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\(E= \frac{F}{O}= \frac{40}{1\times10^{-5}}=4 \times10^{6}\)

\(X_c=\frac{1}{2\pi{fc}}\)=\(\frac{1}{2\times\pi\times\frac{200}{\pi}\times 25\times 10^{-6}}=\frac{10^6}{10^4}\)=100 Ω

E=\(mc^2 = 2\times10^{-6} \times10^{-3}\times(3\times10^{8})^2=1.8\times10^{8}\)

\(E=\lambda v=\frac{hc}{\lambda}\Rightarrow\lambda=\frac{hc}{E}={6.6 \times10^{-34}\times3\times10^8}{1.5\times10^{-19}}\\=1.32\times10^{-6}\)

F = m (v-u) = kg x m x S-2 = M1L1T-2

ma = mgsin\(\theta\) a = \(\frac{5 {\times} 10 {\times} sin30}{t}\) a = 5ms2

M1U1 + M2U2 = (m1 + m2)v = \(\frac{(0.1 {\times} 15) + (0.4 {\times} 0)}{(0.1 + 0.4)}\) v = 3ms-1

In simple harmonic motion (SHM), the force acting on the body is directly proportional to its displacement from the equilibrium position and is in the opposite direction (according to Hooke's Law). This force results in an acceleration of the body, which is also directly proportional to its displacement but in the opposite direction. When the displacement of the body is maximum (either positive or negative), the force acting on it is also maximum, leading to maximum acceleration. Similarly, when the displacement is zero (at the equilibrium position), the force and acceleration are also zero. Thus, the force acting on the body and its acceleration are in phase, meaning they reach their maximum and minimum values simultaneously during the oscillatory motion.

F = mg + ma F = 2 [10 + \(\frac{(40 - 10)}{4}\)] F = 35 N

g = \(\frac{GM}{R^2}\) \(\frac{G_1M_1}{r^2_1}\) = \(\frac{G_2M_2}{r^2_2}\) \(\frac{G_1M_1}{r^2_1}\) = \(\frac{G_2 {\times}10 M_1}{(2r_1)^2}\) \(\frac{G_1}{G_2}\) = \(\frac{2}{5}\)

Using, v\(^2\) = u\(^2\) - 2gx (20)\(^2\) = u\(^2\) - 2 x 10 x 10 u\(^2\) = 400 + 200 = 600(m/s\(^2\)) initial K.E = \(\frac{1}{2}\)mu\(^2\) = \(\frac{1}{2}\) x 0.1 x 600 = 30 J (100g = 0.1kg)

Volume of water displaced = volume of cube therefore, 0.1 x 0.1 x 0.1 = 0.001m3 mass of H2O = 0.001 X 1000 = 1kg upthrust = mg = 1 x 10 =10N

Density of liquid = \(\frac{1 {\times} 10^3}{2.5 {\times} 2 {\times} 4}\) = 50kgm3 pressure heg = 50 x 10 x \(\frac{4}{2}\) = 1 x 103Nm-2

R.d = \(\frac{\text {density of petrol}}{\text{density of water}}\) 0.72 = \(\frac{\text{density of petrol}}{1000}\) Density of petrol =720kg\(m^3\) mass of petrol = 720 x 60 x \(10^{-3}\) =43.2kg The total mass of the tank and fuel = 10 + 43.2kg = 53.2kg.

\(\frac{P_1}{T_1}\) = \(\frac{ P_2}{T_2}\) \(\frac{3 \times\ 10^6}{27 + 273}\) = \(\frac{4 \times\ 10^6}{0 + 273}\) \(\theta\) = 127oC