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ID#5725 HSC Physics 1st MCQ (Jessore 2025)

গতিপথের সমীকরণ y = 10 sin(ωt + δ); আদি সরণ 0.05m হলে আদি দশা (δ) কত?
ক) 0.286°
খ) 0.386°
গ) 0.486°
ঘ) 0.496°

ব্যাখ্যা

$y = A \sin(\omega t + \delta)$; যখন $t=0, y=0.05$
$0.05 = 10 \sin \delta \implies \sin \delta = 0.005$
$\delta = \sin^{-1}(0.005) \approx 0.286^\circ$।
Resource Details
Exam HSC
Subject Physics 1st paper
Chapter 8
Board Jessore
Year 2025

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