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JEE Main 2021 (27 Jul, Shift 2) · Errors in Measurement

Errors in Measurement question from JEE Main 2021 (27 Jul, Shift 2)

Q80JEE Main 2021 · 27 JUL · SHIFT 2Medium

Errors in Measurement

A physical quantity 'y' is represented by the formula y=m2r−4gxl−32y = m^{2} r^{-4} g^{x} l^{-\frac{3}{2}}

If the percentage errors found in y, m, r, l and g are 18, 1, 0.5, 4 and p respectively, then find the value of x and p.

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Answer: (c)

  1. y=m2r−4gxl−3/2⇒Δyy=2Δmm+4Δrr+xΔgg+32Δlly = m^2 r^{-4} g^x l^{-3/2} \Rightarrow \frac{\Delta y}{y} = 2\frac{\Delta m}{m} + 4\frac{\Delta r}{r} + x\frac{\Delta g}{g} + \frac{3}{2}\frac{\Delta l}{l}
  2. 18=2(1)+4(0.5)+xp+32(4)=10+xp18 = 2(1) + 4(0.5) + xp + \frac{3}{2}(4) = 10 + xp
  3. So xp=8xp = 8. Check options: 163×32=8\frac{16}{3} \times \frac{3}{2} = 8; others give 10, 12, 16.
  4. So the answer is (c) 163\frac{16}{3} and ±32\pm\frac{3}{2}.
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