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ID#5714 HSC Physics 1st MCQ (Jessore 2025)

90 Nm⁻¹ বল ধ্রুবকের স্প্রিংকে 4cm সংকুচিত করলে কৃতকাজ কত হবে?
ক) 0.072 J
খ) 0.078 J
গ) 0.082 J
ঘ) 0.086 J

ব্যাখ্যা

$W = \frac{1}{2} kx^2 = \frac{1}{2} \times 90 \times (0.04)^2$
$W = 45 \times 0.0016 = 0.072 J$।
Resource Details
Exam HSC
Subject Physics 1st paper
Chapter 5
Board Jessore
Year 2025

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