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JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2020
- Hard
- MCQ
For the four sets of three measured physical quantities as given below. Which of the following options is correct ? \((i)\) \(\mathrm{A}_{1}=24.36, \mathrm{B}_{1}=0.0724, \mathrm{C}_{1}=256.2\) \((ii)\) \(\mathrm{A}_{2}=24.44, \mathrm{B}_{2}=16.082, \mathrm{C}_{2}=240.2\) \((iii)\) \(\mathrm{A}_{3}=25.2, \mathrm{B}_{3}=19.2812, \mathrm{C}_{3}=236.183\) \((iv)\) \(\mathrm{A}_{4}=25, \mathrm{B}_{4}=236.191, \mathrm{C}_{4}=19.5\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\mathrm{A}_{4}+\mathrm{B}_{4}+\mathrm{C}_{4} > \mathrm{A}_{3}+\mathrm{B}_{3}+\mathrm{C}_{3} = \)\(\mathrm{A}_{2}+\mathrm{B}_{2}+\mathrm{C}_{2} > \mathrm{A}_{1}+\mathrm{B}_{1}+\mathrm{C}_{1}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{A}_{1}+\mathrm{B}_{1}+\mathrm{C}_{1}=24.36+0.0724+256.2\)
\(=280.6324\)
\(=280.6\) (After rounding off) \(\mathrm{A}_{2}+\mathrm{B}_{2}+\mathrm{C}_{2}=24.44+16.082+240.2\)
\(=280.722\)
\(=280.7\) (After rounding off) \(\mathrm{A}_{3}+\mathrm{B}_{3}+\mathrm{C}_{3}=25.2+19.2812+236.183\)
\(=280.6642\)
\(=280.7\) (After rounding off)
\(\mathrm{A}_{4}+\mathrm{B}_{4}+\mathrm{C}_{4}=25+236.191+19.5\)
\(=280.691\)
\(=281\) (After rounding off)
\(\mathrm{A}_{4}+\mathrm{B}_{4}+\mathrm{C}_{4} > \mathrm{A}_{3}+\mathrm{B}_{3}+\mathrm{C}_{3} = \)\(\mathrm{A}_{2}+\mathrm{B}_{2}+\mathrm{C}_{2} > \mathrm{A}_{1}+\mathrm{B}_{1}+\mathrm{C}_{1}\)
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