\(x = v_0 t \implies t = \frac{900}{80} = 11.25 \, \mathrm{s}\) \(\text{এখন, } h = v_0 t + \frac{1}{2} g t^2 \implies h = 0 + \frac{1}{2} \times 9.8 \times (11.25)^2\) \(\therefore h = 620.16 \, \mathrm{m}\) \(\text{Ans. (B)}\)

\(I = Mr^2 \implies r = \sqrt{\frac{I}{M}} = \sqrt{\frac{3.1}{1.5}} = 1.43 \, \mathrm{m}\) \(\text{Ans. (A)}\)

\(y = 4 \sin(3\pi x - 20\pi t) = 4 \sin 3\pi \left(x - \frac{20\pi t}{3\pi}\right)\) \(y = A \sin \frac{2\pi}{\pi}(vt - x) \, \text{এর সাথে তুলনা করে, } v = \frac{20}{3} = 6.67 \, \mathrm{ms^{-1}}\)

\(v = \sqrt{4 + 36} = \sqrt{40}\) \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times \frac{200}{1000} \times 40 = 4 \, \mathrm{J}\)

\( \text{পৃষ্ঠটানের একক } = \frac{\mathrm{N}}{\mathrm{m}} = \frac{\mathrm{Nm}}{\mathrm{m}^2} = \frac{\mathrm{J}}{\mathrm{m}^2} \)

কৌনিক ভরবেগ \( L = \frac{nh}{2\pi} \implies L = n\hbar \, \text{(যেহেতু } \hbar = \frac{h}{2\pi} \text{)} \)

\(\vec{v} = 2\hat{i} + 4\hat{j} + 2\hat{k}; \, |\vec{v}| = \sqrt{2^2 + 4^2 + 2^2} = \sqrt{24}\) \(W = \frac{1}{2}m v_1^2 - \frac{1}{2}m v_2^2 = \frac{1}{2} \times 2 \times (\sqrt{24})^2 - 0 = 24 \, \mathrm{J}\) \([v_1 = |\vec{v}| = \sqrt{24}, v_2 = 0]\)

\(\text{Solve: যেসব তরঙ্গ কঠিনকে ভেদায় না সেসব তরঙ্গ কঠিনের সাথে স্থুল সম্পর্কিত তৈরি করে।} \\ \text{Ans. (E)}\)

\(\text{Solve: প্রতি স্বাধীনতার মাত্রার জন্য শক্তি } \frac{1}{2}KT \\ \text{সুতরাং স্বাধীনতার মাত্রা } 6 \\ \text{তাহলে মোট শক্তি } 6 \times \frac{1}{2}KT = 3KT \\ \text{Ans. (C)}\)

\(\text{Solve: } W = \int_{0}^{2} dw = \int_{0}^{2} F dx = \int_{0}^{2} (6x^2 + 2) dx \\ = \int_{0}^{2} 6x^2 dx + \int_{0}^{2} 2 dx = \frac{6}{3}[x^3]_0^2 + 2[x]_0^2 \\ = 2 \times 8 + 2 \times 2 = 20 \\ \text{Ans. (E)}\)

\(\text{Solve: } I = I_G + Mr^2 = 2.5 + 3 \times (1.2)^2 = 6.82 \\ \text{Ans. (B)}\)