Mathematics in Physics

JEE Main level, single correct, medium

113 of 121

Question 113 of 121

The period of oscillation of a simple pendulum is T=2πlg.T=2\pi \sqrt{\dfrac{l}{g}}. Measured value of ll is 20.0cm20.0cm, known to 1mm1mm accuracy and time for 100100oscillations of the pendulum is found to be 90s90s using a wristwatch of 1s1s resolution. The accuracy in the determination of gg is

Show answer and solution

Answer: (C) 33%

∴T=2πlg⇒g=4π2lT2∴T=2\pi \sqrt{\dfrac{l}{g}}\Rightarrow g=4{\pi }^{2}\dfrac{l}{{T}^{2}}

∴∴ Error in gg can be calculated as

Δgg=Δll+2ΔTT\dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+\dfrac{2\Delta T}{T}.

∴∴ Total time for nnoscillation is t=nTt=nT where TT= time for oscillation.

⇒Δtt=ΔTT\Rightarrow \dfrac{\Delta t}{t}=\dfrac{\Delta T}{T}

⇒Δgg=Δll+2Δtt\Rightarrow \dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+\dfrac{2\Delta t}{t}

Given that Δl=1mm=10−3m,l=20×10−2m\Delta l=1mm={10}^{-3}m,l=20\times {10}^{-2}m

Δt=1s,t=90s\Delta t=1s,t=90s.

%erroringis

Δgg×100=(Δll+2Δtt)×100\dfrac{\Delta g}{g}\times 100=\left(\dfrac{\Delta l}{l}+\dfrac{2\Delta t}{t}\right)\times 100

=(10−320×10−2+2×190)×100=\left(\dfrac{{10}^{-3}}{20\times {10}^{-2}}+\dfrac{2\times 1}{90}\right)\times 100=12+209=\dfrac{1}{2}+\dfrac{20}{9}=0.5+2.22=0.5+2.22=2.72=2.72%

≅3\cong 3%

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Mathematics in Physics (Errors of Measurement), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.