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The force acting on a moving charge in the magnetic field is given by: F= qVBSin \theta sin \theta =F=(8.0x10)-14/1.6x10-19x1.0x106x1.0 = (8.0x10)14/16 x 1013 (8.0x 10)-14/16x10-14 =0.5000 ∴ \theta= sin-1 0.5000 =30^\circ

For liquid W, its mass = vol. x density = 3 x 200 =600kg. similarly, for liquid L, its mass = 7 x 150 = 1050kg ∴ Total mass of mixture = 600kg + 1050kg =1650kg. similarly total vol.of the mixture = 3m3 + 7m3 =10m3. ∴ density of the mixture = (Total Mass)/Total Mixture = (1650kg)/10m3 =165kgm-3

For anybody falling through a viscous liquid, at its terminal velocity. the resultant force acting on its is zero. from the equation: Mg - v - u = 0. where v = viscous drag U = upthrust Mg = weight of the body. ∴ 36 x 10 - 260 - U = 0 ∴ U = 360 - 260 = 100N Upthrust on the body = 100N.

Melting (I): This process involves the breaking of intermolecular forces as a solid turns into a liquid, reflecting the molecular structure of matter. Boiling (II): Similar to melting, boiling involves the transition from liquid to gas, indicating how molecules gain energy and move apart Conduction (IV): This is the process of heat transfer through collisions between molecules, demonstrating how molecular interactions lead to energy transfer.

Radio waves are used to transmit signals between satellites and ground stations because they can travel long distances and effectively penetrate the atmosphere.

For the device rated 2000 W, 250 V, implies that current I, flowing is given by the relation Power = IV I = (power)/v = (200)/250 = 8A. ∴ the most suitable fuse for the device is 8A

The electric force experienced in an electric field of intensity E is given by: F = qE: where q = the test charge. E = F/Q = 0.4/10-10 = 0.4 X 10-10 = 4.0 X 109 NC-1

For the two point charges, the magnitude of the electrostatic force between them is given by the equation: F ∝ 1/r2, and if the distance is increased by a factor of 4, that is new distance = 4r, then, new force F ∝ 1/(4r)2 = ∝ 1/16r2 Thus it will be 1/16 of the former value

Generally, from Hares Experiment, density is inversely proportional to height i.e h ∝1/d; → h1d1 = h2d2 ∴ 8 x 103 = h2 x 800 ∴ h2 = 8 x 103_800 = 10cm

Normal reaction R = 3 kg x 10 = 30N coefficient of friction , μ = 0.5 ∴ acting frictional force = μR = 30 x 0.5 Pulling/sliding force F = 6kg x 10 = 60N. ∴ Net downward force = ma = 60 - 15.0 => (3 + 6)a = 45 ∴ 9a = 45 a = 5m/s2

Since the force is inclined at an angle 60o to the horizontal, the effective component of te force along the horizontal is given by F x cos 60o = 100 x 0.5 = 50N. ∴ work done = F x cos 60 x displacement = 50 x 8.0m = 400.0J

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

V = u + at = 25 = 0 + 5t t = 5s

In SHM, acceleration a = -\(\omega^2\) x , so it varies linearly with displacement from the equilibrium point and is directed towards it.

\(M_1V_1 + M_2U_2 = V(M_1 + M_2)\) 5 x 10 + 6 x 0 = V(5 + 6) V = \(\frac{50}{11} = 4.55\)m/s