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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Physics
Chapter
Mathematics in Physics

JEE Advanced 2006 · Physics · Mathematics in Physics

A student performs an experiment for determination of g(=(4 π^2...

A student performs an experiment for determination of g(=4π2lT2),l≈1 mg\left(=\frac{4 \pi^2 l}{T^2}\right), l \approx 1 \mathrm{~m}, and he commits an error of Δl\Delta l. For TT he takes the time of nn oscillations with the stop watch of least count ΔT\Delta T and he commits a human error of 0.1 s0.1 \mathrm{~s}. For which of the following data, the measurement of gg will be most accurate?

Options

  1. AΔL=0.5,ΔT=0.1,n=20\Delta L=0.5, \Delta T=0.1, n=20
  2. BΔL=0.5,ΔT=0.1,n=50\Delta L=0.5, \Delta T=0.1, n=50
  3. CΔL=0.5,ΔT=0.01,n=20\Delta L=0.5, \Delta T=0.01, n=20
  4. DΔL=0.1,ΔT=0.05,n=50\Delta L=0.1, \Delta T=0.05, n=50

Correct answer

D. ΔL=0.1,ΔT=0.05,n=50\Delta L=0.1, \Delta T=0.05, n=50

Step-by-step solution

Δgg=Δll+2ΔTn\frac{\Delta g}{g}=\frac{\Delta l}{l}+2 \frac{\Delta T}{n} In option (d) error in Δg\Delta g is minimum and number of observations made are maximum. Hence, in this case error in gg will be minimum.
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