यदि Z = \(\frac{A^{2} B^{3}}{C^{4}}\), तो Z में आपेक्षिक त्रुटि कितनी होगी?

  1. \(\frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{\Delta C}{C}\)
  2. \(\frac{2 \Delta A}{A}+\frac{3 \Delta B}{B}−\frac{4 \Delta C}{C}\)
  3. \(\frac{2 \Delta A}{A}+\frac{3 \Delta B}{B}+\frac{4 \Delta C}{C}\)
  4. \(\frac{\Delta A}{A}+\frac{\Delta B}{B}−\frac{\Delta C}{C}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{2 \Delta A}{A}+\frac{3 \Delta B}{B}+\frac{4 \Delta C}{C}\)
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JEE Main 04 April 2024 Shift 1
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Detailed Solution

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व्याख्या:

दिया गया है, Z = \(\frac{A^{2} B^{3}}{C^{4}}\)

या, Z = A2B3C-4   ----- (1)

∵ त्रुटियां हमेशा जुड़ जाती हैं, हम समीकरण (1) को आपेक्षिक त्रुटि के रूप में इस प्रकार लिख सकते हैं-

\(\frac{\Delta Z}{Z}=2\frac{\Delta A}{A} + 3\frac{\Delta B}{B}+4\frac{\Delta C}{C}\)

अतः, विकल्प 3) सही चुनाव है। 

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