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Units and Measurements question

2017 · 9 Apr · Shift 1 · Q66
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Units and Measurements question

2017 · 9 Apr · Shift 1 · Q66

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
A physical quantity P is described by the relation P = a 12^{\frac{1}{2}}21​ b2 c3 d −-− 4 If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then the relative error in P will be :
  1. A
    8%
  2. B
    12%
  3. C
    32%
  4. D
    25%
View written solutionFree

Correct answer: C

  1. Interpret the given relation

    The expression is intended as P=a1/2b2c3d−4P = a^{1/2} b^2 c^3 d^{-4}P=a1/2b2c3d−4

    For a quantity of the form P=axbyczdw,P = a^x b^y c^z d^w,P=axbyczdw, the maximum relative error is ΔPP=∣x∣Δaa+∣y∣Δbb+∣z∣Δcc+∣w∣Δdd.\frac{\Delta P}{P} = |x|\frac{\Delta a}{a} + |y|\frac{\Delta b}{b} + |z|\frac{\Delta c}{c} + |w|\frac{\Delta d}{d}.PΔP​=∣x∣aΔa​+∣y∣bΔb​+∣z∣cΔc​+∣w∣dΔd​.

  2. Write the given relative errors

    Δaa=2%,Δbb=1%,Δcc=3%,Δdd=5%\frac{\Delta a}{a} = 2\%, \quad \frac{\Delta b}{b} = 1\%, \quad \frac{\Delta c}{c} = 3\%, \quad \frac{\Delta d}{d} = 5\%aΔa​=2%,bΔb​=1%,cΔc​=3%,dΔd​=5%

  3. Apply the error formula

    Here the powers are:

    • for aaa: 12\frac{1}{2}21​
    • for bbb: 222
    • for ccc: 333
    • for ddd: −4-4−4

    Therefore, ΔPP=12(2%)+2(1%)+3(3%)+4(5%)\frac{\Delta P}{P} = \frac{1}{2}(2\%) + 2(1\%) + 3(3\%) + 4(5\%)PΔP​=21​(2%)+2(1%)+3(3%)+4(5%)

  4. Calculate each term

    12(2%)=1%\frac{1}{2}(2\%) = 1\%21​(2%)=1% 2(1%)=2%2(1\%) = 2\%2(1%)=2% 3(3%)=9%3(3\%) = 9\%3(3%)=9% 4(5%)=20%4(5\%) = 20\%4(5%)=20%

  5. Add them

    ΔPP=1%+2%+9%+20%=32%\frac{\Delta P}{P} = 1\% + 2\% + 9\% + 20\% = 32\%PΔP​=1%+2%+9%+20%=32%

  6. Choose the correct option

    32%\boxed{32\%}32%​

    So the correct option is C.

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