JEE Advanced Previous Year Questions with Solutions on Error in Measurement Instruments

The error in any instrument is the difference between the value indicated by the instrument and the actual value. The error in a measurement can be due to various factors. The errors can be systematic or random. The systematic error tends to be in a single direction, either positive or negative. The random errors occur irregularly and hence are random with respect to size and sign.

Download JEE Advanced Previous Year Questions with Solutions on Error in Measurement Instruments PDF

Question 1) A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He uses a stopwatch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true?

(A) Error △T in measuring T, the time period is 0.05 seconds

(B) Error △T in measuring T, the time period is 1 second

(C) Percentage error in the determination of g is 5%

(D) Percentage error in the determination of g is 2.5%

Answer: (A) and (C)

Solution:

The relative error in the measurement of time,

Δt/t = 1s/40s =1/40

Time period, T= 40s/20 =2s

Error in the measurement of the time period, ΔT=T× (Δt/t)

=2s × (1/40) =0.05s

The time period of a simple pendulum is, T =2π√(l/g)

Therefore, T²=(4π²l)/g

⇒ g = (4π²l)/T²

Therefore, (Δg/g) = 2ΔT/T

=2× (0.05/2)

= 0.05

(Δg/g) x 100 = 0.05 x 100 = 5%

Question 2)To verify Ohm’s law, a student is provided with a test resistor R_T, a high resistance R₁, a small resistance R₂, two identical galvanometers G₁ and G₂, and a variable voltage source V. The correct circuit to carry out the experiment is

Answer: (C)

Solution:

JEE Advanced Chapter-Wise Solutions on Error in measurements

Question 3) A Vernier Calliper has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier callipers, the least count is

(A) 0.02 mm

(B) 0.05 mm

(C) 0.1 mm

(D) 0.2 mm

Answer: (D) 0.2 mm

Solution:

The least count may be calculated as the ratio between the smallest division on the main scale by the number of divisions on the vernier scale.

So, 20 divisions of vernier scale = 16 divisions of main scale.

Therefore, 1 VSD = 16/20 = 0.8 mm

Also Least count = 1MSD – 1VSD

= 0.2 mm

Question 4) A meter bridge is set up as shown, to determine an unknown resistance ‘X’ using a standard 10 ohm resistor. The galvanometer shows null point when the tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for ends A and B. The determined value of ‘X’ is –

(A) 10.2 ohm

(B) 10.6 ohm

(C) 10.8 ohm

(D) 11.1 ohm

Answer: (B) 10.6 ohm

Solution:

At null point

(X/l₁) = (10/l₂)

Here, l₁ = 52 + end correction

= 52 + 1 = 53 cm

l₂ = 48 + end correction

= 48 + 2 = 50 cm

Therefore, (X/53) = (10/50)

X = 53/5 = 10.6 Ω

Question 5) 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%

Answer: (C) 3.1%

Solution:

Least count of the screw gauge = 0.5/50 = 0.01 mm

Diameter of the ball, D = 2.5 mm + 20(0.01)

= 2.5 + 0.2 = 2.7 mm

Density, ρ = M/vol

\(\begin{array}{l}= \frac{M}{\frac{4}{3}\pi (\frac{D}{2})^{3}}\end{array} \)

The relative percentage error in density

\(\begin{array}{l}\frac{\Delta \rho }{\rho } = \left ( \frac{\Delta M}{M}+\frac{3\Delta D}{D} \right )\end{array} \)

= 2% + (3(0.01/2.7)x 100)

= 3.1%

Question 6)In the determination of Young’s modulus Y = (4MLg/πld²) by using Searle’s method, a wire of length L = 2m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometre, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement.

(A) due to the errors in the measurements of d and l are the same.

(B) due to the error in the measurement of d is twice that due to the error in the measurement of l.

(C) due to the error in the measurement of l is twice that due to the error in the measurement of d.

(D) due to the error in the measurement of d is four times that due to the error in the measurement of l.

Answer: (A) due to the errors in the measurements of d and l are the same.

Solution:

Given,

Y = (4MLg/πld²)

l = 0.25 mm

d = 0.5 mm

Since the pitch and the number of divisions on the circular scale is the same,

Δl = Δd

\(\begin{array}{l}\frac{\Delta Y}{Y}=\frac{\Delta M}{M}+\frac{\Delta L}{L}+\frac{\Delta l}{l}+2\frac{\Delta d}{d}\end{array} \)

\(\begin{array}{l}\frac{\Delta Y}{Y}=\frac{\Delta M}{M}+\frac{\Delta L}{L}+\frac{\Delta l}{0.25}+2\frac{\Delta d}{0.5}\end{array} \)

\(\begin{array}{l}\frac{\Delta Y}{Y}=\frac{\Delta M}{M}+\frac{\Delta L}{L}+ 4\Delta l+4\Delta d\end{array} \)

Question 7)The diameter of a cylinder is measured using Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is :

(A) 5.112 cm

(B) 5.124 cm

(C) 5.136 cm

(D) 5.148 cm

Answer: (B) 5.124 cm

Solution:

The readings on the main scale are 5.10 and 5.15

1 main scale divisions = 0.05 cm

1 vernier scale division, = 2.45/50 = 0.049 cm/div

Least count = 0.05 – 0.049 = 0.001 cm

Diameter = MSR + (L.C x VSD)

= 5.10 + (24 x 0.001)

= 5.124 cm

Question 8)Using the expression 2d sinθ =λ, one calculates the values of d by measuring the corresponding angles θ in the range 0 to 90⁰. The wavelength λ is exactly known and the error in θ is constant for all values of θ. As θ increases from 0⁰

(A) the absolute error in d remains constant

(B) the absolute error in d increases

(C) the fractional error in d remains constant

(D) the fractional error in d decreases

Answer: (D) the fractional error in d decreases

Solution:

2dsinθ = λ

⇒ d = λ/2sinθ

\(\begin{array}{l}\frac{\Delta d}{d}=\frac{\Delta \lambda }{\lambda }-\frac{1}{sin\theta }cos\theta \Delta \theta\end{array} \)

For maximum possible error, all the errors must be added

\(\begin{array}{l}\frac{\Delta d}{d}=\frac{\Delta \lambda }{\lambda }+\frac{cos\theta }{sin\theta } \Delta \theta\end{array} \)

As λ is exactly known, Δλ = 0

⇒ Δd/d = cotθΔθ

As Δθ is a constant, on increasing θ, cotθ will decrease. So, the fractional error (Δd/d) will also decrease.

Question 9)During Searle’s experiment, zero of the Vernier scale lies between 3.20 x 10^-2 m and 3.25 x 10^-2 m of the main scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 x 10^-2m and 3.25 x 10^-2 m of the main scale but now the 45th division of the Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2m and its cross-sectional area is 8 x 10^-7 m². The least count of the Vernier scale is 1.0 x 10^-5 m. The maximum percentage error in Young’s modulus of the wire is.

Answer: 4%

Solution:

Y = (FL/Al)

(△Y/Y) = (△l/l)

Change in length, l = (45 – 20)x 10^-5 m = 25 x 10^-5 m

△l = 10^-5 m

Therefore, (△Y/Y) x 100= 10^-5/25 x 10^-5

= 1/25 = 4%

Question 10)Consider Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:

(A) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.

(B) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

(C) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.

(D) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

Answer: (B) and (C)

Solution:

1 cm on the main scale is divided into 8 equal divisions

1 MSD = ⅛ cm

The least count of the vernier is given as 1M.S.D – 1V.S.D

Given 5 vernier scale divisions coincides with 4 main scale divisions.

5 V.S.D = 4 M.S.D

L.C = 1M.S.D – (⅘) M.S.D

= (⅕) M.S.D = (⅕)(⅛) = 1/40

Checking the options (A) and (B)

Given pitch of screw gauge twice the least count of the Vernier callipers

p = 2 x (1/40) = (1/20) cm

The least count of screw gauge is = pitch/number of divisions on the circular scale

Number of divisions on the circular scale = (1/20)/100 = 0.0005 cm = 0.005 mm

Hence option (B) is correct

Checking the option (C) and (D)

Given least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers

One complete rotation of the circular scale moves it by 2 divisions on the linear scale, so the pitch is 2 times the linear scale division of screw gauge

Pitch = 2 x linear scale division of screw gauge

= 2 x 2(least count of the vernier callipers)

= 4 x (1/40) = 1/10

The least count of the screw gauge = pitch/number of divisions on circular scale

= (1/10)/100

= 0.001 cm = 0.01 mm

Question 11) The energy of a system as a function of time t is given as E(t) = A²e^-αt, where α = 0.2 s^-1 . The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E(t) at t = 5 s is

Answer: 4%

Solution:

E(t) = A²e^-αt, for a small percentage of error we can do differentiation.

dE = 2A(dA)e^-αt+ A²(-αe^-αtdt)

Fractional error, dE/E = [2A(dA)e^-αt+ A²(-αe^-αtdt)]/A²e^-αt

= 2(dA/A) + (-α dt)

Error is always taken as positive

= 2(dA/A) + α dt

Percentage of error in E as a function of t = (2(dA/A) + αtdt/t) x 100

= 2(1.25) + (0.2)(1.5)(5)

= 2.5 + 1.5

= 4%

Question 12) There are two vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of the calipers C₁ has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper C₂ has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C₁ and C₂ respectively are

(A) 2.87 and 2.86

(B) 2.87 and 2.87

(C) 2.87 and 2.83

(D) 2.85 and 2.82

Answer: (C) 2.87 and 2.83

Solution:

MSD = (1/10) cm = 0.1 cm

1st Vernier Caliper,

10 VSD = 9 MSD

⇒ VSD = (9/10) MSD

= 0.9 x 0.1 = 0.09 cm

Least count = 0.1 – 0.09 = 0.01 cm

Reading = Main scale reading + (least count x Vernier scale division coinciding with the main scale)

= 2.8 + (0.01 x 7)

= 2.87 cm

2nd Vernier Scale

10 VSD = 11 MSD

VSD = (11/10) MSD

= 1.1 MSD

= 1.1 x 0.1 = 0.11 cm

Least count = 0.1 – 0.11 = -0.01 mm

Reading = Main scale reading + (least count x Vernier scale division coinciding with the main scale)

= 2.8 + (10-7) x 0.01

= 2.8 + 0.03 = 2.83 cm

Question 13) In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is

\(\begin{array}{l}T =2\pi \sqrt{\frac{7(R-r)}{5g}}\end{array} \)

. The values of R and r are measured to be (60 ± 1) mm and (10 ± 1) , respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true?

(A) The error in the measurement of r is 10%

(B) The error in the measurement of T is 3.57 %

(C) The error in the measurement of T is 2%

(D) The error in the determined value of g is 11%

Answer: (A), (B) and (C)

Solution:

Given,

Values of the time period are found to be 0.52 s, 0.56 s, 0.57s, 0.54 s and 0.59 s

The mean value of the time period, T = ΣTi/5

= 2.78/5

= 0.56 s

The magnitude of the absolute error in each observation,

|△T₁| = 0.56 – 0.52 = 0.04 s

|△T₂| = 0.56 – 0.56 = 0 s

|△T₃| = 0.57 – 0.56 = 0.01 s

|△T₄| = 0.56 – 0.54 = 0.02 s

|△T₅| = 0.59 – 0.56 = 0.03 s

Mean absolute error in time period, ΔT_m = (0.04 + 0+ 0.01+0.02+0.03)/5 = 0.02 s

Error in T, (ΔT_m/T) x 100 = (0.02/0.56) x 100 = 3.57%

Error in the measurement of r, (Δr/r) x 100 = (1/10) x 100 = 10%

\(\begin{array}{l}T =2\pi \sqrt{\frac{7(R-r)}{5g}}\end{array} \)

\(\begin{array}{l}T^{2} =4\pi^{2} [\frac{7(R-r)}{5g}]\end{array} \)

\(\begin{array}{l}\Rightarrow g =\frac{4\pi^{2} 7}{5g} \frac{(R-r)}{T^{2}}\end{array} \)

Therefore,

\(\begin{array}{l}\frac{\Delta g}{g}\times 100 =\frac{\Delta R + \Delta r}{(R-r)}\times 100+ 2\frac{\Delta T}{T}\times 100\end{array} \)

\(\begin{array}{l}\frac{\Delta g}{g}\times 100 =[\frac{1 + 1}{(60-10)}\times 100]+ (2\times 3.57)\end{array} \)

= 4 + 7.14 ≈ 11%

Question 14) A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is δT = 0.01 second and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity g = 10 m² and the velocity of sound is 300 m/s. Then the fractional error in the measurement, δL/L, is closest to

(A) 0.2 %

(B) 5 %

(C) 3 %

(D) 1 %

Answer: (D) 1 %

Solution:

Total time taken,

\(\begin{array}{l}T = \sqrt{\frac{2L}{g}}+\frac{L}{c}\end{array} \)

Now for an error δL in L we will have an error δT in T

So,

\(\begin{array}{l}T+\delta T = \sqrt{\frac{2(L+\delta L)}{g}}+\frac{(L+\delta L)}{c}\end{array} \)

\(\begin{array}{l}T+\delta T = \sqrt{\frac{2L}{g}(1+\frac{\delta L}{L})}+\frac{L}{c}(1+\frac{\delta L}{L})\end{array} \)

Sine δT/T is very small so δL/L is also very small, so taking the binomial approximation

\(\begin{array}{l}T+\delta T = \sqrt{\frac{2L}{g}}(1+\frac{\delta L}{2L})+\frac{L}{c}(1+\frac{\delta L}{L})\end{array} \)

\(\begin{array}{l}T+\delta T = \sqrt{\frac{2L}{g}}+\sqrt{\frac{2L}{g}}(\frac{\delta L}{2L})+\frac{L}{c}+\frac{L}{c}(\frac{\delta L}{L})\end{array} \)

\(\begin{array}{l}T+\delta T = \left [\sqrt{\frac{2L}{g}} +\frac{L}{c} \right ]+\left [ \frac{1}{2}\sqrt{\frac{2L}{g}}+\frac{L}{c}\right ](\frac{\delta L}{L})\end{array} \)

\(\begin{array}{l}T+\delta T = T+\left [ \frac{1}{2}\sqrt{\frac{2L}{g}}+\frac{L}{c}\right ](\frac{\delta L}{L})\end{array} \)

\(\begin{array}{l}\delta T =\left [ \frac{1}{2}\sqrt{\frac{2\times 20}{10}}+\frac{20}{300}\right ](\frac{\delta L}{L})\end{array} \)

\(\begin{array}{l}\delta T =\left [ 1+\frac{1}{15}\right ](\frac{\delta L}{L})\end{array} \)

⇒δL/L = (15/16)δT

= (15/16) x 0.01

(δL/L) x 100 = (15/16) ≈ 1%

Question 15) If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z = x/y. If the errors in x, y and z are △x, △y and △z, respectively, then

\(\begin{array}{l}z\pm \Delta z=\frac{x+\Delta x}{y+\Delta y}=\frac{x}{y}\left ( 1\pm \frac{\Delta x}{x} \right )\left ( 1\pm \frac{\Delta y}{y} \right )^{-1}\end{array} \)

. The series expansion for
\(\begin{array}{l}\left (1\pm \frac{\Delta y}{y} \right )^{-1}\end{array} \)

, to first power in Δy/y, is 1土 Δy/y. The relative errors in independent variables are always added. So the error in z will be

\(\begin{array}{l}\Delta z=z\left ( \frac{\Delta x}{x}+\frac{\Delta y}{y} \right )\end{array} \)

. The above derivation makes the assumption that △x/x ≪ 1,△y/y ≪ 1 . Therefore, the higher powers of these quantities are neglected.

Q. Consider the ratio r = (1-a)/(1+a) to be determined by measuring a dimensionless quantity a. If the error in the measurement of a is △a (△a/a ≪1), then what is the error △r in determining r ?

(A)

\(\begin{array}{l}\frac{\Delta a}{(1+a)^{2}}\end{array} \)

(B)

\(\begin{array}{l}\frac{2 \Delta a}{(1+a)^{2}}\end{array} \)

(C)

\(\begin{array}{l}\frac{2 \Delta a}{\left(1-a^{2}\right)}\end{array} \)

(D)

\(\begin{array}{l}\frac{2 a \Delta a}{\left(1-a^{2}\right)}\end{array} \)

Answer: (B)

Solution:

r = (1-a)/(1+a)

\(\begin{array}{l}\frac{\Delta r}{r}=\frac{\Delta (1-a)}{(1-a)}+ \frac{\Delta (1+ a)}{(1+a)}\end{array} \)

\(\begin{array}{l}\frac{\Delta r}{r}=\frac{\Delta a}{(1-a)}+ \frac{\Delta a)}{(1+a)}\end{array} \)

\(\begin{array}{l}\frac{\Delta r}{r}= \frac{\Delta a(1+a+1-a)}{(1-a)(1+a)}\end{array} \)

\(\begin{array}{l}\frac{\Delta r}{r}= \frac{2\Delta a}{(1-a)(1+a)}\end{array} \)

\(\begin{array}{l}\Delta r= \frac{2\Delta a}{(1-a)(1+a)}\frac{(1-a)}{(1+a)}\end{array} \)

\(\begin{array}{l}=\frac{2 \Delta a}{(1+a)^{2}}\end{array} \)

Also Read:

JEE Main Error in Measurement and Instruments Previous Year Questions with Solutions

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