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JEE Advanced
Exam
JEE Advanced
Subject
Physics
Chapter
Mathematics in Physics

JEE Advanced · Physics · Mathematics in Physics

In a standard meter bridge experiment, to measure the specific...

In a standard meter bridge experiment, to measure the specific resistance of a wire the following data are found:- Length (L)=(1±.01)m(\mathrm{L})=(1 \pm .01) \mathrm{m} measured by a meter scale Radius of wire =1 mm±.01 mm=1 \mathrm{~mm} \pm .01 \mathrm{~mm}. (measured by screw gauge). Resistance of wire (R)=(5±.01)Ω(R)=(5 \pm .01) \Omega The maximum possible error in the measurement of specific resistance is (You may use the formula:- ρ=πr2 LR\rho=\frac{\pi r^2}{\mathrm{~L}} R )

Options

  1. A⋅1μΩm\cdot 1 \mu \Omega m
  2. B⋅2μΩm\cdot 2 \mu \Omega m
  3. C⋅5024μΩm\cdot 5024 \mu \Omega m
  4. D6024μΩm6024 \mu \Omega m

Correct answer

C. ⋅5024μΩm\cdot 5024 \mu \Omega m

Step-by-step solution

ρ=3⋅14×10−6×51=15.7μ\rho=\frac{3 \cdot 14 \times 10^{-6} \times 5}{1}=15.7 \mu ohm metre δρ=ρ[2δrr+δLL+δRR]ohmmetre.\delta \rho=\rho \left[\frac{2 \delta r}{r}+\frac{\delta \mathrm{L}}{\mathrm{L}}+\frac{\delta \mathrm{R}}{\mathrm{R}}\right] ohm metre.=15 ⋅\cdot7 \mu \left[2 \times \frac{.01}{1}+\frac{.01}{1}+\frac{.01}{5}\right]ohmmetreohm metre=15.7μ[⋅02+⋅01+⋅002] ohm − metre =15.7×⋅032μΩm=.5024μΩm\begin{aligned} & =15.7 \mu[\cdot 02+\cdot 01+\cdot 002] \text { ohm }- \text { metre } \\ & =15.7 \times \cdot 032 \mu \Omega \mathrm{m} \\ & =.5024 \mu \Omega \mathrm{m}\end{aligned}$$
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