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Units and Measurements question

2018 · Shift 1 · Q51
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  5. /2018 · Shift 1 · Q51

Units and Measurements question

2018 · Shift 1 · Q51

JEE AdvancedPhysicsUnits and MeasurementsMCQ+3 / −1
If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z=x/y.z = x/y.z=x/y. If the errors in x,yx,yx,y and zzz are Δx,Δy\Delta x,\Delta yΔx,Δy and Δz,\Delta z,Δz, respectively, then z±Δz=x±Δxy±Δy=xy(1±Δxx)(1±Δyy)−1.z \pm \Delta z = {{x \pm \Delta x} \over {y \pm \Delta y}} = {x \over y}\left( {1 \pm {{\Delta x} \over x}} \right){\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}}.z±Δz=y±Δyx±Δx​=yx​(1±xΔx​)(1±yΔy​)−1. The series expansion for (1±Δyy)−1,{\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}},(1±yΔy​)−1, to first power in Δy/y.\Delta y/y.Δy/y. is 1±(Δy/y).1 \pm \left( {\Delta y/y} \right).1±(Δy/y). The relative errors in independent variables are always added. So the error in zzz will be Δz=z(Δxx+Δyy).\Delta z = z\left( {{{\Delta x} \over x} + {{\Delta y} \over y}} \right).Δz=z(xΔx​+yΔy​). The above derivation makes the assumption that Δx/x<<1,Δy/y<<1.\Delta x/x \lt \lt 1,\Delta y/y \lt \lt 1.Δx/x<<1,Δy/y<<1. Therefore, the higher powers of these quantities are neglected. In an experiment the initial number of radioactive nuclei is 3000.3000.3000. It is found that 1000±401000 \pm 401000±40 nuclei decayed in the first 1.0s.1.0s.1.0s. For ∣x∣<<1.ln⁡(1+x)=x\left| x \right| \lt \lt 1.\ln \left( {1 + x} \right) = x∣x∣<<1.ln(1+x)=x up to first power in x.x.x. The error Δλ,\Delta \lambda ,Δλ, in the determination of the decay constant λ,\lambda ,λ, in s−1,{s^{ - 1}},s−1, is
  1. A
    0.040.040.04
  2. B
    0.030.030.03
  3. C
    0.020.020.02
  4. D
    0.010.010.01
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