# 9th Std Science Term 1 Solution in English | Lesson.1 Measurement and Measuring Instruments

## Lesson 1. Measurement and Measuring Instruments

### I. Choose the correct answer :

1. Choose the correct one.

1. mm < cm < m < km
2. mm > cm > m > km
3. km < m < cm < mm
4. mm > m > cm > km

Ans : mm < cm < m < km

2. Rulers, measuring tapes and metre scales are used to measure

1. mass
2. weight
3. time
4. length

Ans : length

3. 1 metric ton is equal to

1. 100 quintals
2. 10 quintals
3. 1/10 quintals
4. 1/100 quintals

Ans : 10 quintals

4. Which among the following is not a device to measure mass?

1. Spring balance
2. Beam balance
3. Physical balance
4. Digital balance

Ans : Spring balance

### II. Fill in the blanks :

1. Metre is the unit of _______________.

Ans : length

2. 1 kg of rice is weighed by _______________.

Ans : beam balance

3. Thickness of a cricket ball is measured by _________.

Ans : vernier caliper

4. Radius of a thin wire is measured by ________.

Ans : screw gauge

5. A physical balance measures small differences in mass up to ___________.

Ans : 1mg or less

### III. State whether true or false. If false, correct the statement :

1. The SI unit of electric current is kilogram. ( False )

Correct statement : The SI unit of electric current is ampere.

2. Kilometre is one of the SI units of measurement. ( True )

3. In everyday life, we use the term weight instead of mass.  ( True )

4. A physical balance is more sensitive than a beam balance.  ( True )

5. One Celsius degree is an interval of 1K and zero degree Celsius is 273.15 K.  ( True )

6. With the help of vernier caliper we can have an accuracy of 0.1 mm and with screw gauge we can have an accuracy of 0.01 mm.  ( True )

### IV. Match the following

1.

 Column I Column II 1. Length Kelvin 2. Mass metre 3. Time kilogram 4. Temperature second Ans : 1 – B, 2 – C, 3 –  D,  4  – A

2.

 Column I Column II 1. Screw gauge Vegetables 2. Vernier caliper Coins 3. Beam balance Gold ornaments 4. Digital balance Cricket ball Ans : 1 – B, 2 – D, 3 –  A,  4  – C

3.

 Column I Column II 1. Temperature Beam balance 2. Mass Ruler 3. Length Digital clock 4. Time Termometre Ans : 1 – D, 2 – A, 3 –  B,  4  – C

### V. Assertion and reason type

1. Assertion (A): The SI systems of units is the improved system of units for measurement.

Reason (R): The SI unit of mass is kilogram

1. Both A and R are true but R is not the correct reason
2. Both A and R are true and R is the correct reason
3. A is true but R is false
4. A is false but R is true

Ans : Both A and R are true and R is the correct reason

2. Assertion (A): Te skill of estimation is important for all of us in our daily life.

1. Both A and R are true but R is not the correct reason
2. Both A and R are true and R is the correct reason
3. A is true but R is false
4. A is false but R is true

Ans : Both A and R are true and R is the correct reason

3. Assertion(A): The scientifically correct expression is “ The mass of the bag is 10 kg”

Reason (R): In everyday life, we use the term weight instead of mass

1. Both A and R are true but R is not the correct reason
2. Both A and R are true and R is the correct reason
3. A is true but R is false
4. A is false but R is true

Ans : Both A and R are true but R is not the correct reason

4. Assertion (A): 0 °C = 273.16 K. For our convenience we take it as 273 K after rounding off the decimal

Reason (R): To convert a temperature on the Celsius scale you have to add 273 to the given temperature

1. Both A and R are true but R is not the correct reason
2. Both A and R are true and R is the correct reason
3. A is true but R is false
4. A is false but R is true

Ans : Both A and R are true and R is the correct reason

5. Assertion (A): The distance between two celestial bodies is measured in the unit of light year

Reason (R): Te distance travelled by the light in one year is one light year

1. Both A and R are true but R is not the correct reason
2. Both A and R are true and R is the correct reason
3. A is true but R is false
4. A is false but R is true

Ans : A is false but R is true

### VI. Comprehensive type

Read the passage and answer the questions given below.

Mass is the amount of matter contained in an object. Measurement of mass helps us to distinguish between a lighter and a heavier body. Beam balance, spring balance and electronic balance are used to measure mass of diferent objects. Te SI unit of mass is the kilogram (kg). But diferent units are used to measure the mass of diferent objects. E.g. weight (mass) of a tablet is measured in milligrams (mg), weight of a student is measured in kilogram (kg) and weight of a truck with goods is measured in metric tons. 1 metric ton is equal to 10 quintals and 1 quintal is equal to 100 kg. 1 gram is equal to 1000 mg.

1. The value of 1 metric ton is equal to

1. 1000 kg
2. 10 quintals
3. 10,00,000 g
4. 100 kg

Ans : 1000 kg

2. How will you measure the weight of a tablet?

1. kg
2. g
3. mg
4. None of these

Ans : mg

### VII. Very short answer type

1. Define measurement.

Measurement is the processes of comparison of the given physical quanity with the know standard quantity of the same nature

2. Define standard unit.

SI System of units is the modernised and improved form of the previous system of units.

3. What is the full form of SI system?

International system of units.

4. Define least count of any device

• The smallest length which can be measured by metre scale is called least count.
• Least count of the instrument = Value of one main scale divison / Total number of vernier scale divison

Least count _ Pitch / No. of head scale divisions

5. What do you know about pitch of screw gauge?

Pitch of the screw gauge is the distance between two successive screw threads. It is measured by the ratio of distance travelled on the pitch scale to the number of rotations of the head scale.

Pitch = Distance travelled on the pitch scale / Number of rotations of the head scale

6. Can you find the diameter of a thin wire of length 2 m using the ruler from your instrument box?
No, I can not find the diameter of a thin wire of length 2 m using the ruler.

### VIII. Short answer type

1. Write the rules that are followed in writing the symbols of units in SI system.

(i) Units named after scientists are written in lower case.

• Eg. joule, kelvin and newton.

(ii) Symbols for the units are always written in lower case.

• Eg. m, kg and s.

(iii) However, the symbols for the units derived from the names of scientists are written in capital letters.

• Eg. C (Celsius), N (newton) and J (joule).

(iv) Symbols are not followed by a full stop.

• Eg. 75 cm and not 75 cm.

(v) Symbols are never written in plural.

• Eg. 100 kg, not as 100 kgs.

2. Write the need of a standard unit

A Standard Unit is needed to maintain uniformity in measurements like length, weight, size and distance.

Eg: Standard Unit of length is metre.

3. Differentiate mass and weight

 Mass Weight 1. Fundamental quantity Derived quantity 2. Has magnitude alone scalar quantity Has magnitude and direction – vector quantity 3. It is the amount of matter contained in a body It is the normal force exerted by the surface on the object against gravitational pull 4, Remains the same Varies from place to place 5. Itis measured using physical balance It is measured using spring balance 6. Its unit is kilogram Its unit is newton

4. How will you measure the least count of vernier caliper?
Least Count or L.C. is the minimum reading or value that can be measured with a measuring tool or device.

### IX. Numerical Problems :

1. Inian and Ezhilan argue about the light year. Inian tells that it is 9.46 x 10/5 m and Ezhilan argues that it is 9.46 x 10? km. Who is right? Justify your answer.

Solution : (Inian is correct)

Light travels 3 x 10° m in one second or 3 Lakhs kilometre in one second.

In one year we have 365 days.

 The total number of second in one year is equal to 365 = 24 x 60 x 60 Distance travelled by light in 1 year = (3.153 x 10” x 3 x 10® = 9,46 x 10% m.

2. The main scale reading while measuring the thickness of a rubber ball using Vernier caliper is 7 cm and the Vernier scale coincidence is 6. Find the radius of the ball.
Solution :

 MSR = 7cm VC = 6cm LC = 0.1mm=0.1 cm Diameter =DR = MSR+(VC xLC) = 7+0.06cm Diameter D = 7.06cm Radius R = D/2 = 7.06/2 =  0.035 m The radius of the ball = 0.0353 m.

3. Find the thickness of a five rupee coin with the screw gauge, if the pitch scale
reading is 1 mm and its head scale coincidence is 68.

Solution :

 PSR = Imm =1 x 10-3 m HSC = 68 LC = 0.01 mm =0.01 x 10-3 m Total reading = PSR+(HSC x LC) Thickness of the five rupee coin = 1 x 10-3 +(68 « 0.01 x 10-3) m Thickness of the five rupee coin = 1.68 x 10-3 m= 1.68 mm

4. Find the mass of an object weighing 98 N.

Solution

 W = mg W = 9N g = 98 m/s2 m W/g = 98 / 9.8 =10 Kg.

### IX. Long answer type

1. Explain a method to find the thickness of a hollow tea cup.

Step 1: The Pitch, Least count and the type of zero error of the screw gauge are determined.
Step 2: The given cup is placed in between two studs.
Step 3: The head screw using the ratchat arrangement is freely rotated until the given cup is held firmly, but not tightly.
Step 4: Pitch scale reading (PSR) by the head scale and head scale coincidence (HSC) with the axis of the pitch scale, are found.
Step 5: The readings are recorded and the experiment for different positions of the given cup is repeated.
Step 6: The thickness of the cup is calculated using the formula P.’S.R+(HSCxL.C )
Step 7: Then the average of the last column of the table. is found.

 Main Scale Reading MSR x 10-4 m Vernier Scale Coincidence Observed Reading OR = MSR + (LC X VC) Corrected Reading = OR + ZC 1. 2. 3. 4.

Hence the thickness of a hollow tea cup = _______________ mm.

2. How will you find the thickness of a one rupee coin?

Step 1: The Pitch, Least count and the type of zero error of the screw gauge are determined.

Step 2: The given coin is placed in between two studs.

Step 3: The head screw using the ratchat arrangement is freely rotated until given one rupee coin is held firmly, but not tightly.

Step 4: Pitch scale reading (PSR) by the head scale and head scale coincidence (HSC) with are axis of the pitch scale are found.

Step 5: The reading are recorded and the experiment for different positions of the given coin is repeated.

Step 6: The thickness of the coin is computed using the formula P.’S.R+(HSCxL.C )

Step 7: Then the average of the last column of the table is found.

 P.S.R.  (mm) HSC (division) CHSC = HSC + ZC (Division) CHSR = CHSC x LC (mm) Total Reading = PSR + CHSR (mm) 1. 2. 3.

mean = ____________________mm
Hence the thickness of a one rupee coin = _______________________mm

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