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JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
For the four sets of three measured physical quantities as given below. Which of the following options is correct ? \((i)\) \(\mathrm{A}_{1}=24.36, \mathrm{B}_{1}=0.0724, \mathrm{C}_{1}=256.2\) \((ii)\) \(\mathrm{A}_{2}=24.44, \mathrm{B}_{2}=16.082, \mathrm{C}_{2}=240.2\) \((iii)\) \(\mathrm{A}_{3}=25.2, \mathrm{B}_{3}=19.2812, \mathrm{C}_{3}=236.183\) \((iv)\) \(\mathrm{A}_{4}=25, \mathrm{B}_{4}=236.191, \mathrm{C}_{4}=19.5\)
- A \(\mathrm{A}_{4}+\mathrm{B}_{4}+\mathrm{C}_{4} < \mathrm{A}_{1}+\mathrm{B}_{1}+\mathrm{C}_{1} < \)\(\mathrm{A}_{3}+\mathrm{B}_{3}+\mathrm{C}_{3} < \mathrm{A}_{2}+\mathrm{B}_{2}+\mathrm{C}_{2}\)
- B \(\mathrm{A}_{1}+\mathrm{B}_{1}+\mathrm{C}_{1} < \mathrm{A}_{3}+\mathrm{B}_{3}+\mathrm{C}_{3} < \)\(\mathrm{A}_{2}+\mathrm{B}_{2}+\mathrm{C}_{2} < \mathrm{A}_{4}+\mathrm{B}_{4}+\mathrm{C}_{4}\)
- C \(\mathrm{A}_{1}+\mathrm{B}_{1}+\mathrm{C}_{1} = \mathrm{A}_{2}+\mathrm{B}_{2}+\mathrm{C}_{2} = \)\(\mathrm{A}_{3}+\mathrm{B}_{3}+\mathrm{C}_{3} = \mathrm{A}_{4}+\mathrm{B}_{4}+\mathrm{C}_{4}\)
- D \(\mathrm{A}_{4}+\mathrm{B}_{4}+\mathrm{C}_{4} > \mathrm{A}_{3}+\mathrm{B}_{3}+\mathrm{C}_{3} = \)\(\mathrm{A}_{2}+\mathrm{B}_{2}+\mathrm{C}_{2} > \mathrm{A}_{1}+\mathrm{B}_{1}+\mathrm{C}_{1}\)
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)…
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