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When an iron rod is moved from the Earth to the Moon, the following properties are affected: > Mass: The mass of the rod remains unchanged regardless of its location. Mass is a measure of the amount of matter in an object and does not vary with location. > Weight: The weight of the rod will change because weight is the force due to gravity acting on the mass of the rod. Since the gravitational force on the Moon is weaker than that on Earth (about 1/6th of Earth's gravity), the weight of the rod will decrease. > Relative Density: Relative density (or specific gravity) is the ratio of the density of a substance to the density of a reference substance (usually water). Since the composition of the iron rod does not change, its relative density remains the same regardless of the location. So the properties that remain unchanged are: < Mass (i) < Relative Density (iii)

Density of liquid = \(\frac{M}{V}\) = \(\frac{20}{13.5}\) = 1.48g/cm\(^3\) Volume = area x length = 0.54 x 25 = 13.5cm\(^3\) = volume of cylinder = volume of liquid. weight of liquid displaced = weight of object = mass of liquid = 20g volume of liquid = 13.5cm\(^3\) Relative density of liquid = \(\frac{\text{ density of liquid}}{\text{density of water}}\) R. d = \(\frac{1.481}{1}\) = 1.48g/cm\(^3\)

Apparent weight = 32N Real weight = 96N Density of water = 1000\(kgm^{-3}\) acceleration (g) = 10\(ms^{-2}\) Upthrust = Real weight - Apparent weight ⇒ 96N - 32N = 64N ρ be the density of liquid and 'V ' be the volume of the body and we know that upthrust force F = ρVg V = \(\frac{F}{ρg}\) = \(\frac{64}{1000 \times10}\) = 6.4 x 10\(^{-3}m^3\)

Limiting frictional force is proportional to normal reaction. but R = normal reaction = mg mass = 1.6kg F\(_{\text{lim}}\) = 8N g = 10m/s\(^2\) R = 1.6 x 10 = 16N coefficient of Friction = \(\frac{\text{F}}{\text{R}}\) = \(\frac{8}{16}\) = 0.5.

Its acceleration decreases before terminal velocity is attained: When the ball bearing is released, it accelerates due to gravity. As it falls, the viscous drag force exerted by the fluid increases. This drag force opposes the weight of the ball bearing, resulting in a decrease in acceleration until terminal velocity is reached.

Let V\(_t\) = velocity of truck, = 20m/s V\(_c\) = velocity of car = 40m/s \(_cV_t\) = V\(_t\) - V\(_c\) = 40 -(-20) = 40 + 20 = 60m/s

The total momentum of the two balls after collision: Can be determined. The total momentum before the collision can be calculated using the mass and velocity of the moving ball, and since momentum is conserved, the total momentum after the collision will be equal to the total momentum before the collision

Work done = area under the triangle Work done = \(\frac{1}{2}\) x base x height = 0.5 x 20 x 100 = 1000J Therefore, the work done, as calculated from the force-displacement diagram, is 1000J

Efficiency = \(\frac{ poweroutput}{powerinput}\) but power output = \(\frac{ f \times s}{ t}\) = \(\frac{75 \times 10 \times 3}{ 30}\) = 75W Efficiency = 80%, power output = 75w then 0.8 = \(\frac{ 75}{powerinput}\) power input = \(\frac{ 75}{0.8}\) = 93.75W.

Using the general gas equation, \(\frac{ P_1 V_1}{ T_1} = \frac{ P_2 V_2}{ T_2}\) \(V_1 = 76cm^3\), \(T_1 = 27^0C\) = 27 + 273K = 300K, \(P_1\) = 80cm, at STP \(T_2 = 0^0C\) = 273K, \(P_2\) = 76cm, \(V_2\) = ? \(V_2 = \frac{ P_1\times V_1\times T_2}{P_2 \times T_1}\) \(V_2 = \frac{ 80 \times 76 \times 273}{76 \times 300}\) \(V_2 = 72.8cm^3\)

The temperature at which the saturated vapour pressure of a liquid is equal to the external atmospheric pressure is known as its BOILING POINT.

The speed of sound depends on the density of the medium in which it travels and increases when the density increases. For example, in solids sound travels faster than in liquid and even faster than in gas, and the density is highest in solids, lower in liquids, and lowest in gas. Hence, sound travels fastest in solids.

The energy(W)stored in an inductor is calculated by the formula: \(\frac{1}{2}\)LI\(^2\) = \(\frac{1}{2}\) x 10 x (0.2)\(^2\) = 0.2J

Bohr's theory of the atom introduced the concept of quantized energy levels for electrons in an atom. It explained how electrons occupy specific energy levels and how they can jump between these levels by absorbing or emitting energy in the form of photons.

The acceleration due to gravity (g) is affected by the distance from the center of the Earth, the latitude, and the rotation of the Earth. However, the mass of the object does not affect g; all objects fall at the same rate in a vacuum, regardless of their mass.