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ID#5685 HSC Physics 1st MCQ (Rajshahi 2025)

লব্ধি বল কত? (m=10⁵kg, θ=65°, বাধা 20N per 10³kg)
ক) 8.9 x 10⁵ N
খ) 8.81 x 10⁵ N
গ) 7.85 x 10⁵ N
ঘ) 7.16 x 10⁵ N

ব্যাখ্যা

উপাংশ $F_g = mg \sin\theta = 10^5 \times 9.8 \times \sin 65^\circ \approx 8.88 \times 10^5$
বাধা $F_f = \frac{10^5}{10^3} \times 20 = 2000$
লব্ধি বল $F = F_g + F_f$ এর হিসেবে অপশন (ক) সঠিক।
Resource Details
Exam HSC
Subject Physics 1st paper
Chapter 4
Board Rajshahi
Year 2025

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