JEE Physics Errors in Measurement MCQs Set 04

Multiple Choice Questions (MCQs) for JEE Physics: Errors in Measurement

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Practice Errors in Measurement MCQs for JEE Physics

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Multiple Choice Questions

Question 1: The value of a resistance is 10.845 Ω and the value of the current is 3.23 A. The potential difference in the correct number of significant figures would be:
(a) 35.03 V
(b) 35 V
(c) 35.0 V
(d) 35.029 V
Show Answer & Explanation

Answer: (c) 35.0 V

Explanation:
1. \( V \) = IR = \( 3.23 \times 10.845 = 35.02935 V \).
2. The current has only 3 significant figures, so the answer is rounded to \( 3 s.f \).: \( 35.0 V \).

Teacher's Note:
1. A product keeps as many significant figures as the least precise factor.
2. 3.23 has three significant figures.

Question 2: The least count of a stop watch is \( \frac15 \) s. The time for 20 oscillations of a pendulum is measured to be 25 s. What is the maximum percentage error in this measurement?
(a) 8%
(b) 0.8%
(c) 16%
(d) 1%
Show Answer & Explanation

Answer: (b) 0.8%

Explanation:
1. Maximum error = least count = \( 0.2 s \).
2. Percentage error = \( (0.2/25) \times 100 \) = 0.8%.

Teacher's Note:
1. Take the least count as the maximum absolute error.
2. Divide by the measured value.

Question 3: The refractive index of water measured by the relation μ = real depth/apparent depth is found to have the values 1.34, 1.38, 1.32 and 1.36. The mean value of the refractive index with percentage error is:
(a) 1.36 ± 0%
(b) 1.35 ± 1.48%
(c) 1.35 ± 0%
(d) 1.36 ± 6%
Show Answer & Explanation

Answer: (b) 1.35 ± 1.48%

Explanation:
1. Mean = \( (1.34 + 1.38 + 1.32 + 1.36)/4 = 1.35 \).
2. Absolute errors: 0.01, 0.03, 0.03, 0.01; mean = 0.02.
3. Percentage error = \( (0.02/1.35) \times 100 \) = 1.48%.

Teacher's Note:
1. Find the mean first, then the mean absolute deviation.
2. Divide by the mean for the percentage error.

Question 4: A wire has a mass 0.3 ± 0.003 g, radius 0.5 ± 0.005 mm and length 6 ± 0.06 cm. The maximum percentage error in the measurement of its density is:
(a) 4
(b) 1
(c) 3
(d) 2
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. ρ = \( m/(\pir^{2}l) \).
2. Percentage error = 1% (mass) \( + 2 \) × 1% (radius) + 1% (length) = 4%.

Teacher's Note:
1. The percentage error in \( r \) counts twice because \( r \) is squared.
2. Add percentage errors for products and quotients.

Question 5: A physical quantity X is given by \( X=\frac{a^3b^2d}{c^{1/2}} \). The percentage errors in the measurement of a, b, c and d are 1%, 3%, 2% and 4% respectively. The maximum percentage error in X is:
(a) 14%
(b) 10%
(c) 11%
(d) 19%
Show Answer & Explanation

Answer: (a) 14%

Explanation:
1. Percentage error = \( 3(1) + 2(3) + (1/2)(2) + 4 = 3 + 6 + 1 + 4 \) = 14%.

Teacher's Note:
1. Multiply each percentage error by the power of its quantity.
2. Add them all.

Question 6: One centimetre on the main scale of a vernier calliper is divided into 10 equal parts. If 10 divisions of the vernier coincide with 8 small divisions of the main scale, the least count of the calliper is:
(a) 0.05 cm
(b) 0.01 cm
(c) 0.02 cm
(d) 0.005 cm
Show Answer & Explanation

Answer: (c) 0.02 cm

Explanation:
1. 1 MSD = 1 mm and 10 VSD = 8 MSD, so 1 VSD = 0.8 mm.
2. Least count = \( 1 - 0.8 = 0.2 \) mm = 0.02 cm.

Teacher's Note:
1. Least count = 1 MSD - 1 VSD.
2. Convert to centimetres at the end.

Question 7: While measuring the length of a rod with a vernier calliper, the reading on the main scale is 6.4 cm and the eighth division on the vernier is in line with a marking on the main scale. If the least count of the calliper is 0.01 cm and the zero error is -0.04 cm, the length of the rod is:
(a) 6.60 cm
(b) 6.48 cm
(c) 6.52 cm
(d) 6.50 cm
Show Answer & Explanation

Answer: (c) 6.52 cm

Explanation:
1. Observed reading = \( 6.4 + 8 \times 0.01 = 6.48 \) cm.
2. Zero correction = \( +0.04 \) cm, so the true length = 6.52 cm.

Teacher's Note:
1. Subtract the zero error (negative zero error means add).
2. Reading = MSR + VSR × LC.

Question 8: A body of mass m = 3.513 kg moves along the x-axis with a speed of 5.00 m/s. The magnitude of its momentum is recorded as:
(a) 17.56 kg m/s
(b) 17.57 kg m/s
(c) 17.6 kg m/s
(d) 17.565 kg m/s
Show Answer & Explanation

Answer: (c) 17.6 kg m/s

Explanation:
1. \( p = 3.513 \times 5.00 = 17.565 \) kg \( m/s \).
2. The speed has only 3 significant figures, so \( p = 17.6 \) kg \( m/s \).

Teacher's Note:
1. Round the product to the least number of significant figures.
2. 5.00 has three significant figures.

Question 9: 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 is:
(a) 0.026 cm
(b) 0.52 cm
(c) 0.005 cm
(d) 0.052 cm
Show Answer & Explanation

Answer: (d) 0.052 cm

Explanation:
1. Least count = 1 mm/100 = 0.01 mm.
2. Diameter = \( 0 + 52 \times 0.01 \) mm = 0.52 mm = 0.052 cm.

Teacher's Note:
1. Convert millimetres to centimetres at the end.
2. Least count = pitch/number of divisions.

Question 10: A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading: 58.5 degree. Vernier scale reading: 09 divisions. Given that 1 division on the main scale corresponds to 0.5 degree and the total divisions on the vernier scale match with 29 divisions of the main scale (30 vernier divisions), the angle of the prism is:
(a) 58.59 degree
(b) 58.77 degree
(c) 58.65 degree
(d) 59 degree
Show Answer & Explanation

Answer: (c) 58.65 degree

Explanation:
1. Least count = \( 0.5^{\circ}/30 = 1/60 \) degree.
2. Reading = \( 58.5^{\circ} + 9 \times (1/60)^{\circ} = 58.5^{\circ} + 0.15^{\circ} = 58.65^{\circ} \).

Teacher's Note:
1. Find the least count from the scale ratio.
2. Add main scale and vernier readings.

Question 11: The current-voltage relation of a diode is given by \( I=(e^{1000V/T}-1)\ \text{mA} \), where the applied voltage V is in volts and the temperature T is in kelvin. If a student makes an error of ± 0.01 V while measuring the voltage at 300 K and the current is 5 mA, what will be the error in the value of the current in mA?
(a) 0.2 mA
(b) 0.02 mA
(c) 0.05 mA
(d) 0.5 mA
Show Answer & Explanation

Answer: (a) 0.2 mA

Explanation:
1. I = 5 mA gives \( e^{1000V/T} = 6 \).
2. dI/dV = \( (1000/T)e^{1000V/T} = (1000/300)(6) = 20 \) mA/V.
3. \( \DeltaI = 20 \times 0.01 = 0.2 \) mA.

Teacher's Note:
1. Differentiate I with respect to \( V \).
2. Use \( e^{1000V/T} = I + 1 = 6 \).

Question 12: The period of oscillation of a simple pendulum is \( T=2\pi\sqrt{L/g} \). The measured value of L is 20.0 cm, known to 1 mm accuracy, and the time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is:
(a) 1%
(b) 5%
(c) 2%
(d) 3%
Show Answer & Explanation

Answer: (d) 3%

Explanation:
1. \( \DeltaL/L = 0.1/20 \) = 0.5%.
2. \( \DeltaT/T = 1/90 \) = 1.1%.
3. \( \Deltag/g \) = 0.5% + 2(1.1%) ≈ 2.7%, which is about 3%.

Teacher's Note:
1. \( g = 4\pi^{2}L/T^{2} \), so the error in \( T \) is doubled.
2. Find the nearest option.

Question 13: A student performs an experiment for the determination of \( g=\frac{4\pi^2l}{T^2} \), l ≈ 1 m, and commits an error Δl. For T he takes the time of n oscillations with a stop watch of least count ΔT and commits a human error of 0.1 s. For which of the following data (see the table) will the measurement of g be most accurate?
OptionΔlΔTnAmplitude of oscillation
(a)5 mm0.2 s105 mm
(b)5 mm0.2 s205 mm
(c)5 mm0.1 s201 mm
(d)1 mm0.1 s501 mm

(a) Option (a): Δl = 5 mm, ΔT = 0.2 s, n = 10
(b) Option (b): Δl = 5 mm, ΔT = 0.2 s, n = 20
(c) Option (c): Δl = 5 mm, ΔT = 0.1 s, n = 20
(d) Option (d): Δl = 1 mm, ΔT = 0.1 s, n = 50
Show Answer & Explanation

Answer: (d) Option (d): Δl = 1 mm, ΔT = 0.1 s, n = 50

Explanation:
1. Δg/g = Δl/l + 2(ΔT + 0.1)/(nT) with \( T \approx 2 s \).
2. \( (a) \): 0.5% \( + 2(0.3)/20 \) = 3.5%. \( (b) \): 0.5% \( + 2(0.3)/40 \) = 2%. \( (c) \): 0.5% \( + 2(0.2)/40 \) = 1.5%. \( (d) \): 0.1% \( + 2(0.2)/100 \) = 0.5%.
3. Option \( (d) \) has the smallest error.

Teacher's Note:
1. A longer total time and smaller least counts reduce the percentage error.
2. Compare \( \Deltal/l \) and \( 2\DeltaT_{total}/(nT) \) for every option.

Question 14: A student performs an experiment to determine the Young's modulus of a wire, exactly 2 m long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. Take g = 9.8 m/s² (exact). The Young's modulus obtained from the reading is close to:
(a) (2.0 ± 0.1) × 10¹¹ N m⁻²
(b) (2.0 ± 0.2) × 10¹¹ N m⁻²
(c) (2.0 ± 0.05) × 10¹¹ N m⁻²
(d) (2.0 ± 0.3) × 10¹¹ N m⁻²
Show Answer & Explanation

Answer: (b) (2.0 ± 0.2) × 10¹¹ N m⁻²

Explanation:
1. \( Y \) = 4FL/(πd²Δl) = \( 4(9.8)(2)/(\pi(0.4 \times 10^{-3})^{2}(0.8 \times 10^{-3})) \approx 1.95 \times 10^{11} \approx 2.0 \times 10^{11} N/m^{2} \).
2. \( \DeltaY/Y = 2(0.01/0.4) + 0.05/0.8 \) = 5% + 6.25% = 11.25%.
3. \( \DeltaY = 0.1125 \times 2.0 \times 10^{11} \approx 0.2 \times 10^{11} N/m^{2} \).

Teacher's Note:
1. Young's modulus depends on \( d^{2} \), so its error counts twice.
2. Add the percentage errors.

Question 15: 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 calliper, the least count is:
(a) 0.05 mm
(b) 0.2 mm
(c) 0.1 mm
(d) 0.02 mm
Show Answer & Explanation

Answer: (b) 0.2 mm

Explanation:
1. 20 VSD = 16 MSD, so 1 VSD = 0.8 mm.
2. LC = \( 1 - 0.8 = 0.2 \) mm.

Teacher's Note:
1. Least count = 1 MSD - 1 VSD.
2. Work in millimetres.

Question 16: A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with a least count of 1 second for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is/are true?
(i) The error ΔT in measuring the time period T is 0.05 second
(ii) The error ΔT in measuring the time period T is 1 second
(iii) The percentage error in the determination of g is 5%
(iv) The percentage error in the determination of g is 2.5%
(a) i, iii
(b) iii, iv
(c) ii, iv
(d) i, ii
Show Answer & Explanation

Answer: (a) i, iii

Explanation:
1. \( T = 40/20 = 2 s \) and \( \DeltaT = 1/20 = 0.05 s \).
2. \( \DeltaT/T \) = 2.5%, so \( \Deltag/g = 2 \) × 2.5% = 5%.
3. Statements \( (i) \) and (iii) are true.

Teacher's Note:
1. The error in the period is the least count divided by the number of oscillations.
2. \( g ∝ 1/T^{2} \), so its percentage error is doubled.

Question 17: 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) 2.4%
(b) 0.9%
(c) 3.1%
(d) 4.2%
Show Answer & Explanation

Answer: (c) 3.1%

Explanation:
1. Least count = \( 0.5/50 = 0.01 \) mm. Diameter = \( 2.5 + 20 \times 0.01 = 2.7 \) mm.
2. Percentage error in \( d = 0.01/2.7 \times 100 \) = 0.37%.
3. ρ \( ∝ m/d^{3} \), so Δρ/ρ = 2% + 3(0.37%) = 3.1%.

Teacher's Note:
1. Density depends on the cube of the diameter.
2. Add the percentage errors with the right powers.

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