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QJEEAdv10Ph5758: JEE Advanced 2010 Physics - Class 11 JEEAdv10Ph5758 Solution: A student uses a simple pendulum of exactly 1m length to…

A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true?

Multiple Choice Options

  • A Error ΔT in measuring T, the time period, is 0.05 seconds
  • B Error ΔT in measuring T, the time period, is 1 second
  • C Percentage error in the determination of g is 5%
  • D Percentage error in the determination of g is 2.5%

Step-by-Step Verified Solution

Relative error in measurement of time,

Δtt=1 s40 s=140{{\Delta t} \over t} = {{1\,s} \over {40\,s}} = {1 \over {40}}

Time period, T=40 s20=2 sT = {{40\,s} \over {20}} = 2\,s

Error in measurement of time period,

ΔT=T×Δtt=2 s×140=0.05 s\Delta T = T \times {{\Delta t} \over t} = 2\,s \times {1 \over {40}} = 0.05\,s

The time period of simple pendulum is

T=2πlgT = 2\pi \sqrt {{l \over g}}

or, T2=4π2lg{T^2} = {{4{\pi ^2}l} \over g}

or, g=4π2lT2g = {{4{\pi ^2}l} \over {{T^2}}}

∴\therefore Δgg=2ΔTT=2×140=120{{\Delta g} \over g} = {{2\Delta T} \over T} = 2 \times {1 \over {40}} = {1 \over {20}} ( ∵\because ΔTT=Δtt{{\Delta T} \over T} = {{\Delta t} \over t} )

Percentage error in determination of g is

Δgg×100=120×100=5%{{\Delta g} \over g} \times 100 = {1 \over {20}} \times 100 = 5\%

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