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Units and Measurements question

2023 · 8 Apr · Shift 1 · Q58
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Units and Measurements question

2023 · 8 Apr · Shift 1 · Q58

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
A cylindrical wire of mass (0.4±0.01)g(0.4 \pm 0.01) \mathrm{g}(0.4±0.01)g has length (8±0.04)cm(8 \pm 0.04) \mathrm{cm}(8±0.04)cm and radius (6±0.03)mm(6 \pm 0.03) \mathrm{mm}(6±0.03)mm. The maximum error in its density will be:
  1. A
    1%
  2. B
    5%
  3. C
    4%
  4. D
    3.5%
View written solutionFree

Correct answer: C

  1. Write the expression for density

For a cylindrical wire,

ρ=mV=mπr2l\rho = \frac{m}{V} = \frac{m}{\pi r^2 l}ρ=Vm​=πr2lm​

where:

  • m=(0.4±0.01) gm = (0.4 \pm 0.01)\,\text{g}m=(0.4±0.01)g
  • l=(8±0.04) cml = (8 \pm 0.04)\,\text{cm}l=(8±0.04)cm
  • r=(6±0.03) mmr = (6 \pm 0.03)\,\text{mm}r=(6±0.03)mm
  1. Use maximum fractional error formula

Since

ρ∝mr2l\rho \propto \frac{m}{r^2 l}ρ∝r2lm​

maximum fractional error is

Δρρ=Δmm+2Δrr+Δll\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}ρΔρ​=mΔm​+2rΔr​+lΔl​
  1. Calculate each fractional error

For mass:

Δmm=0.010.4=0.025=2.5%\frac{\Delta m}{m} = \frac{0.01}{0.4} = 0.025 = 2.5\%mΔm​=0.40.01​=0.025=2.5%

For radius:

Δrr=0.036=0.005=0.5%\frac{\Delta r}{r} = \frac{0.03}{6} = 0.005 = 0.5\%rΔr​=60.03​=0.005=0.5%

So,

2Δrr=1%2\frac{\Delta r}{r} = 1\%2rΔr​=1%

For length:

Δll=0.048=0.005=0.5%\frac{\Delta l}{l} = \frac{0.04}{8} = 0.005 = 0.5\%lΔl​=80.04​=0.005=0.5%
  1. Add the percentage errors
Δρρ=2.5%+1%+0.5%=4%\frac{\Delta \rho}{\rho} = 2.5\% + 1\% + 0.5\% = 4\%ρΔρ​=2.5%+1%+0.5%=4%
  1. Match with the options

The maximum error in density is

4%\boxed{4\%}4%​

So the correct option is C.

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C (4%)

They match.

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