(b) Litre

Question 3

The unit of time is

(a) Light year

(b) Parsec

(c) Leap year

(d) Angstrom.

Answer 3

(c) Leap year

Question 4

1 Å is equal to

(a) 0.1 nm

(b) 10-10cm

(c) 10-8m

(d) 104 µ.

Answer 4

(a) 0.1 nm

Question 5

Light year (ly) is the unit of

(a) Time

(b) Length

(c) Mass

(d) None of these.

Answer 5

(b) Length


       Numericals  EXERCISE-1 (A) 

Solutions of Chapter-1 Measurements and Experimentation Class-9 Revised Concise Physics Selina Publishers

 Question 1

The wavelength of light of a particular colour is 5800 Å. (a) Express it in (i) nanometre and (ii) metre. (b) What is its order of magnitude in metre?

Answer 1

Wavelength of light of particular colour = 5800

(a)

(i) 1 A   = 10-1 nm

5800 A    = 58 00 × 10-1 nm

= 580 nm

(ii) 1 A  = 10-10 m

5800  A= 5800 × 10-10 m

= 5.8 × 10-7 m

(b) The order of its magnitude in metre is 10-6 m because the numerical value of 5.8 is more than 3.2.

Question 2

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

Answer 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

Question 3

The distance of a galaxy is 5·6 × 1025 m.

Assuming the speed of light to be 3 × 108 m s “. (i) Find the time taken by light to travel this distance and (ii) express its order of magnitude.

[Hint : Time taken = distance/speed   ]

Answer 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

(b) Order of magnitude = 100 × 1017 s = 1017 s

(This is because the numerical value of 1.87 is less than the numerical value 3.2)

Question 4

The wave length of light is 589 nm. What its wavelength in   ?

Answer 4

1 nm= 10 A°

589 nm = 10×589A°= 5890 A°

Question 5

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

Answer 5

1 amu= 1.66×10¯27 Kg

16 amu= 16×1.66×10‾27 Kg

Question 6

It takes time 8 min for light to reach form the sun to the earth surface . If speed of light is taken to be 3 × 108 m/s, find the distance from the sun to the earth in km.

Answer 6

1 Min= 60sec

8 Min= 8×60 = 480sec

distance = Speed×Time

= 3×108×480

=1.44×108 Km/s

Question 7

‘The distance of a star form the earth is 8.33 light minutes .’ What do you mean by this statements ? Express the distance in meter.

Answer 7

It means light take 8.33 Minute  to reach the star while the speed of light is 3×108 m/s.

Distance= Speed×Time

= 3×108 × 8.33×60

=1.5×1011Mitre


Exercise – 1(B)

Revised Concise Chapter-1 Measurements and Experimentation Class-9 Selina Solutions 

Question 1

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

Answer 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.

Question 2

A boy makes a scale with graduations in cm on it, i.e. 100 divisions in 1 m. To what accuracy can this scale measure? How can this accuracy be increased?

Answer 2

Total length of the scale = 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 scale can measure with an accuracy of 1 cm.

To increase the accuracy, the total number of divisions on the scale must be increased.

Question 3

A boy measures the length of a pencil by a metre rule and expresses it to be 2.6 cm. What is the least count of the metre rule? Can he write it as 2.60 cm?

Answer 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.

Question 4

Define the least count of vernier callipers. How do you determine it?

Answer 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. ThenValue of n divisions on vernier = (n – 1) x

Alternatively, value of 1 division on vernier =(n – 1) x/n

Hence,

Least count =

L.C. = (Value of one main scale division)/(Total no. of divisions on vernier callipers)

Value of one main scale division = 1 mm

Total no. of divisions on vernier = 10

Therefore, L.C. =1/10 cm=.1 cm=1mm

Question 5

Define the term ‘Vernier constant’.

Answer 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.

Question 6

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

Answer 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.

Question 7

What is meant by zero error of vernier callipers ? How is it determined? Draw neat diagrams to explain it. How is it considered to get the correct measurement?

Answer 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

(i) Positive zero error

(ii) Negative zero error

(i) 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 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

(ii) 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 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)

Question 8

A vernier callipers has a zero error of + 0.06 cm. Draw a neat labelled diagram to represent it.

Answer 8 Measurements and Experimentation Class-9 

 Measurements and Experimentation Class-9 Selina Physics Class-9 Chapter-1 Ans 8 exe 1B

Question 9

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

Answer 9

Diagram of vernier callipers:

 Measurements and Experimentation Class-9 vernier calipers figure with label

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.

Question 10

State three uses of the vernier callipers.

Answer 10

Three uses of vernier callipers are

(a) Measuring the internal diameter of a tube or a cylinder.

(b) Measuring the length of an object.

(c) Measuring the depth of a beaker or a bottle.

Question 11

Name the two scales of a vernier callipers and explain how it is used to measure length correct up to 0.01 cm.

Answer 11

Two scales of vernier calipers are

(a) Main scale

(b) 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.

Question 12

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

Answer 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.

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.)

Question 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.

Answer 13

(a)  Outside jaws

(b)  Inside jaws

(c)  Strip

(d)  Outer jaws

Question 14

Explain the terms (i) Pitch and (ii) Least count of a screw gauge. How are they determined?

Answer 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

Question 15 Measurements and Experimentation Class-9 

How can the least count of a screw gauge be decreased.

Answer 15

The least count of a screw gauge be decreased by

(i) Decreasing pitch

(ii) Increasing total number of division on circular scale.

Question 16

Draw a neat and labelled diagram of a screw gauge.

Name its main parts and state their functions.

Answer 16

Diagram of screw gauge:

Structure of Screwguage Class-9

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.

Question 17

State one use of a screw gauge.

Answer 17

A screw gauge is used for measuring diameter of circular objects mostly wires with an accuracy of 0.001 cm.

Question 18

State the purpose of ratchet in a screw gauge.

Answer 18

Ratchet helps to advance the screw by turning it until the object is gently held between the stud and spindle of the screw.

Question 19

What do you mean by zero error of a screw gauge? How is it accounted for?

Answer 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. so 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.

hence 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)

Question 20 Measurements and Experimentation Class-9 

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

Answer 20

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

Question 21

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

Answer 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.

Question 22

Describe the procedure to measure the diameter of a wire with the help of a screw gauge.

Answer 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.

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

Question 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.

Answer 23

(a) Screw gauge

(b) Screw gauge

(c) Vernier calipers

(d) Screw gauge

Question 24 Measurements and Experimentation Class-9 

Which of the following measures a small length with high accuracy: metre scale, vernier callipers or screw gauge?

Answer 24

Screw gauge measures a length to a high accuracy.

Question  25

Name the instrument which has the least count

(a) 0.1 mm (b) 1 mm (c) 0.01 mm.

Answer 25

(a) Vernier callipers (b) Metre scale (c) Screw gauge.


MCQ  Exe- 1 (B) 

Selina solutions of Chapter-1 Measurements and Experimentation Class-9 Physics

Question 1

The least count of a vernier calipers is :

(a) 1 cm

(b) 0.001 cm

(c) 0.1 cm

(d) 0.01 cm

Answer 1

The least count of a vernier calipers is 0.01 cm

Question 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

Answer 2

0.002 cm

Question 3

The diameter of a thin wire can be measured by

(a) A vernier calipers

(b) A metre rule

(c) A screw gauge

(d) Any of these.

Answer 3

A screw gauge



Numerical s, Exercise -1(B)

Revised Selina Physics solutions of Chapter-1 Measurements and Experimentation Class-9 

Question 1

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

Answer 1

Range of the stop watch = 5s

Total number of divisions = 10

L.C. = 5/10 = 0.5 s

Question 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?

Answer 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

Question 3

A microscope is provided with a main scale graduated with 20 divisions in 1 cm and a vernier scale with 50 division on it of length same as of 49 divisions of main scale. Find the least count of the microscope.

Answer 3

Numericals Chapter-1 Class-9 Physics

Thus, the least count of the microscope is 0.001 cm.

Question 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?

Answer 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

Question 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.

Answer 5 Measurements and Experimentation Class-9 

(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

Question 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.

Answer 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

and = (0.1/20) cm

hence = 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

Question 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 ?

Answer 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

Question 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?

…………………………

Answer 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

Question 9

The pitch of a screw guage is 0.5mm and the head scale is divided in 100 part what is the least count of screw guage.

Answer 9

Least count= Pitch/ Number of division

= 0.5/100 = 0.005 mm = 0.0005 cm

Question 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?

Answer 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

Question 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.

Answer 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

Question 12 Measurements and Experimentation Class-9 

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?

Answer 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

Question 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

Answer 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

Question 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 the 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.

Answer 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

Question 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?

Answer 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 Ex -1(C)

Chapter-1 Measurements and Experimentation Selina Concise Physics Solutions Class-9

Question 1

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

Answer 1 Measurements and Experimentation Class-9 

A simple pendulum is a heavy point mass (known as bob) suspended from a rigid support by a mass less and in extensible 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.

Question 2

Define the terms: (i) oscillation, (ii) amplitude, (iii) frequency and (iv) time period as related to a simple pendulum.

Answer 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).

Question 3

Draw a neat diagram of a simple pendulum.

Show the effective length and one oscillation of the pendulum.

Answer 3

Simple Pendulum:

Simple Pendulam

Question 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

Answer 4

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

(i) Length of pendulum (l)

(ii) Acceleration due to gravity (g)

T=2π √L /√g

Question 5

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

Answer 5

Two factors on which the time period of a simple pendulum does not depend are

(i) Material of the bob

(ii) Amplitude

Question 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.

Answer 6

T=2π √L /√g

We know that,

(a) If length quadruples then,

numericals class-9 Physics ans 6 Chapter-1

Therefore, the time period is doubled.

Question 7

How is the time period T and frequency f of a simple pendulum related to each other?

Answer 7

Time period of a simple pendulum is inversely proportional to its frequency.

f=1/t

Question 8 Measurements and Experimentation Class-9 

How do you measure the time period of a given pendulum? Why do you note the time for more than one oscillation?

Answer 8

Measurement of time period of a simple pendulum:

(i) 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.

(ii) The time ‘t’ for 20 complete oscillations is measured with the help of a stop watch.

(iii)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.

Question 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)?

Answer 9

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

 

ans 9 class-9 physics

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,

 

ans 10

Question 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.

Answer 10

The ratio of their time periods would be 1:1 because the time period does not depend on the weight of the bob.

Question 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

Answer 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.

Question 12 Measurements and Experimentation Class-9 

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.

Answer 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.

Question 13

What is a second’s pendulum?

Answer 13

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

Question 14

State the numerical value of the frequency of oscillation of a second’s pendulum. Does it depend on the amplitude of oscillations?

Answer 14

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


MCQ Exe -1(C) 

Measurements and Experimentation Solutions of Chapter-1 Concise Selina Physics Class-9

Question 1

The length of a simple pendulum is made one-fourth, then its time period becomes

(a)  Four times

(b) One-fourth

(c) Double

(d) Half.

Answer 1

Half

Question 2

The time period of a pendulum clock is

(a) 1 s

(b) 2 s

(c) 1 min

(d) 12 h

Answer 2

2 s

Question 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

Answer 3

The time period of a second’s pendulum is T = 2 s.

The time period is related to the length as…………

mcq ans 3 Chapter-1 concise

 

 

 

 


Numericals  Exe- 1 (C) 

 Chapter-1 Concise Selina Physics Class-9 Measurements and Experimentation Solutions 

Question 1

A simple pendulum completes 40 oscillations in a minute.

Find its (a) Frequency and (b) Time period.

Answer 1

(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

Question 2 Measurements and Experimentation Class-9 

The time period of a simple pendulum is 2 s. What is its frequency? What name is given to such a pendulum ?

Answer 2

Time period = 2 s

Frequency = 1/time period

= (½)s-1

= 0.5 s-1

Such a pendulum is called the seconds’ pendulum.

Question 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?

Answer 3

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

i.e. T∝ 1/√g

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,

ans 8

Question 4

Find the length of a second’s pendulum at a place where g = 10 m s-2 (Take π = 3.14)

Answer 4

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

Let ‘l’ be the length of the seconds’ pendulum.

ans 4 Concise Physics

Question 5

Compare the time periods of the two pendulums of lengths 1 m and 9 m, respectively.

Answer 5

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

ans 5

Question 6

A pendulum completes two oscillations in 5 s. What is its time period? If g = 9.8 m s-2, find its length.

Answer 6

Time period = Total time/total no. of oscillations

= (5/2) s

= 2.5 s

Let ‘l’ be the length. Then,

ans 6

Question 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?

Answer 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.

ans 7

Question 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?

Answer 8

Time period = Time taken to complete 1 oscillation

= (4 × 0.2) s

= 0.8 s

Question 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?

Answer 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.

Return to Concise Selina Physics ICSE Class-9 Solutions


Thanks

 

Please share with your friends

Leave a comment
error: Content is protected !!