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NEET
Exam
NEET
Subject
Physics
Chapter
Mathematics in Physics

NEET · Physics · Mathematics in Physics

In an experiment to determine the specific resistance ( ) of a...

In an experiment to determine the specific resistance (ρ\rho) of a wire, the formula used is ρ=Rπd24l\rho = \dfrac{R \pi d^2}{4l}, where RR is the resistance, dd is the diameter, and ll is the length of the wire. If the maximum permissible percentage error in the determination of ρ\rho is 8%8\%, and the measured percentage errors in RR and ll are 2%2\% and 1%1\% respectively, what is the maximum permissible percentage error in the measurement of the diameter (dd)?

Options

  1. A2.5%2.5\%
  2. B3.5%3.5\%
  3. C5.0%5.0\%
  4. D5.5%5.5\%

Correct answer

A. 2.5%2.5\%

Step-by-step solution

The formula for specific resistance is ρ=Rπd24l\rho = \dfrac{R \pi d^2}{4l}.

The maximum percentage error in ρ\rho is given by the sum of the percentage errors of the individual quantities, taking into account their powers. Errors always add up to give the maximum possible error, regardless of whether the quantity is in the numerator or denominator:

Δρρ×100=ΔRR×100+2(Δdd×100)+Δll×100\dfrac{\Delta \rho}{\rho} \times 100 = \dfrac{\Delta R}{R} \times 100 + 2 \left( \dfrac{\Delta d}{d} \times 100 \right) + \dfrac{\Delta l}{l} \times 100

Given that the maximum percentage error in ρ\rho is 8%8\%, in RR is 2%2\%, and in ll is 1%1\%, we can substitute these values into the equation:

8%=2%+2×(percentage error in d)+1%8\% = 2\% + 2 \times (\text{percentage error in } d) + 1\%

8%=3%+2×(percentage error in d)8\% = 3\% + 2 \times (\text{percentage error in } d)

2×(percentage error in d)=5%2 \times (\text{percentage error in } d) = 5\%

Percentage error in d=2.5%\text{Percentage error in } d = 2.5\%

Answer: 2.5%2.5\%
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