JEE Main 2015PhysicsMathematics in PhysicsEasyMCQ

JEE Main 2015 — Mathematics in Physics Question with Solution

JEE Main 2015 (04 Apr)

Question

The period of oscillation of a simple pendulum is T=2πlg.  Measured value of l is 20.0 cm, known to 1 mm  accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wristwatch of 1 s resolution. The accuracy in the determination of g is 

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Show full solutionCorrect option: C
Correct answer
C3%

Step-by-step explanation

∴T=2πlg⇒g=4π2lT2

∴ Error in g can be calculated as

Δgg=Δll+2ΔTT. 

∴ Total time for n oscillation is t=nT where T= time for oscillation.

⇒Δtt=ΔTT

⇒Δgg=Δll+2Δtt

Given that Δl=1 mm=10-3 m, l=20×10-2 m

Δt=1 s, t=90 s. 

% error in g is

Δgg×100=Δll+2Δtt×100

=10-320×10-2+2×190×100=12+209=0.5+2.22=2.72%

≅3%

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About this question

This is a previous-year question from JEE Main 2015, covering the Mathematics in Physics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.