JEE Main Units, Measurement & Experimental Skills: Physics and Measurement – Chapter Test
JEE Main Units, Measurement & Experimental Skills: Physics and Measurement – Chapter Test is an online chapter test in the JEE Main pattern, covering Units, Measurement & Experimental Skills. Attempt it on SeatBeat with a timer like the real exam, then see a detailed solution for every question and your score and weak-topic analysis.
Topics in this test
| Topic | Questions |
|---|---|
| Units, Measurement & Experimental Skills | 20 |
Difficulty: 9 medium · 11 hard
Physics and Measurement MCQ questions with answers
The first 3 of 20 questions. Unlock the series for the rest, each with a solution.
Q1. Three measured lengths 2.3 cm, 4.56 cm and 0.078 cm are added. Reported to the correct number of significant figures, the sum is:
तीन मापित लंबाइयाँ 2.3 cm, 4.56 cm तथा 0.078 cm को जोड़ा जाता है। सार्थक अंकों की सही संख्या तक व्यक्त करने पर, योग है:
- A. 6.938 cm
- B. 7.0 cm
- C. 6.9 cm
- D. 6.94 cm
Solution
The arithmetic sum is 6.938 cm, but in addition the result is rounded to the least number of decimal places among the terms (one decimal place, from 2.3), giving 6.9 cm.
अंकगणितीय योग 6.938 cm है, परंतु योग में परिणाम को पदों में से न्यूनतम दशमलव स्थानों की संख्या तक (एक दशमलव स्थान, 2.3 से) पूर्णांकित किया जाता है, जिससे 6.9 cm प्राप्त होता है।
Q2. Two temperatures are measured as θ₁ = (50.0 ± 0.5) °C and θ₂ = (20.0 ± 0.3) °C. The temperature difference θ₁ − θ₂, with its maximum absolute error, is:
दो ताप θ₁ = (50.0 ± 0.5) °C तथा θ₂ = (20.0 ± 0.3) °C मापे जाते हैं। तापांतर θ₁ − θ₂, अपनी अधिकतम निरपेक्ष त्रुटि सहित, है:
- A. (30.0 ± 0.5) °C
- B. (30.0 ± 0.2) °C
- C. (30.0 ± 0.3) °C
- D. (30.0 ± 0.8) °C
Solution
In both a sum and a difference, the maximum absolute errors add: Δθ = 0.5 + 0.3 = 0.8 °C. The errors never subtract even though the quantities themselves are subtracted, so (30.0 ± 0.8) °C is correct.
योग और अंतर दोनों में अधिकतम निरपेक्ष त्रुटियाँ जुड़ती हैं: Δθ = 0.5 + 0.3 = 0.8 °C। राशियों को स्वयं घटाने पर भी त्रुटियाँ कभी नहीं घटतीं, अतः (30.0 ± 0.8) °C सही है।
Q3. In the 2019 revision of the SI, the kilogram is defined by fixing the exact numerical value of which fundamental constant?
SI के 2019 के संशोधन में, किलोग्राम को किस मूलभूत नियतांक के यथार्थ संख्यात्मक मान को नियत करके परिभाषित किया जाता है?
- A. Avogadro number / आवोगाद्रो संख्या
- B. Planck constant / प्लांक नियतांक
- C. Boltzmann constant / बोल्ट्ज़मान नियतांक
- D. Elementary charge / प्रारंभिक आवेश
Solution
Since 20 May 2019 the kilogram is defined by fixing the Planck constant h = 6.62607015×10⁻³⁴ J·s. The elementary charge defines the ampere, the Boltzmann constant the kelvin, and the Avogadro number the mole.
20 मई 2019 से किलोग्राम को प्लांक नियतांक h = 6.62607015×10⁻³⁴ J·s को नियत करके परिभाषित किया जाता है। प्रारंभिक आवेश ऐम्पियर को, बोल्ट्ज़मान नियतांक केल्विन को, तथा आवोगाद्रो संख्या मोल को परिभाषित करती है।
Frequently asked questions
How many questions are in the JEE Main Units, Measurement & Experimental Skills: Physics and Measurement – Chapter Test?
20 questions for 20 marks, to be answered in 20 minutes.
Is there negative marking in the JEE Main Units, Measurement & Experimental Skills: Physics and Measurement – Chapter Test?
Yes — 0.25 marks are deducted for each wrong answer, as in the real exam pattern.
Which topics does the JEE Main Units, Measurement & Experimental Skills: Physics and Measurement – Chapter Test cover?
Units, Measurement & Experimental Skills (20 questions).
Is the JEE Main Units, Measurement & Experimental Skills: Physics and Measurement – Chapter Test free?
It is part of the JEE Main 2027 Mock Test Series (104 tests, ₹299). You can first try Basic Concepts & Atomic Structure – Topic Test for free.
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