Any measurement that you make is just an approximation, 100% accuracy is not possible. If you measure the same object two different times, the two measurements may not be exactly the same. The difference between the two measurements is called a variation in the measurements. This variation introduces an unwanted but unavoidable uncertainty in our measurement. This uncertainty is called the Errors in measurement. This ‘error’ should not be confused with a ‘mistake’. Error, unlike mistake, does not mean that you got the wrong answer. It just means you didn’t get as close to the true value as possible. The errors in measurement are a mathematical way to show the uncertainty in the measurement. It is the difference between the result of the measurement and the true value of what you were measuring.
Absolute Error and Relative Error
Two important concepts in errors in measurement are absolute error and relative error. Absolute error is the actual amount of error of the measurement. In other words, it is the difference between the actual or accepted value of measured value and the result of the measurement.
To discuss an example, if the actual distance between Bengaluru and Mumbai is accepted as 980.900 kms but you have measured it as 980.899 kms, then the absolute error in your measurement is 0.001 km (or 1 metre). Similarly, if there is 100 gms of a chemical which has been measured by you as 99 gms, then the absolute error in your measurement is 1 gram. You might note that although the absolute error in both the examples is of the same value (1 metre and 1 gram), the error is more grave in the second example. An error of 1 metre distance when measuring the difference between 2 cities is clearly less significant than an error of 1 gram when measuring 100 grams of a chemical for any other experiment. This brings us to the concept of relative error. Relative error gives us the ratio of the absolute error to the accepted measurement.
Important Formulas:
| Absolute error = Measured value – Accepted value
Relative error = (Measured value – Accepted value)/ Accepted value |
JEE Main Previous Year Solved Questions on Error in Measurement and Instruments
Q1: A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?
(a) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm
(b) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm
(c) A meter scale
(d) A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in 1 cm
Solution
The measured distance by the student is 3.50 cm which means the least count of the measurement done is 0.01 cm.
The least count of a vernier calliper which has 10 divisions in 1cm on the main scale and 10 divisions in vernier scale match with 9 division in the main scale can be given as
1 cm/(10 x 10) = 0.01 cm
Therefore, the student must have used the vernier calliper for the measurement
Answer: (d) A vernier calliper where the 10 divisions in vernier scale match with 9 divisions in the main scale and main scale have 10 divisions in 1 cm
Q2: In an experiment, the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree(= 0.50), then the least count of the instrument is
(a) One degree
(b) Half degree
(c) One minute
(d) Half minute
Solution
30 divisions of the vernier scale coincide with 29 divisions of the main scale.
Therefore, 1 V.S.D = (29/30) M.S.D
Least Count = 1 MSD – 1 VSD
= 1 MSD – (29/30)MSD
= (1/30)MSD
= (1/30) (0.50)
= (1/30) x 30 min = 1 min
Answer: (c) One minute
Q3: A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : 0 mm
Circular scale reading: 52 divisions
Given that 1 mm on the main scale corresponds to 100 divisions of the circular scale
The diameter of the wire from the above date is
(a) 0.026 cm
(b) 0.005 cm
(c) 0.52 cm
(d) 0.052 cm
Solution
Here, the pitch of the screw gauge, p =1 mm
Number of circular divisions, n=100
Thus, least count LC=p/n=1/100=0.01 mm=0.001 cm
Diameter of the wire =MSR+(CSR×LC)=0+(52×0.001cm)=0.052 cm
Answer: (d) 0.052 cm
Q4: The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is
(a) 3.5 %
(b) 4.5 %
(c) 6 %
(d) 2.5 %
Solution
Density ρ = m/V = m/l3
The maximum percentage error in density (Δρ/ρ) x 100 = (Δm/m) x 100 x 3(Δl/l) x 100
⇒(Δρ/ρ) x 100 = 1.5 + 3 x 1 = 4.5%
Answer: (b) 4.5 %
Q5: Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then the error in the value of resistance of the wire is
(a) 3%
(b) 6%
(c) zero
(d) 1%
Solution
R=V/I
(ΔR/R)×100=(ΔV/V) ×100+(ΔI/I)×100=3+3=6%
Answer: (b) 6%
Q6: The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is
(a) 0.9%
(b) 2.4%
(c) 3.1%
(d) 4.2%
Solution
Least count = (pitch/number of division on circular scale)= (0.5/50) = 0.01 mm
Diameter of solid ball =2.5 + (0.5/50) x 20 = 2.7 mm
As density ρ = mass/volume = M/[(4π/3)(D/2)3]
The relative percentage error in density is
(Δρ/ρ) x 100 = {(ΔM/M) + (3ΔD/D)} x 100
Percentage of error = {(ΔM/M) + (3ΔD/D)} x 100 x 100
=(2 + 3 X (0.01/2.7) x 100) = 3.1%
Answer:(c) 3.1%
Q7: Which of the following sets have different dimensions?
(a) Pressure, Young’s modulus, stress
(b) Emf, potential difference, electric potential
(c) Heat, work done, energy
(d) Dipole moment, electric flux, electric field
Solution
Dipole moment = charge x distance = [LTI]
Electric flux = electric field x area = [ML3T-3I-1]
Electric field = [ML1T-3I-1]
∴ Dipole moment, electric flux and electric field have different dimensions
Answer : (d) Dipole moment, electric flux, electric field
Q8: Planck’s constant has dimensions……………
Solution
h = Planck’s constant
h = Energy(E)/Frequency(f) = [ML2T-2 ]/ [T-1] = [ML2T-1]
Answer: [ML2T-1]
Q9: The least count of the main scale of a screw gauge is 1mm. The minimum number of divisions on its circular scale required to measure 5 μm diameter of a wire is
(a) 50
(b) 200
(c) 100
(d) 500
Solution
Least count of the main scale of screw gauge = 1mm
Least count of screw gauge = Pitch/Number of division on the circular scale
5 x 10-6 = 10-3/N
N = 200
Answer: (b) 200
Q10: A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is
(a) 50
(b) 200
(c) 100
(d) 500
Solution
Resistance,R = ρl/A
R = (ρl/A) x (l/l) = (ρl2/v) = (∵ Volume (V) = Al)
Since resistivity and volume remains constant therefore percentage change in resistance
ΔR/R = (2Δl/l) = 2 x (0.5) = 1%
Answer: (c) 100
Also Read:
Error In Measurements Instrumentsf JEE Advanced Previous Year Questions With Solutions
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