Measurement of length MCQ Quiz in मल्याळम - Objective Question with Answer for Measurement of length - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT:

• Micrometer: The unit of spatial measurement or length which is equal to 10-6 m is called one micrometer.
  • A micrometer is denoted by μm.

​1 μm = 10-6 m

• Angstrom: The unit of length which is equal to 10−10 meters is called one Angstrom.
  • Angstrom is denoted by Å.

​1 Å = 10-10 m

CALCULATION:

1 μm = 10-6m        --- (1)

1 Å = 10-10m        --- (2)

Equation (1) divided by (2) gives:

(1 μm/1 Å) = (10-6m /10-10m) = 10⁴ 

So 1 μm = 10⁴ Å

Hence option 4 is correct.

NOTE:

MEMORY TIPS:

The correct answer is option 1) i.e. Distance

CONCEPT:

• Light year: It is the distance travelled by light in one year. 
  • A light-year is a convenient unit of measurement of large distances.
  • Generally, a light-year is used to represent astronomical distances. 
  • One light-year is approximately 9 trillion (9 × 10¹²) km.

EXPLANATION:

• Large distances are expressed conveniently using light-years. Hence, a light-year is a unit of distance.

• A screw gauge is used to measure small lengths such as the diameter of the wire, the thickness of the plate, etc.
• it is also called a micrometer screw gauge.
• micro screw gauge is introduced by William Gascoigne. 

Least count:

L . C = \( \frac{{P}}{{D}}\)

Where,

P = pitch.

D = no of division on a circular scale.

If the pitch is 1 mm and there is 100 division on the circular scale, then

Least count = \( \frac{{1\ mm}}{{100}} = 0.01\ mm = \frac{{0.01}}{{10}} = 0.001\;cm = \frac{{0.001}}{{100}} = 0.00001\ m = {10^{ - 5}}\ m\)

Spherometer:

• A spherometer is used to measure the radius of curvature of the curved surface.
• spherometer is used by an optician.
• spherometer is used for precise measuring.
• it has a disk-like shape.

Explanation:

Least count:

L . C = \( \frac{{P}}{{D}}\) 

Where,

P = pitch

D = no of division on a circular scale.

If the pitch is 1 mm and there is 100 division on the circular scale, then

Least count = \( \frac{{1\ mm}}{{100}} = 0.01\ mm = \frac{{0.01}}{{10}} = 0.001\;cm = \frac{{0.001}}{{100}} = 0.00001\ m = {10^{ - 5}}\ m\) 

The least count is the same for both the screw gauge and spherometer.

A screw gauge and a spherometer can measure distances up to 10^-5 m.

CONCEPT:

Screw Gauge:

• It's an instrument used to measure the diameter, radius, or thickness of a thin wire or a thin metal sheet.
• It is also known as a micrometre screw gauge.

It has two scales:

• Pitch scale
• Circular scale

→Observed reading in screw gauge = Main scale reading (M.S.R) + Circular scale reading (C.S.R) + Least count of circular scale (L.C)

→Observed reading = Correct reading + Zero error (with sign)

CALCULATION:

Correct reading = 2.675 mm, L.C = 0.005 mm

Let, Main scale division = x,

∴ M.S.R = a × 0.5, C.S.R = b

∴ (a + 0.5) + (b × 0.005) = 2.675 + 0.02

Now if, a = 5,  M.S.R = 2.5

[This is the maximum value "a" can have. For any other values e.g: a = 6, M.S.R will be 3, which is not according to the correct reading]

So, 2.5 + b × 0.005 = 2.695

b = \(\frac{2.695 - 2.5}{0.005}\) = 35

So, the correct answer is option (2).

CONCEPT:

A screw gauge is defined as an instrument that measures the diameter or thickness of the wire accurately. The least count of the screw gauge is 0.01 mm and the total reading in the screw gauge is calculated as

Total reading = Linear scale reading + circular scale reading \(×\) Least count

CALCULATIONS:

Given : Pitch = 1 mm

Number of divisions, N = 100

Least count, LC = \(\frac{pitch}{N}\)

Now, on putting the given values we have;

Least count, LC = \(\frac{1 mm}{100}\)

= 0.01 mm

Here the instrument is in positive zero error, therefore we have;

e = + n × LC

⇒ e = + 8 × 0.01

⇒ e =+ 0.08 mm

The linear scale reading = 1 × 1 mm

= 1 mm

The circular scale reading = 72 × 0.01 mm

= 0.72 mm

Therefore, the measured reading is = 1 + 0.72

= 1.72 mm

The true reading is written as = 1.72 - 0.08

= 1.64 mm

Hence option 1) is the correct answer.

Measurement is the comparison of an unknown quantity with a known quantity.

• Standard Unit of Length: When a unit is accepted by the majority of people, it is called a standard unit of measurement. Examples, kilogram, metre, etc.
• Non-Standard Unit of Measurement: Those units which are not accepted by the majority of people are called non-standard units of measurement. For example,
  • Fathom: It is used for specification in marine depth, usually, 1 fathom (ftm) = 6 feet = 1.8288 metres
  • Handspan: When your fingers are stretched out, the length between your little finger and thumb is called a handspan.
  • Cubit: The length from your elbow to your middle finger is used for measurement.
  • Finger: Smaller objects can be measured using finger

Hence, we conclude When your fingers are stretched out, the length between your little finger and thumb is known as handspan.

Measurement is the comparison of an unknown quantity with a known quantity.

• Standard Unit of Length: When a unit is accepted by the majority of people, it is called a standard unit of measurement. Example, kilogram, metre, etc.
• Non-Standard Unit of Measurement: Those units which are not accepted by majority of people are called non-standard unit of measurement. For example,
  • Fathom: It is used for specification in marine depth, usually, 1 fathom (ftm) = 6 feet = 1.8288 metres
  • Handspan: When your fingers are stretched out, the length between your little finger and thumb is called a handspan.
  • Cubit: The length from your elbow to your middle finger is used for measurement
  • Finger: Smaller objects can be measured using finger

Hence, we conclude that fathom, hand, cubit, and finger are used as non-standard units of length.

Calculation:

Given: 

The pitch of the screw gauge is 1 mm.

The circular scale has 100 divisions.

The zero error of the gauge is that the zero of the circular scale is 5 divisions below the reference line.

During the measurement, 4 linear scale divisions are visible, and 60 divisions on the circular scale coincide with the reference line.

Calculation: 

Least count = \(\frac{1}{100}\) mm = 0.01mm

zero error = +0.05 mm

Final Reading = 4 × 1 mm + 60 × 0.01 mm – 0.05 mm

⇒ 4.55 mm

∴ The diameter of the wire is 4.55 mm

CONCEPT:

→The least value that can be measured by the vernier calipers, is called the least count.

→M.S.D = Main scale division, V.S.D = Vernier scale division, L.C = least count

EXPLANATION:

Given:

1 M.S.D = a cm

n V.S.D = (n-1) M.S.D

∴ 1 V.S.D = \( \frac{n-1}{n}\)M.S.D

L.C = 1 M.S.D - 1 V.S.D

= 1 M.S.D -  \( \frac{n-1}{n}\)M.S.D

= \( (1-\frac{n-1}{n}) = \frac{n-n+1}{n} = \frac{1}{n}\) M.S.D

= \( \frac{a}{n}\) cm = \( \frac{10a}{n}\) mm 

So, the correct answer is option (3).

CONCEPT:

• A cube is a cuboid that has all the sides equal.
• Surface area of a cube of side 'a' = S = 6a²
• The volume of a cube of side 'a' = V = a³

EXPLANATION:

Given that the surface area and volume of a cube are numerically equal.

S = V

6a2 = a3

6 = a

So the side of the cube is 6 units.

Hence the correct answer is option 4.