Answer the following questions

PHYSICS

MEASUREMENT

TEXT BOOK EXERCISES

VI. Answer very briefly:

1. Define measurement.

Answer: Measurement is the processes of comparison of the given physical quantity with the known standard quantity of the same nature.

2. Define standard unit.

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

3. What is the full form of SI system?

Answer: International System of Units.

4. Define least count of any device.

(i) The smallest length which can be measured by metre scale is called least count.

(ii) Least count of the instrument = Value of one main scale division / Total number of vernier scale division

Least count = [Pitch / No. of head scale divisions]

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

 Answer: 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?

Answer: No, I can not find the diameter of a thin wire of length 2 m using the ruler.

VII. Answer briefly:

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

Answer: (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.

Answer: 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.

Answer:

Mass

1. Fundamental quantity

2. Has magnitude alone - scalar quantity

3. It is the amount of matter contained in a body

4 .Remains the same

5. It is measured using physical balance

6. Its unit is kilogram

Weight

1. Derived quantity

2. Has magnitude and direction - vector quantity

3. It is the normal force exerted by the surface on the object against gravitational pull

4. Varies from place to place

5. It is measured using spring balance

6. Its unit is newton

4. How will you measure the least count of vernier caliper?

Answer: Least Count or L.C. is the minimum reading or value that can be measured with a measuring tool or device.

VIII.Answer in detail:

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

Answer:

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 s: 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+ (HSC × L.C)

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

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

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

Answer:

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+ (HSC × L.C)

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

Mean = _______mm

Hence the thickness of a one rupee coin = _________ mm

IX. Numerical Problems:

1. Inian and Ezhilan argue about the light year. Inian tells that it is 9.46 × 1015 m and Ezhilan argues that it is 9.46 × 1012 km. Who is right? Justify your answer.

Solution:

(Inian is correct)

Light travels 3 × 108 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 × 60 × 60

Distance travelled by light in 1 year = (3.153 × 107) × (3 × 108)

 = 9.46 × 1015 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 = 7 cm

 VC = 6 cm

 LC = 0.1 mm = 0.01 cm

Diameter = DR = MSR + (VC × LC)

 = 7 + 0.06 cm

Diameter D = 7.06 cm

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 = 1mm = 1× 10- 3m

 HSC = 68

 LC = 0.01mm = 0.01×10-3 m

 Total reading = PSR+ (HSC×LC)

∴ Thinkness of the five rupee coin = 1×10-3 + (68×0.01×10-3) m

∴ Thinkness of the five rupee coin = 1.68× 10-3 = 1.68mm

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

 Solution:

W = mg

 W = 98N

 g = 9.8m/s2

 m = W/g = (98/9.8) = 10kg.

Intext Activities

ACTIVITY - 1

Using Vernier caliper find the outer diameter of your pen cap .

Aim: To find the outer diameter of the pen cap.

Materials required: Vernier caliper, pen cap.

Solution:

Result:

The outer diameter of the pen cap = 9.35 cm

 ACTIVITY - 2

Determine the thickness of a single sheet of your science textbook with the help of a Screw gauge.

LC = Least Count

PSR = Pitch Scale Reading

HSC = Head Scale Coincidence

HSR = Head Scale Reading

TR = Total Reading

Result:

The thickness of the single sheet is = 0.30mm.

Numerical Problems

1. A piece of iron of volume 40cm3 whose density is 6.8g/cm3. Find the mass of iron. Solution:

Given, density of iron, D = 6.8g/cm3

volume of iron, V = 40 cm3

 mass of iron, M = V × D           [ ∴ mass = volume × density]

 = 40 cm3× 6.8g / cm3

 m =272.0g.

2. Solve: The mass of 40 apples in a box is 5 kg.

(i) Find the mass of a dozen of them.

(ii) Express the mass of one apple in gram.

Solution:

(i) 40 apples = 5kg = 5000 g

 1 apple = 5000/40

 1 apple = 125 g

∴ 1 dozen = 12 apples

12 apple = 125 × 12g

 12 apple = 1500 g.

(ii) 40 apples = 5000 g

1 apple = 5000/40 g

1apple = 125 g

The mass of 1 apple = 125 g.