Accuracy, precision of instruments and errors in measurement MCQ Quiz - Objective Question with Answer for Accuracy, precision of instruments and errors in measurement - Download Free PDF

Calculation:

In the given Vernier callipers, each 1 cm is equally divided into 8 main scale divisions (MSD). Thus,

1 MSD = 1 / 8 = 0.125 cm.

Further, 4 main scale divisions coincide with 5 Vernier scale divisions (VSD), i.e.,

4 MSD = 5 VSD → 1 VSD = (4 / 5) × 0.125 = 0.1 cm.

The least count of the Vernier callipers is given by:

LC = 1 MSD - 1 VSD = 0.125 - 0.1 = 0.025 cm.

In the screw gauge, let l be the distance between two adjacent divisions on the linear scale. The pitch p of the screw gauge is the distance travelled on the linear scale when it makes one complete rotation.

Since circular scale moves by two divisions on the linear scale when it makes one complete rotation, we get:

p = 2l

The least count (lc) of the screw gauge is defined as the ratio of the pitch to the number of divisions on the circular scale (n), i.e.,

lc = p / n = 2l / 100 = l / 50

If pitch p = 2 × LC = 2 × 0.025 = 0.05 cm, then l = p / 2 = 0.025 cm

Substitute l in the equation to get least count:

lc = 0.025 / 50 = 5 × 10⁻⁴ cm = 0.005 mm

If l = 2 × LC = 2 × 0.025 = 0.05 cm, then again from the equation:

lc = 0.05 / 50 = 1 × 10⁻³ cm = 0.01 mm

Calculation:

\(\mathrm{L}=\frac{\mathrm{pn}^{4}}{\mathrm{~pq}}\)

\(\Rightarrow \frac{\Delta \mathrm{L}}{\mathrm{L}} \times 100=\left[\frac{\Delta \mathrm{p}}{\mathrm{p}}+4 \frac{\Delta \mathrm{n}}{\mathrm{~n}}+\frac{\Delta \mathrm{p}}{\mathrm{p}}+\frac{\Delta \mathrm{q}}{\mathrm{~q}}\right] \times 100 \)

\(\Rightarrow \frac{\mathrm{11x}}{10}=\left[\frac{3}{60}+4\left(\frac{0.1}{20}\right)+\left(\frac{0.2}{40}\right)+\frac{0.1}{50}\right] \times 100\)

⇒ x = 7

Calculation:

Least count = 1 MSD − 1 VSD

1 MSD − (9 / 10) MSD

= (1 / 10) MSD

= (1 / 10) × 0.1 cm = 0.01 cm

Zero error = +0.1 cm

Main scale reading = 5 cm

Vernier scale reading = 8 × 0.01 = 0.08 cm

Final measurement of diameter

= 5 + 0.08 − 0.1 = 4.98 cm

Correct option is: (3) 4.98 cm

Concept:

Vernier Caliper

• Vernier Caliper is a precision instrument that can be used to measure internal and external distances accurately.
• A vernier caliper consists of two main parts:
• the main scale engraved on a solid L-shaped frame and the vernier scale that can slide along the main scale. 
• It works on the principle of vernier and can measure the dimensions to an accuracy of 0.02 mm.

Parts of a Vernier caliper:

• Outside jaws: Used to measure the external diameter or width of an object
• Inside jaws: Used to measure the internal diameter of an object
• Locking screw: to lock the jaws
• Adjusting screw: to take an accurate measurement of the workpiece.

Explanation:

In general one vernier scale division is smaller than one main scale division but in some modified cases it may be not correct.

Also least count is given by one main scale division / number of vernier scale division for normal vernier calliper.

Calculations:

Given:

Main scale readings: 5.10 cm and 5.15 cm

Vernier scale reading: 24 divisions

1. Main Scale Division (MSD):

The main scale can measure from 5.10 cm to 5.15 cm, indicating a total length of 0.05 cm (5.15 cm - 5.10 cm).

Since there are 50 divisions on the main scale, each main scale division (MSD) is: 0.05 cm / 50 = 0.001 cm.

2. Vernier Scale Division (VSD):

Each vernier scale division (VSD) corresponds to: 1 MSD - (1 VSD) = 0.001 cm.

Therefore, VSD = 0.001 cm.

3. Least Count (LC):

The least count is the difference between one main scale division and one vernier scale division: LC = MSD - VSD = 0.001 cm - 0.001 cm = 0.000 cm.

However, considering the measurement precision, the least count is effectively: LC = 0.001 cm.

4. Diameter Measurement:

Using the least count, the measured diameter is: 5.10 cm + (24 divisions × 0.001 cm/division) = 5.10 cm + 0.024 cm = 5.124 cm.

The measured diameter is approximately 5.124 cm.

CONCEPT:

• Percentage error is the difference between theoretical value and an experimental, divided by the theoretical value, multiplied by 100 to give a percent.

To calculate percentage error in the expression A = xmyn / zp . We use formula

\(\frac{{\Delta A}}{A} × 100 = m\frac{{\Delta x}}{x}× 100 + n\frac{{\Delta y}}{y} × 100+ p\frac{{\Delta z}}{z}× 100\)

where \(\frac{{\Delta A}}{A} × 100\) is the percentage error in A and \(\frac{{\Delta x}}{x} × 100\) is the percentage error in x.

CALCULATION:

Given in the question:

\(\frac{{\Delta x}}{x}× 100 = 1\%\)

\(\frac{{\Delta y}}{y}× 100 = 2\%\)

\(\frac{{\Delta z}}{z} × 100= 4\% \)

now we have to find the error in

 \(Q = \frac{{x^3{y^2}}}{{{z}}}\)

\(\frac{{\Delta Q}}{Q} × 100\)

\( = 3\frac{{\Delta x}}{x}× 100 + 2\frac{{\Delta y}}{y} × 100\)

\(+ \frac{{\Delta z}}{z}× 100\)

\(\frac{{{{\Delta Q}}}}{{\text{Q}}} = 3\frac{{{{\Delta x}}}}{{\text{x}}} + \frac{{2{{\Delta y}}}}{{\text{y}}} + \frac{{{{\Delta z}}}}{{\text{z}}}\)

= 3 × 1 + 2 × 2 + 4 = 11

∴ % error in t = ΔQ/Q × 100 = 11%

• So the correct answer will option 3.

CONCEPT:

• Error: The difference in the true value and the measured value of a quantity is called an error of measurement. 
  • The true value can be x ± Δx where x is the measured value and Δx is the error. 

While doing any experiment due to faulty equipment, carelessness, or any other random cause the final results get affected.

• To resolve the percentage change we use the formula:

\(Y = \frac{{{A^a}\;.\;\;{B^b}}}{{{C^c}}} \Leftrightarrow \frac{{{\bf{Δ }}Y}}{Y} = \; \pm \;\left( {a\;\frac{{{\bf{Δ }}A}}{A} + b\;\frac{{{\bf{Δ }}B}}{B} + c\;\frac{{{\bf{Δ }}C}}{C}\;} \right)\)

[Where ΔY = change in value, Y = original value, a = power of first element, again change in A and follows...]

CALCULATION:

The area is given as 

A = l × b

Given length l = 5, error Δ l = 0.1

Breadth b = 2,  error Δ b = 0.01

Area on the measured value 

The error expression of this relation is given as:

\(\frac{Δ A}{A} = \frac{Δ l}{l} + \frac{Δ b}{b}\)

\(\implies \frac{Δ A}{10} = \frac{0.1}{5} + \frac{0.01}{2}\) = 0.025

⇒ Δ A = 0.25

So, the error in measurement of the area is 0.25

The area will be A ± Δ A  = 10 ± 0.25

So, the correct option is 10 ± 0.25

CONCEPT:

• The instrument which is used to measure the length is called length measuring device.
  • The Screw gauge, meter scale, and Vernier caliper are generally the lengths measuring instrument.
• Screw gauge: A length measuring device that is mainly used to measure the diameter of a thin wire is called screw gauge.

EXPLANATION:

• The screw gauge is used to measure the length(Diameter is also a type of length). So option 1 is correct.
• We can't measure mass and density by using a screw gauge.

Explanation:

Given, Z = \(\frac{A^{2} B^{3}}{C^{4}}\)

or, Z = A²B³C^-4   ----- (1)

∵ Errors always get added, we can write equation (1) in terms of relative error as-

\(\frac{\Delta Z}{Z}=2\frac{\Delta A}{A} + 3\frac{\Delta B}{B}+4\frac{\Delta C}{C}\)

Hence, Option 3) is the correct choice.

CONCEPT:

• Error: While doing any experiment due to faulty equipment, carelessness, or any other random cause the final results get affected.
  • To resolve the percentage change we use the formula:
    • \(Y = \frac{{{A^a}\;.\;\;{B^b}}}{{{C^c}}} \Leftrightarrow \frac{{{\bf{\Delta }}Y}}{Y} = \; \pm \;\left( {a\;\frac{{{\bf{\Delta }}A}}{A} + b\;\frac{{{\bf{\Delta }}B}}{B} + c\;\frac{{{\bf{\Delta }}C}}{C}\;} \right)\) [Where ΔY = change in value, Y = original value, a = power of first element, again change in A and follows...]

CALCULATION:

Given - % change in length = ± 6% and % change in gravity = ± 2%, the time period of simple pendulum is

\( \Rightarrow T = \sqrt {\frac{l}{g}} \; \Leftrightarrow \frac{{{\rm{\Delta }}T}}{T}\; = \frac{{\sqrt l }}{{\sqrt g }}\; = \;\frac{{{l^{\frac{1}{2}}}}}{{{g^{\frac{1}{2}}}}}\)

\( \Rightarrow \frac{{{\rm{\Delta }}T}}{T} = \pm \left( {\frac{1}{2}\;\frac{{{\rm{\Delta }}L}}{L}\; + \;\frac{1}{2}\;\frac{{{\rm{\Delta }}g}}{g}} \right)\)

\( \Rightarrow \% \frac{{{\rm{\Delta }}T}}{T} = \; \pm \left( {\frac{1}{2}\left( {\frac{{{\rm{\Delta }}L}}{L} \times 100} \right) + \frac{1}{2}\left( {\frac{{{\rm{\Delta }}g}}{g} \times 100} \right)} \right)\)

\(\Rightarrow \% \frac{{{\rm{\Delta }}T}}{T} = \; \pm \left( {\left( {\frac{1}{2} \times 6} \right) + \left( {\frac{1}{2} \times 2} \right)} \right)\; \Rightarrow \; \pm 4\% \)

CONCEPT:

• Least Count: The smallest value that an instrument can measure accurately is called least count.
  • All the readings or measured values are accurate only up to this value.

EXPLANATION:

• The smallest value that an instrument can measure accurately is called least count.
• This is the least value that can be measured by the instrument.
• The least count error is the error associated with the resolution of the instrument.
• The least count of a vernier caliper is 0.01 cm.
• The least count of a spherometer is 0.001 cm.
• So the correct answer is option 4.

CONCEPT:

• Percentage error is the difference between theoretical value and an experimental, divided by the theoretical value, multiplied by 100 to give a percent.
• To calculate percentage error in the expression A = xmyn / zp . We use formula

\(\frac{{\Delta A}}{A} × 100 = m\frac{{\Delta x}}{x}× 100 + n\frac{{\Delta y}}{y} × 100+ p\frac{{\Delta z}}{z}× 100\)

where \(\frac{{\Delta A}}{A} × 100\) is the percentage error in A and \(\frac{{\Delta x}}{x} × 100\) is the percentage error in x.

• The Kinetic Energy (E) of an object due to its linear speed is given by:

​​E = 1/2 (m × v2) 

where m is the mass of a body and v is the speed.

CALCULATION:

Given in the question:

\(\frac{{\Delta m}}{m}× 100 = 1\%\)

\(\frac{{\Delta v}}{v}× 100 = 2\%\)

Now we have to find the error in ​​E = 1/2 (m × v2) 

\(\frac{{\Delta E}}{E} × 100 = \frac{{\Delta M}}{M}× 100 + 2\frac{{\Delta v}}{v} × 100\)

\(\frac{{\Delta E}}{E} × 100 =1 + 2\times 2=5 \%\)

∴ % error in the estimation of the kinetic energy of the body is = 5%

• So the correct answer will option 4.

CONCEPT:

• Error: The result of every measurement of experiments by any measuring instrument contains some uncertainty. This uncertainty is called error. 
• Systematic errors: The errors that tend to be in one direction only, either positive or negative, and occur due to a systematic problem
• Systematic errors can occur due to many reasons:

1. Instrumental error: Error due to instruments itself
2. Imperfection in experimental technique or procedure: When we did not use an instrument correctly
3. Personal errors: Due to a person's carelessness

EXPLANATION:

• Instrumental error is one of the systematic errors.
• Instrumental errors occur due to errors in the imperfect calibration of the measuring instrument, zero error in the instrument, or imperfect design, etc.
• If we take an example of a thermometer, the temperature graduations of a thermometer may be inadequately calibrated (it may read 104 °C at the boiling point of water at STP whereas it should read 100 °C);
• In another example of vernier calipers:  the zero mark of the vernier scale may not coincide with the zero marks of the main scale.
• Since the reason for the error can be both imperfect design and imperfect calibration of the measuring instrument, So both options 1 and 2 are correct.
• Hence the correct answer is option 3.

CONCEPT:

To get the maximum percentage error in the given physical quantities we have to differentiate the given equation in terms of the error that occurred in the measurement. In a more general way, it can be written as;

\(\frac{{\Delta X}}{X} = \frac{{\Delta A}}{A} + \frac{{\Delta B}}{B} + \frac{{\Delta C}}{C} + \frac{{\Delta D}}{D}\)

The maximum percentage error is

\(\frac{{\Delta X}}{X}\times 100 = \frac{{\Delta A}}{A}\times 100 + \frac{{\Delta B}}{B}\times 100 + \frac{{\Delta C}}{C} \times 100+ \frac{{\Delta D}}{D}\times 100\)

Here, X, A, B, C, and D are the measurement of physical quantities.

CALCULATION:

Given:\(X = \frac{{{A^2}{B^{\frac{1}{2}}}}}{{{C^{\frac{1}{3}}}{D^3}}}\) ----(1)

Now, differentiate the equation (1) we have;

\(\frac{{\Delta X}}{X} =2 \frac{{\Delta A}}{A} +\frac{1}{2} \frac{{\Delta B}}{B} +\frac{1}{3} \frac{{\Delta C}}{C} + 3\frac{{\Delta D}}{D}\)

Now,maximum percentage error is,

 \(\frac{{\Delta X}}{X} \times 100 = 2\frac{{\Delta A}}{A} \times 100 + \frac{1}{2}\frac{{\Delta B}}{B} \times 100 + \frac{1}{3}\frac{{\Delta C}}{C} \times 100 + 3\frac{{\Delta D}}{D} \times 100\)

                  \( = 2 \times 1\% + \frac{1}{2} \times 2\% + \frac{1}{3} \times 3\% + 3 \times 4\% \)

                  = 2% + 1% + 1% + 12%

                  = 16%

Hence, option 2) is the correct answer.

CONCEPT:

• The difference in the true value and measured value of a quantity is called error of measurement.
• Error in the sum of two quantities: If ΔA and ΔB are the absolute errors in the two quantities A and B respectively. Then,

The measured value of A = A ± ΔA

The measured value of B = B ± ΔB

Consider the sum, Z = A + B

The error ΔZ in Z is then given by

ΔZ = Z ± (ΔA + ΔB)  

CALCULATION:

Given - R1 = 36 Ω ± 1.89 Ω and R2 = 75 Ω ± 3.75 Ω

In series combination,

⇒ Req = R1 + R2

⇒ Req = (36 ± 1.89) + (75 ± 3.75) = 111 ± (1.89 + 3.75) = (111 ± 5.64) Ω