# Selina Concise Class 9 Physics Chapter 1 Measurements and Experimentation Solution

## Selina Concise Class 9 Physics Solution Chapter No. 1- ‘Measurements and Experimentation’ For ICSE Board Students.

Exercise: 1(A) Solution

1.) Solution:

Ans:

Measurement is the basic requirement for measuring the physical quantities.

Measurement is the process in which given physical quantity is compared with the known standard quantity having same nature.

For example:

If the length of the wire is 14meter long then it means that the length of wire is measured in meter and it is 14 times in length of the given part of wire.

2.) Solution:

Ans:

While measuring the physical quantities we requires units. Unit is the quantity of constant magnitude which can be used to measure the magnitude of other physical quantities of the same nature.

The SI system of units is universally accepted, according to which length is measured in meter, mass in kilogram, time in second, temperature in Kelvin, luminous intensity in candela, electric current in ampere and amount substance in mole.

3.) Solution:

Ans:

The units which are used to measure physical quantities must be universally accepted and that are used over all the world at all times. So, the units used for measuring the physical quantities should follow following properties:

• The unit must be of convenient size.
• The unit should be defined without ambiguity.
• It must be reproducible.
• And the value of unit should not be changed with time and space i.e. it remains same everywhere.

If above all properties are satisfied by a particular unit then it is universally accepted.

4.) Solution:

Ans:

According to SI system, the following are the fundamental quantities with their standard units.

• Length which is measured in meter.
• Mass which is measured in kilogram.
• Time which is measured in second.

5.) Solution:

Ans:

In mechanics, length, time and mass are the fundamental quantities for which following system of units are developed.

1) C.G.S. system or French system:

In this system of unit, the length is measured in centimetre (cm), mass in gram (gm) and time in second (s).

2) F.P.S. system or British system:

In this system, length is measured in foot (ft), mass is measured in pond (lb) and time is measured in second (s).

3) M.K.S. system or Metric system:

In this system of units, the length is measured in meter (m), mass in kilogram (kg) and time is measured in second (s).

6.) Solution:

Ans:

Fundamental unit is the basic unit which doesn’t depends on any other quantity for its measurement and which cannot be changed or cannot be related to any other fundamental quantities.

For example:

Length, mass, time, temperature, electric current and amount of substance are the fundamental quantities having fundamental units which cannot depends on any other physical quantities for their measurement.

7.) Solution:

Ans:

According to SI system, the units of fundamental quantities length, mass, time, temperature, luminous intensity, current and amount of substance are considered as fundamental units and the units of angle and solid angle are considered as complementary fundamental units.

Thus, in SI system there are seven fundamental quantities, two complementary fundamental quantities with their fundamental units are explained as below.

• Length is measured in meter (m)
• Mass in kilogram (kg)
• Time in second (s)
• Temperature in Kelvin (K),
• Electric current in ampere (A)
• Amount of substance in mole (mol*)
• Luminous intensity in candela (cd)
• Solid angle in steradian (st-rd)

8.) Solution:

Ans:

Derived units are the units which depends on the fundamental units for their measurement i.e. derived units are expressed in terms of fundamental units.

For example:

If we have to measure the area of rectangle, then we must know length and it’s breadth in meters which are fundamental units and then we find the area of rectangle as area= length*breadth which is measured in meter2.

Similarly, if we have measure volume we have to know length, breadth and height and hence volume = length* breadth*height.

Thus, volume is measured in meter3.

Thus, derived units are expressed in terms of fundamental units.

9.) Solution:

Ans:

Meter is defined in terms of speed of light, according to which one meter is the distance travelled by the light in air or vacuum during the time interval of 1/299,792,458 of a second.

10.) Solution:

Ans:

The units which are bigger than meter are astronomical unit, light year and parsec.

This units are used to measure the distance between two heavenly bodies.

1) astronomical unit:

The value of one astronomical unit is the mean distance between the earth and the sun and it is given in terms of meter as,

1A.U. = 1.496*1011 meter

2) Light year:

A light year is the distance travelled by the light in vacuum in one year and it is given by,

1 light year= speed of light*time in 1 year

1 light year = 3*108 m/s*(365*24*60*60) s

1 light year= 9.46*1015 m

11.) Solution:

Ans:

For small length measurement the units micrometre and nanometres are used.

1) Micrometre:

Micrometre is one millionth part of meter and denoted by u.

It is also called as micron and it is related with meter as,

1micron = 10-6 meter

=10-4 cm

= 10-3 mm

2) Nanometres:

It is the one billionth part of a meter and it is expressed in terms of meter as follows.

1 nanometer = 10-9m

12.) Solution:

Ans:

Nanometre is the one billionth part of a meter and it is expressed as,

1 nanometer = 10-9m

And, angstrom is the 10-10th part of meter which is expressed as,

1angstrom A0= 10-10m

= 10-8cm

1angstrom = 10-1nm

And hence, 1nm = 10A0

In this way, Nanometre is expressed in terms of angstrom.

13.) Solution:

Ans:

The length is measured in meter in SI units.

While millimetre and nanometres are the small length units and they related to meter as follows.

1mm = 10-3m = 10-1cm

And, 1nm = 10-9m = 10-4cm = 10-3mm

For measuring the distance between two heavenly bodies large length units are measured they are astronomical unit and light year which are expressed in terms of meter as follows.

1 A. U. = 1.496*1011m

And,

1 light year= speed of light*time 1 year

= 3*108m/s*(365*24*69*60) s

=9.46*1015m

14.) Solution:

Ans:

The SI unit of mass is kilogram.

It is the mass of 1 liter =1000ml of water at 40C.

15.) Solution:

a) 1 light year = 9.46*1015m

b) 1 m = 1010 Ao

c) 1 m = 106 u

d) 1 micron = 104 Ao

e) 1 fermi = 10-15 m

16.) Solution:

Ans:

The smaller units used to measure mass are gram and milligrams.

1) gram (g):

1gram is the one thousandth part of a kilogram.

That is, 1g =10-3kg

1kg = 1000g

2) milligrams (mg):

1 mg is the one millionth part of a kilogram or one thousandth part of a gram.

That is, 1mg = 10-6 kg

1mg = 10-3g

16.) Solution:

Ans:

The bigger units used to measure mass are quintal and metric tonne.

1) quintal:

1 quintal is one hundred times a kilogram.

That is, 1 quintal = 100kg

2) metric tonne:

Metric tonne is one thousand times a kilogram.

That is, 1 metric tonne= 1000kg = 10 quintal

18.) Solution:

a) 1g = 10-3kg

b) 1 mg= 10-6kg

c) 1 quintal= 100kg

d) 1 a.m.u. = 1.66*10-27 kg

19.) Solution:

Ans:

The SI unit of time is second and it is denoted by s.

1 second is defined as 1/86400th part of a mean solar day.

That is,

1s= 1/86400 * one mean solar day

Where, one solar day is the time required by the earth to complete one rotation about its own axis of rotation.

20.) Solution:

Ans:

There are many bigger units of time than a second that are minutes, hour, day, lunar month, month, year, leap year, decade, century and millennium.

1) Minute:

1 minute has 60 seconds.

That is 1min = 60 seconds

2) Hour:

1hour has 60 minutes.

That is, 1hr= 60 min= 60*60s= 3600 s

3) Day:

The time required by the earth to rotate about its own axis in order to complete one rotation is called as a day.

1day has 24 hours.

That is, 1day = 24hrs= 24*60*60= 86400 s

21.) Solution:

Ans:

Leap year is that year in which the February month has 29 days.

Hence, 1leap year= 366 days

22.) Solution:

Ans:

Leap year is that year in which the February month has 29 days.

Hence, 1 leap year= 366 days

And, the year which is completely divisible by 4 is the leap year.

As, 2020 is completely divisible by 4 hence it is the leap year.

23.) Solution:

Ans:

The Hindu and Muslim’s calendars are based on the phases of the moon as seen from the earth and hence one month is the time of one lunar cycle which is nearly 29.5 days.

And hence, the period of 12 lunar month is 354.37 days.

24.) Solution:

a) 1 nano second = 10-9 s

b) 1 us = 10-6 s

c) 1 mean solar day = 86400 s

d) 1 year = 3.5*107 s

25.) Solution:

a) u

Since, mass is measured in a.m.u. or in u.

b) ly

Since, distance or length is measured in light year or ly.

c) ns

Since, time is measured in nanoseconds or ns.

d) nm

Since, length is measured in nanometres or nm.

26.) Solution:

a) Speed

Since, speed = distance/time

Hence, speed is measured in meter per second or m/s.

b) Force

Since, force = mass*acceleration

Hence, force has unit kg m s-2.

c) Work

Since, work = force* displacement

Hence, work has unit kg m2 s-2.

d) pressure

Since, pressure = force/ area

Hence, pressure has unit kg m-1 s-2.

27.) Solution:

a) newton

Since, newton is the unit of force.

And, force = mass *acceleration

Hence, newton is related to fundamental unit as kg m s-2.

b) watt

Since, power has the unit watt.

And, power = work/ time

Hence, watt is related to fundamental unit as kg m2 s-3.

c) joule

Since, work or energy is measured in joule.

And work = force*displacement

Hence, joule is related to fundamental unit as kg m2 s-2.

d) pascal

Since, pressure is measured in pascal.

And, pressure= force/ area

Hence, pascal is related to fundamental unit as kg m-1 s-2

28.) Solution:

Ans:

a) km2

Since, area is measured in km2.

b) newton

Since, newton is the unit of force.

c) joule

Work or energy is measured in joule.

d) pascal

Pressure is measured in pascal.

e) watt

Power is measured in watt.

Multiple choice type:

1) The fundamental unit is

Ans: d) second

Explanation: Because all other units are derived units.

2) Which of the following is not the fundamental unit.

Ans: b) litre

Explanation: Because litre is not the fundamental unit.

3) The unit of time is

Ans: c) leap year

Explanation: Because leap year is the bigger unit of time which has 366 days.

4) 1 angstrom is equal to

Ans: a) 0.1 nm

Because, 1angstrom = 10-10 m

And, 1nm = 10-9 m

Hence, 1angstrom = 0.1 nm = 10-1 nm

5) ly is the unit of

Ans :b) length

Explanation: Because, ly means light year and it is the bigger unit of length used to measure the distance between two heavenly bodies.

Numerical:

1) Solution:

Ans:

Given that, the wavelength of light of particular colour is 5800A0.

We know that, 1Ao = 10-10m

And 1Ao = 10-1nm

Thus,

a) 5800 Ao = 5800*10-1nm = 580nm

b) 5800Ao = 5800*10-10m = 5.8* 10-7 m

2) Solution:

Ans:

Given that, the size of bacteria is 1u.

And we know that, 1u = 10-6m

Hence, 1m = 106 u

Thus, the number of bacteria in one meter length will be 106.

3) Solution:

Ans:

Given that, the distance of galaxy from the earth is 5.6*1025m

And speed of light is 3*108m/s.

Thus, time required to travel this distance is given by,

Time= distance travelled/ speed of light

Time= 5.6*1025m/s*3*108m/s

Time= 1.87*1017 s

Thus, the time take by light to travel from galaxy to earth is 1.87*1017s.

4) Solution:

Ans:

We know that,

1angstrom = 10-10m = 10-1nm.

And, 1nm = 10-9m = 10 A0

Hence, 589nm = 589*10Ao = 5890 A0.

5) Solution:

Ans:

We know that, 1a.m.u. or 1u = 1.66*10-27 kg

Hence, 16 u = 16*1.66*10-27 kg = 2.656*10-26 kg

6) Solution:

Ans:

Given that, time required to reach the light from sun to earth’s surface is 8 minutes= 8*60 seconds = 480s

And speed of light is 3*108m/s

Thus, distance= speed of light*time

= 3*108m/s*480s

= 1440*108 m

= 1.44*103*108 m

= 1.44*108 km

7) Solution:

Ans:

Since, one light minute is the distance travelled by light in free space in one second and 1light minute = 1.8*1010 m

Hence, here the star is at a distance of 8.33 light minute means,

8.33 light minutes = 8.33*1.8*1010 m = 14.99*1010 m = 1.5*1011 m

Thus, the star is at a distance of 1.5*1011 m from the earth.

Exercise B Solution:

1.) Solution:

Ans:

For measuring the physical quantities the instruments are required like Verniercaliper is used for measuring length, thermometer for measuring temperature, ammeter for measuring current etc. All these instruments are having definite limit for accuracy measurement and that is expressed in least count of an instrument.

The least count is the smallest measurement which is accurately taken with instrument.

The least count of different measuring instruments is different.

For example:

• The least count of stop watch is 0.5 seconds only if there are 10 divisions between 0 and 5 s marks.
• The ammeter with 5 divisions between 0 and 1A mark is having least count as 0.2 A.
• Thus, if the smaller is the least count then the measurement will be more precise of that instrument.

2.) Solution:

Ans:

• The given ruler has length 1m and there are 100 divisions on it.
• Hence, 1m= 100cm
• To find the length of each division, we write as
• Length of one division on ruler = total length of ruler/ total no. of divisions on ruler
• Length of one division on ruler= 1m/ 100 = 0.01m

= 1 cm

• Thus, this ruler can measure the length with an accuracy up to 1cm.
• If we have to increase the accuracy of this ruler, we have to increase the total no. of divisions on ruler so that we can measure the length up to millimetre accurately.

3.) Solution:

Ans:

• We already know that, length is measured in meter, centimetre, millimetre also.
• Here, the length of pencil measured by meter scale is 2.6 cm, which may be accurate but not precise.
• And he can write 2.6cm as 2.60 cm also, as there is no difference in between 2.6 cm and 2.60 cm because the zero is written at the extreme end of decimal part which is not more significant.

4.) Solution:

Ans:

The least count of Vernier caliper is defined as, it is the difference between values of one main scale division and one Vernier scale division. And it is determined by using following equation,

L.C. = value of one main scale division/ total number of divisions on Vernier scale.

We know that, there 10 divisions on Vernier scale and the smallest division on main scale has value 1mm.

Hence, L.C. = 1mm/10 = 0.1mm = 0.01cm

Thus, the L.C. if Vernier caliper is 0.1mm or 0.01cm.

In this way, the least count of Vernier caliper is increased by increasing the number of divisions in Vernier scale and by decreasing the value of one division on main scale.

5.) Solution:

Ans:

Vernier constant is also called as the least count of Verniercaliper. And it is the smallest length that can be measured accurately using Vernier caliper. Also, it is the difference between the value of one main scale division and one Vernier scale division.

And Vernier constant is the specific name given to the least count of Vernier caliper.

6.) Solution:

Ans:

In case of Vernier caliper, when zero mark of the Vernier scale should coincide with the zero mark of the main scale. In this position, if the tenth division of Vernier caliper coincides with the ninth division of main scale then Vernier caliper is said to be free from zero error.

7.) Solution:

Ans:

Due to mechanical error in Vernier caliper, sometimes the zero mark of Vernier scale does not coincide with the zero mark of the main scale, when the two jaws j1 and j2 are in contact. Then it is said that the Vernier caliper has zero error. In this case, the zero error is the length between the zero mark of the main scale and zero mark of the Vernier scale.

There are two types of zero errors.

1) Positive zero error:

• When the two jaws are bring together, if zero mark of the Vernier scale is on right of zero mark of the main scale, then the zero error is said to be positive error.
• The following figure shows the two scales of Verniercaliper with positive zero error.
• In figure, the 6th division of Vernier scale coincides with the main scale division.
• And, we know that least count of Verniercaliper is 0.1mm or 0.01cm.
• Hence, zero error= +6*L.C.

= +6*0.01 cm

= + 0.06 cm 2) Negative zero error:

• When the two jaws are bring together, if the zero mark of Vernier scale is to the left of zero mark of the main scale then the zero error is said to be negative.
• The following figure shows the two scales of Vernier caliper with negative zero error.
• In figure, the 6th division of main scale coincides with a certain division on main scale. And there are total 10 divisions on Vernier scale.
• Hence, zero error= – (10-6)*L.C.

= -4*0.01cm

= -0.04 cm Thus, to get the correct reading the zero error with its proper sign is subtracted from the observed reading.

8.) Solution:

Ans:

• When the two jaws are bring together, if zero mark of the Vernier scale is on right of zero mark of the main scale, then the zero error is said to be positive error.
• The following figure shows the two scales of Verniercaliper with positive zero error.
• In figure, the 6th division of Vernier scale coincides with the main scale division.
• And, we know that least count of Verniercaliper is 0.1mm or 0.01cm.
• Hence, zero error= +6*L.C.

= +6*0.01 cm

= +0.06 cm 9.) Solution:

Ans:

The following figure shows the neat labelled diagram of Vernier calipers. The following are the main parts of Verniercalipers with their functions.

1) Outside jaws:

Outside jaws are used to measure the length of a rod, diameter of sphere, external diameter of a hollow cylinder.

2) Inside jaws:

Inside jaws are measured to use the internal diameter of hollow cylinder or pipe.

3) Strip:

Strip is used to measure the depth of the bottle or beaker.

4) Main scale:

Main scale is used to measure length correct up to 1 mm.

5) Vernier scale:

Vernier scale is used to measure length correct up to 0.1 mm.

10.) Solution:

Ans:

• Vernier caliper is used to measure the length of a rod.
• It is also used to measure diameter of a sphere.
• The internal and external diameter of hollow cylinder are measured by using Vernier caliper.
• The depth of small beaker or bottle is also measured by using Vernier caliper.

11.) Solution:

Ans:

• In Vernier caliper there are two scales one is the main scale which is fixed and the other is Vernier scale which slides along the main scale.
• The main scale is graduated such that one division on it is equal to 1mm. While Vernier scale has 10 divisions and the total length of these 10 divisions is (10-1) =9 divisions of the main scale which is equal to 9mm. Thus, each division on Vernier scale has length 0.9mm.
• To understand in detail we take an example.

The following figure shows a main scale graduated to read up to 1mm and a Vernier scale on which the length of 10 divisions is equal to the length of 9 divisions on main scale. Let, we determine the least count.

From fig, value of 1 division of main scale =1mm

Total number of divisions on Vernier = 10

Hence, least count of Vernier caliper is given by,

L.C. = value of one main scale division/ total number of divisions on Vernier

L.C. = 1mm/10= 0.1mm= 0.01cm

In this way, the Vernier caliper is made to measure length correct up to 0.01 cm.

12.) Solution:

Ans:

The following are the steps described to measure the length of small rod using Vernier caliper.

• First we have to find the least count and zero error of the Vernier caliper.
• Then we have to move the jaws J2 away from the jaw J1 and we have to fix the rod between jaws J1 and J2 whose length we have to measure. Now, we move the jaw J2 towards the jaw J1 so that it touches the rod. And we tight the screw S to fix the Vernier scale along its position.
• Now we will note the main scale readings. Let us suppose, p is the division on Vernier scale which coincides with any division on main scale. Thus, we get the Vernier scale reading as
• Vernier scale reading = p*L.C.
• Then, we add this Vernier scale reading to the main scale reading which gives us the observed length of the rod.
• In this way, we repeat the procedure two times and we have recorded the observations and tabulated as below.

Observations:

Total number of divisions on Vernier scale = n =….

Value of one division on main Scale x = ….cm

Least count = x/n = …cm

Zero error = …cm

Observation table:

 Sr. No. Main scale reading a Vernier division coinciding (p) Vernier scale reading b = p*L.C. Observed length = a + b 1. 2. 3.

• In this way, by constructing the above table we find the mean observed length.
• Finally, we subtract the zero error with its proper sign from mean observed length to get true measurement of length of the rod.
• Thus, observed length= main scale reading + p*L.C.
• True length = observed length – zero error with proper sign.

13.) Solution:

Ans:

a) External diameter of the tube:

Outside jaws are used to measure the external diameter of the tube.

b) Internal diameter of a mug:

Inside jaws are used to measure the internal diameter of a mug.

c) Depth of a small bottle:

Strip is used to measure the depth of a small bottle.

d) Thickness of a pencil:

Main scale and Vernier scale are used to measure the thickness of a pencil.

14.) Solution:

Ans:

Pitch:

The pitch of the screw is defined as it is the distance moved along its axis by screw in one complete rotation of its head.

Normally, the pitch of screw is 1mm or 0.5 mm.

Least count:

The least count of the screw is defined as it is the distance moved along the axis by it in rotating the circular scale by one division.

And it is determined by,

L.C. = pitch is a screw/ total number of divisions on circular scale.

If a screw is moved by 1mm in one rotation and it has 100 divisions on its circular scale then pitch of the screw is 1mm and least count of the screw is given by,

L.C. = 1mm/ 100 = 0.01mm = 0.001cm

15.) Solution:

Ans:

The least count of screw gauge can be decreased by decreasing the pitch and by increasing the total number of divisions on the circular scale.

16.) Solution:

Ans:

The following figure shows the neat labelled diagram of screw gauge. The following are the main parts of screw gauge with their functions.

1) Ratchet:

Ratchet is used to advance the screw by turning it till the object is gently hold between the stud and spindle of the screw.

2) Sleeve:

Sleeve is used to mark the main scale and base line.

3) Thimble:

Thimble is used to mark the circular scale.

4) Main scale:

Main scale is used to measure the length correct up to 1mm.

5) Circular scale:

It helps to read length correct up to 0.01 mm.

17.) Solution:

Ans:

Screw gauge is generally used to measure the diameter of a wire or thickness of a paper.

18.) Solution:

Ans:

In case of screw gauge, the thimble is attached to a ratchet by a spring and by turning the ratchet the screw is always rotated. As the flat end B of the screw comes in contact either  with stud A or with the object in between A and B, thus further rotation of the ratchet does not move the screw linearly to press B against A. In this way, ratchet helps in holding the given object gently between stud and the end B of the screw.

19.) Solution:

Ans:

Sometimes, due to mechanical error when the stud A is brought in contact with stud B, the zero mark of the circular scale is either above or below the base line of the main scale then the screw gauge is said to have a zero error.

There are two types of errors which are explained as follows.

1) Positive zero error:

When the flat end B of the screw is brought in contact with the stud A, then the zero mark on the circular scale is below the base line of main scale. This error is called as positive zero error.

In figure, the 5th division of circular scale coincides with the base line and if the least count of screw gauge is 0.001cm, then positive zero error become

= +5*0.001 = +0.005 cm 2) Negative zero error:

If the flat end B of the screw gauge is brought in contact with the stud A, if the zero mark on the circular scale is above the base line of a main scale then the zero error is said to be negative zero error.

In figure, 95th division of circular scale coincides with the baseline and total number of divisions on the circular scale are 100. If least count of screw gauge is 0.001cm then negative zero error is given by,

= – (100-95)*0.001= -0.005 cm. • Thus, to find the correct reading the zero error with its proper sign is subtracted from the observed reading.
• Hence, Correct reading= observed reading – zero error with proper sign.

21.) Solution:

Ans:

When we move the thimble in a reverse direction then it remains steady and doesn’t moves in one rotation directly. This is due to the continuous handling or use of the screw by rotating it. And this cause the error in observations also, this type of error is called as backlash error.

This types of error are occurring mostly in screw gauge due to its rotation in different direction. In order to decrease the error we have to rotate the thimble of the screw in one direction always. And we can’t change its direction of rotation in one rotation. We have to take all observations by rotating screw in one direction only to avoid the backlash error.

If we have to rotate screw in reverse direction then first we rotate it and then we stop at that point and then further we rotate it in reverse direction. In this way, we can avoid the backlash error and our observations becomes true observations.

22.) Solution:

Ans:

Following are the steps which are performed while measuring the diameter of wire with screw gauge.

• First we have to find the least count and zero error of the screw gauge properly.
• Next, we have to turn the ratchet anticlockwise so that gap between stud A and the flat end B is obtained.
• Then we keep the wire between stud A and flat end B and again we turn the ratchet clockwise such that wire is hold gently in between the stud A and the flat end B of the screw gauge.
• Then we take the main scale readings.
• Let us suppose p be the division of the circular scale which coincides with the base line of main scale. Thus, circular scale reading is given by,
• Circular scale reading = p*L.C.
• In this way, the process is repeated by putting the wire in perpendicular direction also. And also by putting wire at different positions take the readings.

Observations:

• Pitch of the screw = …cm
• Total number of divisions on circular scale = …
• Hence, least count of screw gauge is given by,
• C. = Pitch/ total number of divisions on circular scale = …cm
• Zero error =…

Observation table:

The following table shows the observation table for measuring diameter of wire with screw gauge.

 Sr. No. Main scale reading a Circular scale reading b = p*L.C. Observed diameter = a + b 1.      In one direction 2.      In perpendicular direction

• Thus, the observed diameter is determined as,
• Observed diameter = main scale reading + p*L.C.
• True diameter = observed diameter – zero error with proper sign
• In this way, we can measure the diameter of the wire by using screw gauge.

23.) Solution:

Ans:

a) The diameter of the needle

The diameter of the needle is measured accurately by using screw gauge.

b) The thickness of a paper

The thickness of a paper is measured by using screw gauge.

c) The internal diameter of the neck of a water bottle

The internal diameter of the neck of the water bottle is measured by using Verniercaliper.

d) The diameter of pencil

The diameter of pencil is measured by using screw gauge.

24.) Solution:

Ans:

As the least count of screw gauge is smaller as compared to meter rule and Vernier caliper, hence screw gauge measures small length to high accuracy.

Because, smaller the least count of an instrument more precise is the measurement taken by it.

25.) Solution:

Ans:

a) 0.1mm

0.1mm is the least count of Verniercaliper.

b) 1mm

1mm is the least count of meter rule

c) 0.01mm

0.01 mm is the least count of screw gauge.

Multiple choice type:

1) The least count of Verniercaliper is

Ans: d) 0.01 cm

2) A microscope has its main scale with 20 divisions in 1cm and Vernier scale with 25 divisions, the length of which is equal to the length of 24 divisions on main scale. The least count of microscope is

Ans: a) 0.002 cm

Because, least count is given by,

L.C. = smallest division on main scale/ total number of divisions on Vernier scale = 0.5mm / 25 = 0.1/5 = 0.02 mm = 0.002 cm

3) The diameter of a thin wire can be measured by

Ans: c) a screw gauge

Numerical:

1.) Solution:

Ans:

Given that, stop watch has 10 divisions between 0 and 5.

Hence, least count of stop watch is given by,

L.C. = smallest division/ total number of divisions = 5/10 = 0.5 s

2.) Solution:

Ans:

Given that, the main scale is calibrated in mm, hence smallest division on main scale = 1mm

Number of divisions on Vernier scale = 10

Thus, least count of Vernier caliper is given by,

L.C. = smallest division on main scale /total number of divisions on Vernier scale

L.C. = 1mm/ 10 = 0.1mm = 0.01cm

3.) Solution:

Ans:

Given that, Vernier scale has 50 divisions of length same as of 49 divisions of main scale.

Hence, smallest division on Vernier scale= 49/50 = 0.98 main scale division

But, least count is the difference between the main scale reading and the Vernier scale reading,

Hence, L.C. = 1-0.98 = 0.02 main scale division

But, here there are 20 divisions on main scale in 1cm.

Hence, 1m.s.d. = 1/20=0.05cm

Thus, least count = 0.02* m.s.d = 0.02*0.05= 0.001cm

4.) Solution:

Ans:

Given that, observed length of pencil is 1.4mm.

And the positive zero error is = 0.02cm = 0.2 mm

We know that, the corrected value or true value is given by,

Corrected value= observed value – zero error with proper sign

Corrected value = 1.4mm – 0.2mm = 1.2 mm

5.) Solution:.

Ans:

Given that, the main scale is graduated in mm.

Hence, smallest division on main scale = 1mm

Total number of divisions on Vernier scale= 10

Thus, least count of Vernier caliper is given by

L.C. = smallest division on main scale/ total number of divisions in Vernier scale

= 1mm/10

= 0.1 mm = 0.01 cm

As, when two jaws are in contact the zero of the Vernier scale is ahead of zero of main scale, which gives the positive zero error.

Also, given that 3rd division of main scale coincides with the main scale division.

Thus, positive zero error = +3*L.C.

=+3*0.01cm = +0.03cm

6.) Solution:

Ans:

Given that, the main scale of Vernier caliper is calibrated in mm.

Hence, smallest division on main scale = 1mm = 0.1cm

And, 19 divisions of main scale are equal in length to 20 divisions of Vernier scale.

Thus, total number of divisions on Vernier scale = 20

Hence, least count of Vernier caliper is given by,

L.C. = smallest division on main scale / total number of divisions on Vernier scale

= 0.1cm/ 20 = 0.005 cm

While measuring the diameter of cylinder

Main scale reading = 35mm = 0.35cm

And, here 4th division of Vernier scale coincides with the main scale division.

Hence, p= 4

Thus, v.s.r. = 4*L.C. = 4*0.005 = 0.02 cm

Thus, the total reading is given by,

Total reading = m.s.r + v.s.r = 3.5 + 0.2 = 3.52 cm

Thus, we get the diameter of cylinder = 3.52 cm

Now, radius of cylinder = diameter/2 = 3.52/2 = 1.76 cm

Thus, least count = 0.005 cm

Radius of cylinder = 1.76 cm

7.) Solution:

Find the length.

If zero error of Vernier caliper is -0.02 cm, what is the correct length?

Ans:

Given that, there are 10 divisions on the Vernier scale and 1 cm on the main scale.

L.C. = Smallest division on main scale/ total number of divisions on Vernier scale

L.C. = 1cm/ 10 = 0.1 cm

Given that, m.s.r. = 1.8 cm, and here 4th division of Vernier coincides with the main scale division.

Hence, p= 4

Thus, v.s.r. = p*L.C. = 4*0.01cm = 0.04 cm

Now, we find the total reading as,

Total reading = m.s.r. + v.s.r. = 1.8 + 0.04 = 1.84 cm

Thus, the observed length is found to be 1.84 cm

Now, given that zero error of Vernier caliper is -0.02 cm.

Then, corrected length is given by,

Corrected length = observed length – zero error with proper sign

= 1.84 – (-0.02) = 1.86 cm

Thus, the corrected length found to be 1.86 cm

8.) Solution: Ans:

We know that, least count of Vernier caliper is 0.01cm.

From the figure, m.s.r. = 3.3.cm

And, from figure we can say that, the 6th division on Vernier scale coincides with the main scale division.

Hence, here p = 6

Thus, v.s.r. = p*L.C. = 6*0.01= 0.06 cm

Now, we find the length of the rod by finding the total reading,

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

Thus, the length of the rod found to be 3.36 cm.

9.) Solution:

Ans:

Given that, pitch of the screw gauge is = 0.5 mm = 0.05 cm

And head scale is divided into 100 parts.

Hence, least count of screw gauge is given by,

L.C. = 0.5mm/ 100 = 0.005 mm = 0.0005 cm

10.) Solution:

Ans:

We know that, the pitch of the screw gauge is defined as, it is the distance moved by spindle in one revolution.

Hence, pitch = 1/2 = 0.5 mm = 0.05cm

Now, given that the thimble of screw gauge has 50 divisions in one rotation, which are the circular scale divisions.

Least count of screw gauge is given by,

L.C. = pitch/ number of divisions on circular scale = 0.5mm / 50 = 0.01 mm = 0.001 cm

11.)  Solution:

Ans:

Given that, the pitch of the screw gauge = 1 mm

And circular scale has 100 divisions on it.

Hence, the least count of screw gauge is given by,

L.C. = pitch/ circular scale divisions = 1mm/ 100= 0.01 mm = 0.001 cm

Now, also given that, m.s.r. = 2mm

And, p = 45

Hence, c.s.r. =p*L.C. = 45* 0.001cm = 0.045 cm

Thus, the diameter of the wire is given by,

Total reading = m.s.r. + c.s.r. = 2mm + 0.45 mm = 2.45 mm = 0.245 cm

Hence, the diameter of the wire is 0.245 cm.

12.) Solution:

If the zero error is +0.005 cm, what is the correct diameter?

Ans:

Given that, least count of screw gauge is 0.01 mm.

Sleeve reading = m.s.r. = 1mm = 0.1 cm

Reading on thimble is 27 divisions i.e. p = 27.

Hence, c.s.r. = p*L.C. = 27*0.01 mm = 0.27 mm = 0.027 cm

Thus, diameter of the wire is given by,

Total reading = m.s.r. + c.s.r. = 0.1+ 0.027 = 0.127 cm

Now, if the zero error is +0.005 cm then the corrected diameter is given by,

Corrected diameter = observed diameter – the zero error with proper sign = 0.127 cm – 0.005 cm = 0.122 cm

Thus, the corrected diameter of the wire is found to be 0.122 cm.

13.) Solution:

Ans:

Given that, the screw gauge has 50 divisions on its circular scale.

And screw is moved by 1mm on turning circular scale by two rotations.

Hence, pitch = 1mm/2 = 0.5 mm

Now, least count of screw gauge is given by,

L.C. = pitch/ circular scale divisions = 0.5mm / 50 = 0.01mm

Also, given that m.s.r. = 4

And when the fat end of the screw gauge is in contact with the stud, the zero of the circular scale leas below the base line.

Hence, the zero error is positive here.

Thus, zero error = + 4*L.C. = +4*0.01 mm = +0.04 mm

14.) Solution: Ans:

Given that, the circular scale divisions are 50.

And the screw advances by 1 division on main scale when circular head is rotated once.

Pitch = 1mm/ 1 = 1mm = 0.1cm

The least count of screw gauge is given by,

L.C. = pitch/ circular scale divisions = 1mm/ 50 = 0.02 mm = 0.002 cm

From fig. m.s.r. = 4mm

And p= 47, as 47th division on circular scale coincides with the division on main scale.

Hence, c.s.r. = p*L.C. = 47*0.02mm = 0.94 mm

Thus, now we can determine the diameter of the wire as,

Total reading = m.s.r. + c.s.r. = 4+ 0.94= 4.94 mm = 0.494 cm

Thus, the diameter of the wire is 4.94 mm

15.) Solution:

Ans:

Given that, pitch of the screw gauge is 0.5 mm

And its least count is 0.001 mm

We know that,

L.C. = pitch / number of divisions on circular scale

Hence, number of divisions on circular scale = pitch/ L.C. = 0.5mm/ 0.001mm = 500

Thus, the number of divisions on the head of screw gauge so as to read correct up to 0.001 mm are 500.

Exercise 1- C Solution:

1.) Solution:

Ans:

Simple pendulum:

• It is heavy bob which is having a point mass and it is suspended through inextensible and a weightless string from the rigid support.
• And the pendulum used in pendulum clock is not the simple pendulum.
• Because, simple pendulum is the ideal case in which the Bob which is heavy point mass suspended by weightless string.
• And in real life, it is not possible to found a point heavy mass and such type of weightless string.
• Hence, pendulum used in pendulum clock is not the simple pendulum.

2.) Solution:

Ans:

a) Oscillation:

• The Bob of simple pendulum moves to and fro continuously. Thus, the one complete to and fro motion of that Bob of simple pendulum is called as oscillation.

b) Amplitude:

• Amplitude is the maximum displacement of the Bob from its mean position on either side of the mean position.

c) Frequency:

• The number of oscillations performed by the bob in one second is called as the frequency of oscillation of simple pendulum.
• It is measured in hertz or per second.

d) Time period of oscillation:

• The time required by the bob to complete one oscillation is called as the time period of oscillation.
• It is denoted by T.
• It is measured in second.

3.) Solution:

Ans:

The following figure shows the neat labelled diagram of simple pendulum. Effective length of simple pendulum:

• Effective length is the distance between the centre of gravity of the bob i.e. point O and the point of suspension S.
• It is denoted by l.

Oscillation:

• The one complete to and fro motion of the bob is called as oscillation.
• In fig. The bob moves from O to A, A to O, O to B and then B to O. This complete motion is called as oscillation.
• Or when bob moves from A to B and again moves back from B to A, then it completes one oscillation.

4.) Solution:

Ans:

• The time period of oscillation depends on the effective length and acceleration due to gravity.
• The relation between time period T, effective length l and acceleration due to gravity g is given by,

T =2π√l/g

Thus,

• The time period of oscillation of a simple pendulum is directly proportional to the square root of the effective length.
• And, time period of oscillation of simple pendulum is inversely proportional to the square root of acceleration due to gravity.

5.) Solution:

Ans:

The time period of simple pendulum does not depend on following two factors.

• The mass of the bob suspended
• And the amplitude of oscillation

6.) Solution:

Ans:

a) The length is made four times.

We know that, the time period of simple pendulum is directly proportional to the square root of effective length.

Thus, T α √l

If length is made four times greater then,

T α √4l

Hence, in this case the time period of oscillation is doubled.

b) The acceleration due to gravity is reduced to its one fourth.

We know that, the time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity.

Thus, T α 1/√g

If the acceleration due to gravity is reduced to its one fourth then,

T α 1/√g/4 =2/√g

Hence, in this case also the time period of simple pendulum is doubled.

7.) Solution:

Ans:

Time period T and the frequency of oscillation f of an oscillation of the simple pendulum are related by the following equation.

T = 1/ f

Or f= 1/T

8.) Solution:

Ans:

The following is the process for measuring the time period of simple pendulum.

• In a simple pendulum the point mass i.e. bob is suspended from a rigid support through the weightless in extensible string as shown in figure.
• O is the mean position of the bob. Now, to measure the time period of one oscillation we slightly displaced the bob and then we release it.
• After releasing the bob, it performs to and fro motion about its mean position O.
• Now, we measure the time t by stop watch required for 20 oscillations.
• And then, by dividing t with 20 we get the time period T.
• Since, T = t/20 • We repeat the same process many times and we find the time period T for each observation.
• In this way can find the time period of an oscillation of simple pendulum by considering time for 20 oscillations.

• The time for more than one oscillation is recorded to minimise the uncertainty and to get accurate observations.
• Also, the least count of stop watch is in second and we can’t measure oscillations in fractions, so that we take a particular set of oscillations and for that we find the time t.

9.) Solution:

Ans:

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

Hence, T α √l

• The following graph shows the variation of T2 with respect to variation in effective length of pendulum l.
• It is a straight line inclined to l-axis, hence it shows that T2 is directly proportional to l.
• The slope of this graph gives the value of acceleration due to gravity g.
• For that, we take two points A and B, and we draw the normal from this points on the X and Y axis respectively.
• Let the time period at p is T12 and at q it is T22. Fig. Graph of T2 vs. l

• Again let l1 and l2 be the effective length at point r and s respectively.
• Thus, slope from the graph is given by,

Slope = AC/BC = pq/rs = (T12 – T22)/ (l1– l2)

• And the value of slope at that point is found to be constant and which is equal to 4π2/g, where g is the acceleration due to gravity at that point.

Thus, slope = 4π2/ g

And hence, g = 4π2 / slope of graph T2vs. l.

• In this way, we can find the acceleration due to gravity at a point from the slope of the graph T2 l.

10.) Solution:

Ans:

We know that, the time period of oscillation of simple pendulum depends on effective length and acceleration due to gravity only.

It does not depend on the mass of the bob.

And here the length of the pendulum A and B is same, so the ratio of their time period is 1:1 only.

Because there is no effect of mass of bob on time period of oscillation of simple pendulum, as it does not depend on the mass of bob.

11.) Solution:

Ans:

Given that, the pendulum A has length 1m and pendulum B has 4 m.

And we know that, time period of oscillation of simple pendulum is directly proportional to the square root of effective length.

Hence, T α √l

Thus, for A, T α √1

And for B, T α √4

Thus, time period for pendulum B is more than A.

And hence pendulum A makes more oscillations in one minute than the pendulum B.

12.) Solution:

Ans:

a) Length of pendulum:

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

b) Mass of bob:

The time period of oscillation of simple pendulum does not depend on the mass on bob.

c) Amplitude of oscillation:

The time period of oscillation of simple pendulum does not depend on the amplitude of oscillation.

d) Acceleration due to gravity:

The time period of oscillation of simple pendulum is inversely proportional to the square root of acceleration due to gravity.

13.) Solution:

Ans:

The pendulum whose time period of oscillation is 2 second is called as seconds pendulum.

In our house, we use the clock whose pendulum is the second’s pendulum.

14.) Solution:

Ans:

Second’s pendulum is the pendulum whose time period is 2 second.

Hence, T = 2 second

We know that, frequency of oscillation is given by,

T= 1/f

Hence, f = 1/T = 1/2= 0.5 s-1

And the frequency of oscillation does not depend on the amplitude of oscillation. It depends only on the time period of oscillation of simple pendulum.

Multiple choice type:

1) The length of simple pendulum is made one fourth. It’s time period becomes.

Ans: c) half

Because, the time period of oscillation of simple pendulum is directly proportional to the square root of effective length.

2) The time period of pendulum clock is

Ans: b) 2 s

Because, in pendulum clock, the pendulum used is the second’s pendulum whose time period is 2 second.

3) The length of seconds’ pendulum is nearly

Ans: c) 1.0 m

Because, to make the time period of oscillation of pendulum 2 s, it is made of 1.0 m.

Numerical:

1.) Solution:

Ans:

Given that, the simple pendulum completes 40 oscillations in one minute.

Hence, the time period of oscillation is given by,

T= 60s/ 40 = 1.5 s

And frequency of oscillation is given by,

f = 1/T = 1/ 1.5 = 0.67 s-1

2.) Solution:

Ans:

Given that, the time period of oscillation of simple pendulum is 2 second.

Hence, its frequency is given by,

f = 1/ T = 1/2 = 0.5 s-1.

And, the pendulum whose time period of oscillation is 2 second is called as seconds pendulum.

3.) Solution:

Ans:

If the second’s pendulum is placed at a position where acceleration due to gravity falls to one fourth, then it’s time period at that place is also doubles.

Because, we know that the time period of oscillation of simple pendulum is inversely proportional to the square root of acceleration due to gravity.

Hence, T α 1/√g

And its new time period is given by,

When acceleration due to gravity is g, its time period is 2 second. Since it is a second’s pendulum.

When acceleration due to gravity falls to one fourth, its time period doubles.

Hence, its new time period is 4 s.

4.) Solution:

Ans:

The time period of second’s pendulum is 2 second.

We know that, the time period of pendulum is given by,

T= 2π √l/g

On squaring,

T2= 4π2*l/g

Thus, l = T2*g/ 4*π2 = 4*10/4*9.86 = 10/9.86 = 1.0141m.

Thus, the length of second’s pendulum is found to be 1.0141 m.

5.) Solution:

Ans:

We know that, time period of simple pendulum is directly proportional to the square root of effective length.

Hence, for first pendulum

T1 α √1

And for second pendulum,

T2 α √9

Hence, their ratio T1/T2 = √1/√9 = 1/3.

Thus, T1:T2 = 1:3

6.) Solution:

Ans:

Given that, the pendulum completes 2 oscillations in 5 s.

Hence, its time period is given by,

T = 5/2 = 2.5 s

Now, if g = 9.8m/s2 then it’s length is given by,

T = 2π*√l/g

2.5 = 2*3.14*√l/9.8

2.5 = 6.28*√l/9.8

0.398 = √l/9.8

Hence, l= (0.398)2*9.8 = 1.55 m

Thus, the length of pendulum is found to be 1.55 m.

7.) Solution:

Ans:

We know that, the time period of oscillation of simple pendulum is directly proportional to the square root of effective length.

Hence, T1 α √l1

And T2 α √l2

Given that, T1:T2 = 2:1

Thus, l1: l2 = 4:1

8.) Solution:

Ans:

Given that, the bob takes 0.2 s for moving from mean position to its one end.

We know that, the amplitude of oscillation is 4*distance of mean position from one end.

Hence, its time period is given by,

T = 4*0.2 = 0.8 s

9.) Solution:

Ans:

We know that, the time period of second’s pendulum is 2 s.

That means it completes one oscillation in 2 s.

And the distance between two extremes positions of the bob is half of its amplitude of oscillation.

Hence, time required to complete half oscillation will be, T = 2/2 = 1 s

Updated: July 22, 2021 — 11:01 pm