Measurement of length MCQ Quiz - Objective Question with Answer for Measurement of length - Download Free PDF

Calculation:

Since least count of the instrument can be calculated as

Least count = \(\frac{\text { pitch length }}{\text { No. of division on circular scale }}\)

= \(\frac{0.75}{15}=0.05 \mathrm{~mm}\)

Here we are provided L = 5 mm & W = 2.5 mm

L  = 5 mm & W = 2.5 mm

∵ We know that

A = L.W

For calculating fractional error, we can write

\(\frac{\mathrm{dA}}{\mathrm{~A}}=\frac{\mathrm{dL}}{\mathrm{~L}}+\frac{\mathrm{dW}}{\mathrm{~W}}\)

Here dL = dW = 0.05 mm

\(\frac{\mathrm{dA}}{\mathrm{~A}}=\frac{0.05}{5}+\frac{0.05}{2.5}\)

⇒ \(\frac{\mathrm{dA}}{\mathrm{~A}}=\frac{1}{100}+\frac{2}{100}=\frac{3}{100}\),

So, x = 3 

Concept:

• Screw gauge: A screw gauge is used to measure the thickness or diameter of small objects with high precision. It consists of a main scale and a circular scale.
• Pitch: The pitch of a screw gauge refers to the distance moved by the spindle per complete rotation. In this case, the pitch is given as 1 mm.
• Least count: The least count is the smallest length that can be measured using the screw gauge. It is given by the formula:
  • Least Count = Pitch / Number of divisions on the circular scale
• Formula for thickness: The thickness of an object is calculated by adding the main scale reading and the circular scale reading. The circular scale reading is obtained by multiplying the number of coinciding divisions by the least count.

Calculation:

L.C = 1/100=0.01 mm

Error = 60× LC =0.6 mm

Main scale reading = 2 mm

Circular scale reading = 21n× 0.01=0.21n mm

Total thickness = Main scale reading + Circular scale reading = 2 mm + 0.21n mm = 2..42 mm.

Thus, The value of n is 2.

• 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.

Solution:

Given:

Pitch = 0.5 mm

Circular scale divisions = 50

Main scale reading = 0.5 mm

Zero error = -5 × 0.01 mm

We know that:

Least count (LC) = (Pitch) / (Circular scale divisions)

LC = 0.5 / 50 = 0.01 mm

Negative zero error = -5 × 0.01 mm = -0.05 mm

The measured value is:

Measured value = 0.5 mm + 25 × 0.01 - (-0.05) mm = 0.8 mm

The correct answer is: option 2) 0.8 mm

Given:

Screw gauge readings

Pitch = 0.5 mm

Circular scale divisions = 50

Main scale reading = 2.5 mm

Least count = 0.5 / 50 = 0.01 mm

Circular scale reading = 20 divisions

Relative error = 2%

Then, the dimension reading from the screw gauge gives the diameter of the solid ball.

Screw Gauge Reading = Main scale reading + (Pitch / Circular scale divisions) × Circular scale reading

= 2.5 + (0.5 / 50) × 20 = 2.7 mm

Density formula: ρ = Mass / Volume = M / [(4π / 3) × (D / 2)³]

The relative percentage error in density is given by:

(Δρ / ρ) × 100 = [ΔM / M + 3ΔD / D] × 100

Substituting the values:

= [2 / 100 + 3 × (0.01 / 2.7)] × 100

= (2 + 3 × 0.0037) = 2 + 0.111 ≈ 3.1%

Final relative error in density = 3.1%

The correct answer is option 4) i.e. average distance between Earth and Sun

CONCEPT:

• Range of lengths: The sizes of different objects vary over a very wide range.
  • They can be as small as 10^–14 m or as large as 10²⁶ m which is the extent of the observable universe.
  • So, certain terms and units are used to represent smaller and larger distances or lengths.

EXPLANATION:

• An astronomical unit is defined as the average distance between the Sun from the Earth.
• The astronomical unit is used for measuring distances within the solar system.

CONCEPT:

• Fermi meter: The unit of spatial measurement or length which is equal to 10-15 m is called one fermi meter.
  • A fermi meter is denoted by fm.

​1 fm = 10-15 m​

EXPLANATION:

1 fm = 10-15 m​

Hence option 1 is correct.

NOTE:

CONCEPT:

• Pico-metre = 10^-12 metre
• Angstrom = 10^-10 metre
• Nanometre = 10^-9 metre
• Kilometre = 10³ metre

EXPLANATION:

• Arranging in decreasing order Kilometre > Nanometre > Angstrom > Pico-metre

CONCEPT:

• A physical quantity is a property of a material. It can be expressed in number by measurement.
  • A physical quantity is expressed by a numerical value and a unit. For example, the physical quantity length can be expressed as 4 meters, where 4 is the numerical value and meter is the unit.
• The SI units are the standard units of measurement defined by the International System of Units (SI).

EXPLANATION:

• Following are the SI units of the physical quantity:

• The length is a physical quantity and the rest all (meter, gram, kelvin) are the units of physical quantities. Therefore option 4 is correct.

• The name and symbol of SI units are written in lowercase.
• Except for the symbols of those SI units, named after a person, which are written with an initial capital letter.
• For example, the second has the symbol s, but the kelvin has the symbol K because it is named after Lord Kelvin.

The correct answer is option 2) i.e. 1.47∘ 

CONCEPT:

• Parallax method: This method is used to determine large distances such as the distance of a planet or a star from the earth.
  • Parallax is the approximate shift in the position of an object with respect to another when we shift the point observation sideways.
  • The distance between two points of observation is called basis (b).
  • The distance of the object from the two viewpoints is D.
  • The angle between the two directions along which the object is viewed is the parallax angle or parallactic angle (θ). 

If S is the position of the object and A B are the two points of observation,

\(\theta =\frac{b}{D}\)

• Similarly, for a planet, if d is the diameter of the planet and α is the angular size of the planet

\(α =\frac{d}{D}\)

CALCULATION:

Given that:

Distance of a planet from Earth, D = 78R_earth

Diameter of earth, d = 2R_earth

The diameter of the earth as seen from the other planet will be α.

Using, \(α =\frac{d}{D}\)

\(⇒ α =\frac{2R_{earth}}{78R_{earth}} = 0.025\: rad\)

\(⇒ α = 0.025 \times \frac{180}{\pi}=1.47^∘ \)

CONCEPT:

• Meter: The distance, light travels through a vacuum in exactly 1/299792458 seconds is called 1 meter.
•  The meter is the basic unit of length in the SI system of units. 
• The unit of length which is equal to 10⁻⁶ m is called one micrometer. Micron is denoted by μ.

EXPLANATION:

1 μm = 10^-6 m 

1 m = 10⁶ m = 1000000 µm.

So option 2 is correct.

The correct answer is option 3) i.e. 1 angstrom = 1010 m

CONCEPT:

• Range of lengths:
  • The sizes of different objects vary over a very wide range.
  • They can be as small as 10–14 m or as large as 1026 m which is the extent of the observable universe.
  • So, certain terms and units are used to represent smaller and larger distances or lengths.

EXPLANATION:

• The incorrect option is 1 angstrom = 1010 m.

The correct answer is option 2) i.e. Length

CONCEPT:

• Length is a measure of the size or distance between two points. The SI unit of length is the meter (m).
• 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:

• A light-year is a unit of length.

CONCEPT:

• 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

• Fermi: It is a small practical unit of distance used for measuring nuclear size. It is also known as a femtometer. A femtometer is denoted by fm.

⇒ 1 fermi = 1 fm = 10^-15 m

CALCULATION:

• Ratio of (1 angstrom / 1 femtometer).

\( ⇒ \frac{{{{10}^{ - 10}}}}{{{{10}^{ - 15}}}} = {10^{\left( { - 10} \right) - \left( { - 15} \right)}} = {10^5}\)

Additional Information

The correct answer is option 4) i.e. effects of the Earth's atmosphere

CONCEPT:

• Parallax method: This method is used to determine large distances such as the distance of a planet or a star from the earth.
  • Parallax is the approximate shift in the position of an object with respect to another when we shift the point observation sideways.

EXPLANATION:

• The parallax method is affected by the path of light that travels from the distant planet to the points of observation on Earth.
• The light from a distant object reaching the observer goes through multiple layers of the atmosphere of Earth.
• These layers of atmosphere subject the light to undergo multiple refractions and decreases the accuracy of the method.
• Hence, extremely small parallax angles less than 0.01 arcsec are very difficult to measure due to the effects of the Earth's atmosphere.