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JEE Main Previous Year Questions (2021-2026): Units & Measurements
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Page 1 JEE Main Previous Year Questions (2021-2026): Units & Measurements (January 2026) Q1. The time period of a simple harmonic oscillator is . Measured value of mass (m) of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant (k) is ________%. (a) 3.43 (b) 7.60 (c) 3.35 (d) 6.76 Ans. D Sol. The formula for the time period of simple harmonic oscillator is Using propagation of error, when quantities are multiplied or divided, their relative errors (percentage errors) are added. The constants have no error. The measured value of mass is, m=10g Accuracy/Error in measurement of mass is, ?m = 10mg = 0.01g Page 2 JEE Main Previous Year Questions (2021-2026): Units & Measurements (January 2026) Q1. The time period of a simple harmonic oscillator is . Measured value of mass (m) of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant (k) is ________%. (a) 3.43 (b) 7.60 (c) 3.35 (d) 6.76 Ans. D Sol. The formula for the time period of simple harmonic oscillator is Using propagation of error, when quantities are multiplied or divided, their relative errors (percentage errors) are added. The constants have no error. The measured value of mass is, m=10g Accuracy/Error in measurement of mass is, ?m = 10mg = 0.01g So, percentage error in measurement of mass The percentage error in the time period (T) is the same as the percentage error in the total time (t) measured for n oscillations. T otal time measured is, t=60s Resolution/Error in measurement of time is, ?t=2s So, percentage error in measurement of time is, So, using propagation formula: Percentage error in k = (% error in m) + 2 × (% error in t) = 0.1% + 2 × 3.33% = 6.76% Therefore, the correct option is (d). Q2. The speed of a longitudinal wave in a metallic bar is 400 m/s. If the density and Young's modulus of the bar material are increased by 0.5% and 1%, respectively then the speed of the wave is changed approximately to ______ m/s. (a) 398 (b) 402 (c) 401 (d) 399 Ans. (c) Sol. The formula for the speed of a longitudinal wave ( v ) in a solid metallic bar is given as Where, Y = Young's modulus and its ? is its density Page 3 JEE Main Previous Year Questions (2021-2026): Units & Measurements (January 2026) Q1. The time period of a simple harmonic oscillator is . Measured value of mass (m) of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant (k) is ________%. (a) 3.43 (b) 7.60 (c) 3.35 (d) 6.76 Ans. D Sol. The formula for the time period of simple harmonic oscillator is Using propagation of error, when quantities are multiplied or divided, their relative errors (percentage errors) are added. The constants have no error. The measured value of mass is, m=10g Accuracy/Error in measurement of mass is, ?m = 10mg = 0.01g So, percentage error in measurement of mass The percentage error in the time period (T) is the same as the percentage error in the total time (t) measured for n oscillations. T otal time measured is, t=60s Resolution/Error in measurement of time is, ?t=2s So, percentage error in measurement of time is, So, using propagation formula: Percentage error in k = (% error in m) + 2 × (% error in t) = 0.1% + 2 × 3.33% = 6.76% Therefore, the correct option is (d). Q2. The speed of a longitudinal wave in a metallic bar is 400 m/s. If the density and Young's modulus of the bar material are increased by 0.5% and 1%, respectively then the speed of the wave is changed approximately to ______ m/s. (a) 398 (b) 402 (c) 401 (d) 399 Ans. (c) Sol. The formula for the speed of a longitudinal wave ( v ) in a solid metallic bar is given as Where, Y = Young's modulus and its ? is its density T aking the natural log on both sides of Differentiating both sides gives the fractional change: Initial speed = v = 400m/s Increase in Young's modulus Increase in density So, the percentage change in speed, % change in Since the result is positive, the speed increases by 0.25% So, new speed (v’) Initial Speed + Change in Speed Hence, the correct option is (C). Q3. Match List - I with List - II. Page 4 JEE Main Previous Year Questions (2021-2026): Units & Measurements (January 2026) Q1. The time period of a simple harmonic oscillator is . Measured value of mass (m) of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant (k) is ________%. (a) 3.43 (b) 7.60 (c) 3.35 (d) 6.76 Ans. D Sol. The formula for the time period of simple harmonic oscillator is Using propagation of error, when quantities are multiplied or divided, their relative errors (percentage errors) are added. The constants have no error. The measured value of mass is, m=10g Accuracy/Error in measurement of mass is, ?m = 10mg = 0.01g So, percentage error in measurement of mass The percentage error in the time period (T) is the same as the percentage error in the total time (t) measured for n oscillations. T otal time measured is, t=60s Resolution/Error in measurement of time is, ?t=2s So, percentage error in measurement of time is, So, using propagation formula: Percentage error in k = (% error in m) + 2 × (% error in t) = 0.1% + 2 × 3.33% = 6.76% Therefore, the correct option is (d). Q2. The speed of a longitudinal wave in a metallic bar is 400 m/s. If the density and Young's modulus of the bar material are increased by 0.5% and 1%, respectively then the speed of the wave is changed approximately to ______ m/s. (a) 398 (b) 402 (c) 401 (d) 399 Ans. (c) Sol. The formula for the speed of a longitudinal wave ( v ) in a solid metallic bar is given as Where, Y = Young's modulus and its ? is its density T aking the natural log on both sides of Differentiating both sides gives the fractional change: Initial speed = v = 400m/s Increase in Young's modulus Increase in density So, the percentage change in speed, % change in Since the result is positive, the speed increases by 0.25% So, new speed (v’) Initial Speed + Change in Speed Hence, the correct option is (C). Q3. Match List - I with List - II. Choose the correct answer from the options given below : (a) A-I, B-II, C-IV , D-III (b) A-IV , B-III, C-I, D-II (c) A-IV , B-I, C-II, D-III (d) A-I, B-III, C-II, D-IV Ans. (b) Sol. From Newton's law of viscosity, the force is , where A is area and is the velocity gradient. Surface tension is de?ned as the force acting per unit length. Page 5 JEE Main Previous Year Questions (2021-2026): Units & Measurements (January 2026) Q1. The time period of a simple harmonic oscillator is . Measured value of mass (m) of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant (k) is ________%. (a) 3.43 (b) 7.60 (c) 3.35 (d) 6.76 Ans. D Sol. The formula for the time period of simple harmonic oscillator is Using propagation of error, when quantities are multiplied or divided, their relative errors (percentage errors) are added. The constants have no error. The measured value of mass is, m=10g Accuracy/Error in measurement of mass is, ?m = 10mg = 0.01g So, percentage error in measurement of mass The percentage error in the time period (T) is the same as the percentage error in the total time (t) measured for n oscillations. T otal time measured is, t=60s Resolution/Error in measurement of time is, ?t=2s So, percentage error in measurement of time is, So, using propagation formula: Percentage error in k = (% error in m) + 2 × (% error in t) = 0.1% + 2 × 3.33% = 6.76% Therefore, the correct option is (d). Q2. The speed of a longitudinal wave in a metallic bar is 400 m/s. If the density and Young's modulus of the bar material are increased by 0.5% and 1%, respectively then the speed of the wave is changed approximately to ______ m/s. (a) 398 (b) 402 (c) 401 (d) 399 Ans. (c) Sol. The formula for the speed of a longitudinal wave ( v ) in a solid metallic bar is given as Where, Y = Young's modulus and its ? is its density T aking the natural log on both sides of Differentiating both sides gives the fractional change: Initial speed = v = 400m/s Increase in Young's modulus Increase in density So, the percentage change in speed, % change in Since the result is positive, the speed increases by 0.25% So, new speed (v’) Initial Speed + Change in Speed Hence, the correct option is (C). Q3. Match List - I with List - II. Choose the correct answer from the options given below : (a) A-I, B-II, C-IV , D-III (b) A-IV , B-III, C-I, D-II (c) A-IV , B-I, C-II, D-III (d) A-I, B-III, C-II, D-IV Ans. (b) Sol. From Newton's law of viscosity, the force is , where A is area and is the velocity gradient. Surface tension is de?ned as the force acting per unit length. Pressure is de?ned as force acting per unit area. Surface energy (U s ) is de?ned as the work done by surface tension in changing the surface area Hence, the correct option is (b)Read More
FAQs on JEE Main Previous Year Questions (2021-2026): Units & Measurements
| 1. How do I convert between different unit systems in JEE physics problems? |  |
Ans. Unit conversion in JEE requires using conversion factors systematically-multiply by the ratio of units to cancel unwanted dimensions. For example, converting 1 km/h to m/s involves dividing by 3.6. Master dimensional analysis alongside conversion to solve complex problems efficiently. Practice with previous year questions to identify patterns in unit transformations commonly tested.
| 2. What's the difference between fundamental and derived units, and why does it matter for JEE? | |
Ans. Fundamental units are independent base quantities (metre, kilogram, second, ampere, kelvin, mole, candela), while derived units combine them (velocity = m/s, force = kg·m/s²). Understanding this distinction helps solve dimensional analysis problems where examiners test whether you can break complex quantities into SI base units-a skill crucial for JEE scoring.
| 3. Can dimensional analysis really help solve physics problems I haven't studied yet? | |
Ans. Yes-dimensional analysis allows verification of formula correctness and sometimes predicts relationships between physical quantities without solving equations. By checking if dimensions match on both sides, students can eliminate incorrect formulas in multiple-choice questions. This technique frequently appears in JEE Main and Advanced exams to test conceptual understanding beyond rote memorization.
| 4. Why do JEE questions focus so much on significant figures and measurement precision? | |
Ans. Significant figures determine answer accuracy in experimental physics and practical calculations. JEE examiners test this because real measurements have limitations-reporting excessive decimal places suggests false precision. Understanding significant figure rules prevents common mistakes in numerical problems. Refer to EduRev's detailed notes and MCQ tests on measurement uncertainty to master this frequently tested concept.
| 5. What measurement errors and unit mistakes appear most often in JEE Main previous year questions? | |
Ans. Common errors include confusing absolute and relative errors, incorrect conversion between CGS and SI units, and mishandling scientific notation. Previous year questions frequently test confusion between milligrams and micrograms or misapplication of unit prefixes (kilo-, centi-, nano-). Reviewing actual JEE papers reveals these patterns-flashcards on unit prefixes and error calculation formulas strengthen retention effectively.
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