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Retardation = \(\frac{u - v}{t}\) = \(\frac{15 -0}{20}\) = 0.75ms-1

Energy is constant: Since the angular velocity (\(\omega\)) and radius (r) are constant, the speed remains constant. Kinetic energy depends only on mass and speed, so it remains constant. Linear momentum varies: Linear momentum is a vector quantity. Although its magnitude is constant, its direction changes continuously in circular motion, so the momentum varies.

P.E = K.E = mgh = 50 x 10 x 2 = 1000J

No expanded working has been attached for this question yet. The verified answer is Option B.

F = \(\mu\)mg = 0.2 x 20 x 10 = 40N

V.R = \(\frac{2\pi L}{p}\) = \(\frac{2 \times \pi \times 2}{0.3}\) V.R = 140\(\pi\)

\(\frac{15}{30 - L}\) = \(\frac{10}{5}\) L = 22.5cm

No expanded working has been attached for this question yet. The verified answer is Option A.

Masses of kerosene = 24.8 - 15.2 = 9.6 mass of H2O = 27.2 - 15.2 = 12 R.d = \(\frac{9.6}{12}\) = 0.8

R.d \(\alpha\) weight of liquid displaced \(\frac{\text {R.d of P}}{\text {R.d of Q}}\) = \(\frac{\text{weight of p displaced}}{\text{weight of Q displaced}}\) \(\frac{2}{1.5}\) = \(\frac{\text {weight of p displaced}}{\text {weight of Q displaced}}\) A solid floats when the weight of the liquid it displaces is equal to the weight of the solid. The relative density indicates that liquid P is denser than liquid Q. Conclusion: Since liquid P is denser, for the same volume of the solid X displaced, the weight of the liquid displaced will be greater in liquid P than in liquid Q.

T = 78 + 273 = 351k

\(\frac{p_1}{T_1}\) = \(\frac{p_2}{T_2}\) \(\frac{2 \times 10^5}{27 + 273}\) = \(\frac{p_2}{87 + 273}\) P2 = 2.4 x 105Nm-2

Heat lost by L1 = heat gained by L2 cL1 = 2cL2 100 x 2cL2 x (78 - 66) = xg x cL2 x (66 - 50) xg = 150g

p x t = heat energy P x 1 x 60 = 100 x 336 p = 560W

No expanded working has been attached for this question yet. The verified answer is Option D.