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Units and Measurements question

2025 · 29 Jan · Shift 2 · Q71
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Units and Measurements question

2025 · 29 Jan · Shift 2 · Q71

JEE MainPhysicsUnits and MeasurementsNumerical+4 / −1
 A physical quantity Q is related to four observables a,b,c,d as follows : Q=ab4cd\text { A physical quantity } Q \text { is related to four observables } a, b, c, d \text { as follows : }Q = \frac{ab^4}{cd} A physical quantity Q is related to four observables a,b,c,d as follows : Q=cdab4​ where, a=(60±3)Pa;b=(20±0.1)m;c=(40±0.2)Nsm−2\mathrm{a}=(60 \pm 3) \mathrm{Pa} ; \mathrm{b}=(20 \pm 0.1) \mathrm{m} ; \mathrm{c}=(40 \pm 0.2) \mathrm{Nsm}^{-2}a=(60±3)Pa;b=(20±0.1)m;c=(40±0.2)Nsm−2 and d=(50±0.1)m\mathrm{d}=(50 \pm 0.1) \mathrm{m}d=(50±0.1)m, then the percentage error in Q is x1000\frac{x}{1000}1000x​, where x=‾x=\underline{\hspace{2cm}}x=​ .
Numerical answer
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Correct answer: 7700

  1. We are given Q=ab4cdQ=\frac{ab^4}{cd}Q=cdab4​ with a=(60±3),b=(20±0.1),c=(40±0.2),d=(50±0.1)a=(60\pm 3),\quad b=(20\pm 0.1),\quad c=(40\pm 0.2),\quad d=(50\pm 0.1)a=(60±3),b=(20±0.1),c=(40±0.2),d=(50±0.1)

  2. For multiplication/division and powers, the maximum fractional error adds as ΔQQ=Δaa+4Δbb+Δcc+Δdd\frac{\Delta Q}{Q}=\frac{\Delta a}{a}+4\frac{\Delta b}{b}+\frac{\Delta c}{c}+\frac{\Delta d}{d}QΔQ​=aΔa​+4bΔb​+cΔc​+dΔd​

  3. Compute each fractional error:

  • For aaa: Δaa=360=0.05=5%\frac{\Delta a}{a}=\frac{3}{60}=0.05=5\%aΔa​=603​=0.05=5%

  • For bbb: Δbb=0.120=0.005=0.5%\frac{\Delta b}{b}=\frac{0.1}{20}=0.005=0.5\%bΔb​=200.1​=0.005=0.5% Since b4b^4b4 is present, 4Δbb=4(0.005)=0.02=2%4\frac{\Delta b}{b}=4(0.005)=0.02=2\%4bΔb​=4(0.005)=0.02=2%

  • For ccc: Δcc=0.240=0.005=0.5%\frac{\Delta c}{c}=\frac{0.2}{40}=0.005=0.5\%cΔc​=400.2​=0.005=0.5%

  • For ddd: Δdd=0.150=0.002=0.2%\frac{\Delta d}{d}=\frac{0.1}{50}=0.002=0.2\%dΔd​=500.1​=0.002=0.2%

  1. Add the percentage errors: ΔQQ=5%+2%+0.5%+0.2%=7.7%\frac{\Delta Q}{Q}=5\%+2\%+0.5\%+0.2\%=7.7\%QΔQ​=5%+2%+0.5%+0.2%=7.7%

  2. The percentage error is given as x1000\frac{x}{1000}1000x​ This means x1000=7.7\frac{x}{1000}=7.71000x​=7.7 So, x=7700x=7700x=7700

Hence, the required integer is 7700\boxed{7700}7700​

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